MA1605 Complex Numbers Lecture Notes PDF

MA1605 Algebra Lecture notes Andreas Fring 1 Complex numbers 1.1 Definitions and notation If c is real number, then regardless of whether c is positive, negative or zero, the square c2...

MA1605 Algebra Lecture notes Andreas Fring 1 Complex numbers 1.1 Definitions and notation If c is real number, then regardless of whether c is positive, negative or zero, the square c2 is greater than or equal to 0 (a negative number times a negative number is a positive number). In other words, negative numbers do not have real square roots. Put yet another way, if d is a negative real number, then the quadratic equation x2 = d has no real solutions. Complex numbers provide the way around this problem. √ Notation. We set i = −1, an imaginary solution of the equation x2 = −1. Definition 1.1.1. A complex number is a number of the form z = a + bi where a and b are real numbers. The real number a is known as the real part of z and the real number b is known as the imaginary part of z. We write a = Rez and b = Imz. Two complex numbers are equal precisely if their real and imaginary parts are equal; that is, a + bi = c + di if and only if a = c and b = d. Real numbers are regarded as complex: a real number is simply a complex number with zero imaginary part. The complex numbers whose real part is zero, i.e., numbers of the form 0 + bi are called purely imaginary. We write a + 0i simply as a; 0 + bi as bi; a + (−b)i as a − bi; a + 1i as a + i, 0 + 0i as 0. Notation. We write R for the set of all real numbers and C for the set of all complex numbers. Thus C = {a + bi : a, b ∈ R}. 1 1 COMPLEX NUMBERS 2 Definition 1.1.2. Let z = a + bi be a complex number, where a, b ∈ R. The complex conjugate of z is the number a − bi and it is denoted by z (in some books, the complex conjugate is denoted by z ∗ ). 1.1.3. Let z = a + bi be a complex number, where a, b ∈ R. The modulus of z is Definition √ the number a2 + b2 and it is denoted by |z|. Note that |z| is a real and non-negative number and |z| = 0 if and only if z = 0. √ √ Example 1.1.4. (i) If z = 1 + 2i, then z̄ = 1 − 2i, |z| = 12 + 22 = 5. (ii) If z = −i, then z̄ = i, and |z| = 1. (iii) If z = −3, then z̄ = −3 and |z| = 3. (iv) If z = a is a real number, then the modulus of a is just the absolute value of a. Hence modulus extends the notion of absolute value to the complex numbers. 1.2 Operations on complex numbers We add, subtract, multiply and divide complex numbers as we would expect, the key relation to keep in mind is the equation i2 = −1. We add (and subtract) complex numbers by adding (and subtracting) their real and imaginary parts: (a + bi) + (c + di) = (a + c) + (b + d)i, (a + bi) − (c + di) = (a − c) + (b − d)i. We multiply complex numbers by expanding brackets, using the equation i2 = −1, and collect- ing real and imaginary parts: (a + bi)(c + di) = ac + bci + adi + bdi2 = ac + bci + adi − bd = (ac − bd) + (ad + bc)i. Let us see what happens when we multiply a complex number z = a + bi with its complex conjugate: z z̄ = (a + bi)(a − bi) = a2 + − abi  − b2 i2 = a2 + b2 = |z|2. abi (1) We can use the above to divide complex numbers. Simply multiply the numerator and denom- inator by the complex conjugate of the denominator, the point being that this converts the denominator into the square of its modulus which is a real number: a + bi (a + bi)(c − di) ac + bd + bci − adi  ac + bd   bc − ad  = = = 2 + 2 i. c + di (c + di)(c − di) c2 + d2 c + d2 c + d2 Note that the above is only defined as long as c + di 6= 0 (just as for division in the real numbers). √ √ Example 1.2.1. (a) (3 + 74 i) + ( 2 − i) = (3 + 2) − 37 i 1 COMPLEX NUMBERS 3 (b) (1 + 4i)(7 − 3i) = 19 + 25i (c) (a + bi)2 = (a2 − b2 ) + 2abi. (d) (2 + 6i)(2 − 6i) = 40 1+i (1+i)(2+3i) −1+5i 1 5 (e) 2−3i = (2−3i)(2+3i) = 13 = − 13 + 13 i. 1 2−i 2−i 2 (f) 2+i = (2+i)(2−i) = 5 = 5 − 15 i. If z = a + bi is a non-zero complex number, then 1 a − bi a − bi a b 1/z = = = 2 2 = 2 2 − 2 i. a + bi (a + bi)(a − bi) a +b a +b a + b2 Let us check that 1/z is indeed the reciprocal of z: a − bi 1 a2 + b 2 (1/z).z = ( )(a + bi) = (a − bi)(a + bi) + = 1. a2 + b 2 a2 + b 2 a2 + b 2 We often use the notation z −1 for z1 to emphasise that the two numbers z and z −1 mutiply to give 1. Since a2 + b2 = |z|2 , the above equation gives us the following formula: If z 6= 0, then z̄ z −1 =. |z|2 1.3 Properties of conjugation and the modulus The next results list useful properties of complex conjugation and the modulus. You should understand and be able to reproduce the proofs given and also be able to do the proofs which have been left as exercises. 1.3.1. Let z and w be complex numbers. (i) z + w = z + w. (ii) z − w = z − w. (iii) zw = z w. (iv) z/w = z/w if w 6= 0. (v) z = z. Proof. Let z = a + bi, w = c + di. 1 COMPLEX NUMBERS 4 (i) Left hand side: (z + w) = (a + bi) + (c + di) = (a + c) + (b + d)i = (a + c) − (b + d)i. Right hand side: z + w = (a + bi) + (c + di) = (a − bi) + (c − di) = (a + c) − (b + d)i. Left hand side= Right hand side. (ii) Exercise. (iii) LHS: (zw) = (a + bi)(c + di) = (ac − bd) + (ad + bc)i = (ac − bd) − (ad + bc)i. RHS: z w = (a + bi)(c + di) = (a − bi)(c − di) = ac − bd − (ad + bc)i. LHS=RHS. (iv) Exercise. (v) Exercise. 1.3.2. Let z and w be complex numbers. (i) |zw| = |z||w|. (ii) |w−1 | = 1/|w| if w 6= 0. (iii) |z/w| = |z|/|w| if w 6= 0. Proof. Let z = a + bi, w = c + di. (i) LHS: |zw| = |(a + bi)(c + di)| = |(ac − bd) + (ad + bc)i| p = (ac − bd)2 + (ad + bc)2 √ = a2 c2 + b2 d2 − 2abcd + a2 d2 + b2 c2 + 2abcd √ = a2 c 2 + b 2 d 2 + a2 d 2 + b 2 c 2 RHS: √ √ |z| |w| = a2 + b 2 c 2 + d 2 p = (a2 + b2 )(c2 + d2 ) √ = a2 c2 + b2 c2 + a2 d2 + b2 d2 LHS=RHS. 1 COMPLEX NUMBERS 5 (ii) This can be directly by calculating the left and right sides separately. Here we give a different argument. Consider the equation w.w−1 = 1. Taking modulus of both sides |w.w−1 | = |1| = 1. By (i), |w.w−1 | = |w||w−1 |. So, the above equation becomes |w||w−1 | = 1 Dividing through with |w|, 1 |w−1 | = |w| as required. (iii) Follows from (i) and (ii): z 1 (i) 1 (ii) 1 |z| | | = |z. | = |z|| | = |z| =. w w w |w| |w| Proposition 1.3.3. (i) z z̄ = |z|2. (ii) |z̄| = |z|. Proof. (i) This is done in Equation (1). (ii) Exercise. 1.4 Quadratic equations with real coefficients. The imaginary numbers i and −i are the two square roots of −1, that is i and √ −i are the √ two 2 solutions of x = −1. Similarly, if d < 0 is any negative real number, then −di and − −di are the two square roots of d, e.g. 2i and −2i are the square roots of −4. The quadratic equation ax2 + bx + c = 0, a 6= 0, a, b, c ∈ R. can be solved by the method of completing squares. The method is as follows: Multiply both sides by 4a: 4a2 x2 + 4abx + 4ac = 0. 1 COMPLEX NUMBERS 6 Add b2 to both sides: 4a2 x2 + 4abx + b2 + 4ac = b2 4a2 x2 + 4abx + b2 = b2 − 4ac. Rewrite LS as a square (2ax + b)2 = b2 − 4ac. Take square roots of both sides: √ 2ax + b = ± b2 − 4ac Solving for x gives the quadratic formula: √ b2 − 4ac −b ± x= (2) 2a Notation. The number b2 − 4ac is called the discriminant of the polyniomial ax2 + bx + c. We have three cases depending on whether the discriminant is positive, zero, or negative.: b2 − 4ac > 0. The quadratic equation has two real, distinct solutions (also called roots), namely √ −b ± b2 − 4ac. 2a b2 − 4ac = 0. The equation has exactly one solution, namely −b. 2a This is called a repeated solution. b2 −4ac < 0. Then b2 − 4ac has an imaginary square root, p −(b2 − 4ac) i. The quadratic equation has two distinct complex solutions, namely p −b ± −(b2 − 4ac) i. 2a The two solutions are complex conjugates of each other. Example 1.4.1. 1. Consider x2 + 2x − 7 = 0. So, a = 1, b = 2, c = −7, b2 − 4ac = 32 > 0. Two real solutions: √ √ −2 + 32 −2 − 32 ,. 2 2 1 COMPLEX NUMBERS 7 2. Consider x2 + 2x + 5 = 0. So, a = 1, b = 2, c = 5, b2 − 4ac = −16 < 0. Pair of complex conjugate solutions: √ √ −2 + 16 i −2 − 16 i , 2 2 or −1 + 2i, −1 − 2i. Graphical interpretation. The graph of the function y = ax2 + bx + c is a parabola. The three cases for the quadratic formula correspond to different types of graphs. If b2 − 4ac > 0 then the graph crosses the x-axis at two points corresponding to the two real soultions of the equation. If b2 − 4ac = 0, then the x-axis is tangent to the graph at the point corresponding to the unique solution. If b2 − 4ac < 0, then the graph does not cross the x-axis, and the equation has non-real soultions. y = ax2 + bx + c, a > 0 y y y x x x b2 − 4ac > 0, b2 − 4ac = 0, b2 − 4ac < 0, two real roots repeated real root non-real complex conjugate roots 1.5 Properties of Complex Addition and Multiplication. 1.5.1. Properties of Addition: (i) Associativity. For all complex numbers u, w, z, we have (z + w) + u = z + (w + u). (ii) Identity. There is a complex number 0 such that for any complex number z, z + 0 = z and 0 + z = z. (iii) Inverse For any complex number z, there is a complex number −z such that z + (−z) = 0 = (−z) + z, where −z = −a − bi. 1 COMPLEX NUMBERS 8 (iv) Commutativity. For all complex numbers z, w, z + w = w + z. Properties of Multiplication: (v) Associativity. For all complex numbers z, w, u, we have (zw)u = z(wu). (vi) Distributivity. For all complex numbers z, w and u. u(z + w) = uz + uw and (z + w)u = zu + wu (vii) Identity. There is a complex number 1 such that for any complex number z, 1z = z = z1. (viii) Inverse. For any non-zero complex number z, there is a complex number z −1 = 1/z such that z −1 z = 1 = zz −1. (ix) Commutativity. For all complex numbers z, w, zw = wz. The characteristics of addition and multiplication listed above can be summarized by the fol- lowing sentence: The set of complex numbers is a field under addition and multiplication. We will give rigorous sense to this sentence later in the course. Any number system with an addition and multiplication which satisfies properties (i)–(ix) is a field. The set of real numbers is also a field. However, the set Z of integers is not a field. This is because Property (viii) fails: The inverse of 2 is 1/2 but 1/2 is not an integer. 1 COMPLEX NUMBERS 9 1.6 The Complex Plane. Complex plane. The real numbers are represented as points on a line. Complex numbers have two components, their real and imaginary parts, so can be represented as points on a plane which is called the complex plane or Argand plane or Argand diagram. The point (a, b) represents the complex number a + bi. The horizontal axis contains all real numbers and is called the real axis. The vertical axis contains all the purely imaginary axis; we call the vertical axis the imaginary axis. We say that (a, b) are the Cartesian coordinates of z = a + bi. Im(z) −2 + 3i The point with coordinates (−2, 3) represents the com- plex number z = 2 + 3i. 1 The point with coordinates Re(z) (0, −2) represents the com- plex number −2i. The point with coordinates (1, 0) rep- resents the real number 1. −2i Figure 1: Complex Plane (a, b) √ r r = a2 + b2 = |z| cos θ = a/r, sin θ = b/r a = r cos θ, b = r sin θ. θ (a, 0) Figure 2: Polar Form Let z = a + bi be a complex number represented by the point (a, b) on the complex plane. Let r be the distance of (a, b) from the origin. If z 6= 0, we let θ be the angle that the positive real axis makes with the line connecting z to the origin (measured in the anticlockwise direction). If we drop a perpendicular from (a, b) to the x-axis and consider the right-angled triangle with corners (0, 0), (a, 0) and (a, b), we obtain the above picture. 1 COMPLEX NUMBERS 10 1.7 Polar Form For any real number θ, set eiθ = cos θ + i sin θ. If z = a + bi is a complex number and r and θ are as in Figure 2, we obtain a = r cos θ, b = r sin θ, hence a + bi = r cos θ + r sin θi = r(cos θ + i sin θ) = reiθ. Thus, every non-zero complex number can be written in two ways: Cartesian Form: a + bi, a, b ∈ R. Polar Form: reiθ , r, θ ∈ R , r > 0. Note that if z = 0, then r = 0 and θ is not defined. The polar form of 0 is just 0. Example 1.7.1. √ Let z = 1 + i, represented √ iπ by the point (1, 1). The line joining (1, 1) has length r = 2 and θ = π/4. Hence z = 2e 4 in polar form. The next result answers the following question: In how many ways can a complex number be written in polar form? Proposition 1.7.2. Suppose that the complex number z has Cartesian form a + bi and has polar form reiθ , r > 0. Then √ r = |z| = a2 + b2. cos θ = √ a a2 +b2 , sin θ = √ b a2 +b2. Proof. a + bi = z = reiθ = r(cos θ + i sin θ) = r cos θ + ir sin θ. Comparing real and imaginary parts on both sides a = r cos θ, b = r sin θ. (3) Squaring a and b: a2 = r cos2 θ, b2 = sin2 θ a2 + b2 = r2 cos2 θ + r2 sin2 θ = r2 (cos2 θ + sin2 θ) = r2. Taking square roots of the left and right sides gives: √ a2 + b 2 = r √ proving the first statement. Dividing through by r in Equation (3) and using r = a2 + b 2 gives a a b b cos θ = = √ and sin θ = = √ r a2 + b 2 r a2 + b 2 as required. 1 COMPLEX NUMBERS 11 The proposition shows that if z = reiθ , with r > 0, then the number r is uniquely determined, namely r = |z|. However, the number θ can be replaced by θ + 2kπ for any integer k. This is because for any two numbers θ and θ0 , cos θ0 = cos θ and sin θ0 = sin θ ⇐⇒ θ0 = θ + 2πk for some k ∈ Z Definition 1.7.3. If z = reiθ , r > 0, then θ is called an argument of z. If θ is an argument of z, then so is θ + 2kπ for any integer k. The principal argument of z is the unique argument in the interval (−π, π] and is written Arg (z). The argument of 0 is not defined (0 = 0eiθ for all θ ∈ R). Example 1.7.4. Arg (1 + i) = π/4 whereas any of the numbers π/4, π/4 ± 2π, π/4 ± 6π, π/4 ± 8π etc. is an argument of z. 1.8 Changing between polar and cartesian form √ Suppose we want to write a + bi in polar form reiθ. By Proposition 1.7.2 r = a2 + b2 so is easy to write down. Also, a cos θ = √ a + b2 2 and a sin θ = √. a2 + b 2 We solve the equation cos θ = √a2a+b2 for θ using the inverse cosine function. We have to be careful as the range of the inverse cosine function is [0, π], which only covers the upper half circle. We have two cases: Case 1. b ≥ 0. Then z lies above or on the real axis which means that θ belongs to the closed interval [0, π], hence √ θ = cos−1 (a/ a2 + b2 ). Case 2. b < 0. Then z lies below the real axis which means that θ belongs √ to the open interval (−π, 0), hence −θ belongs to (0, π). Since cos(−θ) = cos θ = a/ a2 + b2 , we obtain √ −θ = cos−1 (a/ a2 + b2 ). In other words, √ θ = − cos−1 (a/ a2 + b2 ). √ √ To decide whether θ is cos−1 (a/ a2 + b2 ) or − cos−1 (a/ a2 + b2 ), it is a good idea to plot z on the complex √ plane. √ √ In case a/ a2 + b2 is one of the standard trignometric values 0, ±1/2, ±1/ 2, ± 3/2, ±1, then θ should be expressed explicitly, not in terms of cos−1. 1 COMPLEX NUMBERS 12 Example 1.8.1. Find the modulus, principal argument and polar form for each of the following. (i) 2 + 3i. (ii) 2 − 3i. (iii) −1. √ (iv) − 3 − i. Solution. It always helps to draw a picture! Let z = reiθ. p √ √ (i) a = 2, b = 3, so r = 22 +√(−3)2 = 13, cos θ = 2/ 13 and z is above the real axis (as b > 0), hence θ = cos−1 (2/ 13). √ √ √ −1 √ |2 + 3i| = 13, Arg (2 + 3i) = cos−1 (2/ 13), polar f orm : 2 + 3i = 13ei cos (2/ 13) p √ √ (ii) a = 2, b = −3, so r =√ 22 + (−3)2 = 13, cos θ = 2/ 13, and z is below the real axis, hence θ = − cos−1 (2/ 13). √ √ √ −1 √ |2−3i| = 13, Arg (2−3i) = − cos−1 (2/ 13), polar f orm : 2−3i = 13e−i cos (2/ 13) , (iii) | − 1| = 1 and Arg z = π (Could also use the formulas as above but this would be overkill). Polar form: −1 = eiπ Check out this equation! √ √ (iv) a = − 3, b = −1, so r = 2, cos θ = − 3/2, and z is below the real axis, hence √ 5π θ = − cos−1 (− 3/2) = − 6 √ √ 5π √ 5πi | − 3 − i| = 2, Arg (− 3 − i) = − , polar f orm : − 3 − i = 2e− 6. 6 Example 1.8.2. Suppose |z| = 7, Arg z = π/3. Write z in √Cartesian form. Solution. z = 7eiπ/3 = 7(cos(π/3) + i sin cos(π/3)) = 72 + 7 23i. 1.9 Multiplication in polar form. Revision (trignometric identities): cos(θ + α) = cos θ cos α − sin θ sin α. sin(θ + α) = sin θ cos α + cos θ sin α. 1 COMPLEX NUMBERS 13 The trignometric identities can be rewritten as: eiθ eiα = (cos θ + i sin θ)(cos α + i sin α) = (cos θ cos α − sin θ sin α) + i(sin θ cos α + cos θ sin α) = cos(θ + α) + i sin(θ + α) = ei(θ+α). The identity eiθ eiα = ei(θ+α) explains the exponential notation. If one takes the exponential property as a starting point, then comparing real and imaginary parts one obtains the trigno- metric identities. In future courses this connection will be explored further. For now, the above gives us a neat way to multiply and divide complex numbers in polar form. 1.9.1. Let z = reiθ and w = seiα be complex numbers. Then (i) zw = rsei(θ+α). (ii) If w 6= 0, then w−1 = s−1 ei(−α). (iii) z w = rs−1 ei(θ−α). (iv) z̄ = rei(−θ). Proof. (i) zw = reiθ seiα = rseiθ eiα = rsei(θ+α). by (i) (ii) ws−1 ei(−α) = seiα s−1 ei(−α) = ei0 = cos 0 + i sin 0 = 1. Dividing through by w gives (ii). (iii) Combine (i) and (ii). (iv) z = reiθ = r cos θ + ir sin θ, z̄ = r cos θ − ir sin θ = r cos(−θ) + ir sin(−θ) = rei(−θ). In words, to multiply in polar form multiply the r’s and add the θ’s. To obtain the inverse of a complex number: take the reciprocal of the r and negate the θ. To obtain the conjugate of a complex number: take the same r and negate the θ. Addition is easier in Cartesian form. 2eiπ/4 Example 1.9.2. (a) Write (2eiπ/4 )(5eiπ/2 ) and 5eiπ/2 in polar form. (b) Suppose that Arg z = π/6 and |z| = 2. Find Arg z̄ and |z̄|. Answer: 2eiπ/4 (a) (2eiπ/4 )(5eiπ/2 ) = 10eiπ/4 eiπ/2 = 10ei3π/4 , 5eiπ/2 = 25 ei(−π/4). (b) z = 2eiπ/6 , so z̄ = 2ei(−π/6) , |z| = 2 and Arg (z) = −π/6. 1 COMPLEX NUMBERS 14 Example 1.9.3. Write the following in the form cos α + i sin α. (a) (cos(7θ) + i sin(7θ))/(cos(2θ) + i sin(2θ)). (b) (cos(3θ) + i sin(3θ))(cos(6θ) − i sin(6θ)). (c) (cos(2θ) + i sin(2θ))2 (cos θ + i sin θ)3. Answer: (a) (cos(7θ) + i sin(7θ))/(cos(2θ) + i sin(2θ)) = ei7θ /ei2θ = ei5θ = cos(5θ) + i sin(5θ). (b) (cos(3θ) + i sin(3θ))(cos(6θ) − i sin(6θ)) = ei3θ ei(−6θ) = ei(−3θ) = cos(−3θ) + i sin(−3θ). (c) (cos(2θ) + i sin(2θ))2 (cos θ + i sin θ)3 = (ei(2θ) )2 (eiθ )3 = ei(4θ) ei(3θ) = ei(7θ) = cos(7θ) + i sin(7θ). Theorem 1.9.4. (De Moivre’s theorem) For all integers n, we have (eiθ )n = einθ. In other words, (cos θ + i sin θ)n = cos(nθ) + i sin(nθ). 1.10 Roots Definition 1.10.1. Let z and w be complex numbers and let n be a positive integer. We say that w is an n-th root of z if wn = z. In particular, a square root of z is a 2-th root of z and a cube root of z is a 3-th root of z. Let z = reiθ , w = seiα be non-zero complex numbers. Then w is an n-th root of z if and only if wn = z ⇐⇒ (seiα )n = reiθ ⇐⇒ sn einα = reiθ ⇐⇒ sn = r and nα = θ + 2kπ, k ∈ Z (By Proposition 1.7.2) θ + 2kπ ⇐⇒ s = r1/n and α = , k ∈ Z. n Thus the n-th roots of z are the complex numbers of the form θ 2kπ w = r1/n ei( n + n ) , k ∈ Z. Now it seems that since we can choose k to be any integer, the equation has infinitely many solutions. But this is not the case since replacing k by k + n, k − n, k + 2n , k − 2n, k + 3n, k − 3n etc. yields the same w. Also, as k runs through the integers 0, 1, 2, · · · , n − 1, we obtain different values of w. In other words, the solutions are repeating with a period of n, and k = 0, 1, 2, · · · , (n − 1) will give all the distinct solutions. Since 0n = 0, 0 is not an n-th root of any non-zero number. Thus, we have proved: 1 COMPLEX NUMBERS 15 Proposition 1.10.2. Let z = reiθ be a non-zero complex number and let n be a positive integer. Then z has n distinct roots of n-th roots, namely θ θ 2π θ 4π θ 2(n−1)π r1/n ei n , r1/n ei( n + n ) , r1/n ei( n + n ) ,..., r1/n ei( n + n ). In other words, the n-th roots of z are the n complex numbers with modulus |z|1/n = r1/n and argument one of θ θ 2π θ 4π θ 2(n − 1)π , + , + ,..., +. n n n n n n n Let us note two special cases of the above proposition. n = 2. The square roots of z are √ iθ √ θ √ θ re 2 and rei( 2 +π) = − rei 2. n = 3. The cube roots of z are θ θ 2π θ 4π r1/3 ei 3 , r1/3 ei( 3 + 3 ) , and r1/3 ei( 3 + 3 ). Notation. If in the above θ is the principal argument of z, i.e. −π < θ ≤ π, then we set θ z 1/n := r1/n ei n. √ √ Example 1.10.3. (a) Solve w3 = −4 3 + 4i, that is find the third roots of −4 3 + 4i. √ (b) Find the fourth roots of −4 3 + 4i (c) Suppose z is a complex number such that |z| = 81 and let w = 5eiπ/4. Is w a cube root of z? Answer: (a) We need the polar form of the right hand side. Modulus: √ q √ √ | − 4 3 + 4i| = (−4 3)2 + 42 = 64 = 8. √ Principal argument: −4 3 + 4i lies above the real axis, so √ ! √ ! 4 3 3 θ = cos−1 − = cos−1 − = 5π/6. 8 2 So √ −4 3 + 4i = 8ei5π/6. 1 COMPLEX NUMBERS 16 The solutions are the numbers with modulus 81/3 = 2 and argument one of 5π 5π 2π 17π 5π 4π 29π , + = , + =. 18 18 3 18 18 3 6 Thus the solutions are 5π 17π 29π 2ei 18 , 2ei 18 , 2ei 18. Note that the argument of the last root is not the principal argument as 29π18 is greater than 29π −7π i −7π 29π π. So we may replace it by 18 − 2π = 18 and write 2e 18 instead of 2ei 18. This last step is optional and may be skipped unless explictly asked for. √ (b) By (a) −4 3 + 4i = 8ei5π/6. Hence the fourth roots are: 5π 17π 29π 41π −7π 81/4 ei 24 , 81/4 ei 24 , 81/4 ei 24 , 81/4 ei 24 = 81/4 ei 24. (c) |w| = 5 ⇒ |w|3 = 125 6= 81 = |z|. Thus w is not a cube root of z. 1.11 Geometric interpretation. As explained in Figure 2, if z = a + bi, then |z| is the distance between (a, b) and (0, 0) and the complex plane. We obtain: The complex numbers z which satisfy |z| = r are exactly the points which lie on the circle of radius r and centre (0, 0). Example 1.11.1. (i) Describe on the complex plane the set of all complex numbers z satis- fying |z − 2|2 = 25. (ii) Describe all complex numbers z satisyfying (2 − i)z + (2 + i)z̄ = 4. Answer: (i) |z − 2|2 = 25 ⇐⇒ |a + bi − 2|2 = 25 ⇐⇒ (a − 2)2 + b2 = 25. Thus the complex numbers satisfying |z − 2|5 = 25 are the points of the circle with radius 5 and center (2, 0). (ii) (2 − i)z + (2 + i)z̄ = 4 ⇐⇒ (2 − i)(a + bi) + (2 + i)(a − bi) = 4 ⇐⇒ 4a + 2b = 4 ⇐⇒ b = 2 − 2a. Hence the solution consists of all points on the line y = 2 − 2x. If w1 and w2 are complex numbers with the same modulus and argument α and α + β respectively, then on the complex plane w2 can be obtained from w1 by rotating anti- clockwise by the angle β. 1 COMPLEX NUMBERS 17 The n-th roots of a non-zero complex number z with argument θ all have modulus |z|1/n and their arguments are : θ θ 2π θ 4π θ 2(n − 1)π , + , + ,..., +. n n n n n n n Each successive number in the list is obtained from the previous by rotating by an angle of 2π n. Thus the nth roots of z form the vertices of a regular n-sided polygon on the circle of radius |z|1/n. Example 1.11.2. Find the cube roots of 27 and draw them on the complex plane. Answer: In polar form 27 = 27ei0 , hence the cube roots of 27 are 0 0 2π 2π 0 4π 4π 3ei 3 = 3, 3ei( 3 + 3 ) = 3ei 3 , 3ei( 3 + 3 ) = 3ei 3. The three roots, called w0 , w1 and w2 on the Argand diagram: As another example, the three solutions, w0 , w1 , w2 of Example 1.10.3 (a) can be depicted as follows on the complex plane: If we join up the three points representing the three 3-rd roots we will obtain an equilateral triangle. In general, if we join up the points representing the n-th roots of a positive number r we will obtain a regular polygon with n sides inscribed inside a circle of radius r1/n. Remark. In this section we have seen that given any comple number z, the equation wn = z has a solution in the complex numbers. In fact, much more is true. The equation wn = z may be replaced by any polynomial equation with complex coefficients. 1 COMPLEX NUMBERS 18 Theorem 1.11.3. (The fundamental theorem of Algebra) Let α0 , α1 ,... , αn−1 be complex num- bers. The equation wn + αn−1 wn−1 + · · · + α1 w + α0 = 0 has a complex solution. 2 MATRICES 19 2 Matrices 2.1 Definitions and Notation. Definition 2.1.1. An m × n matrix over is an array of elements of numbers (real, complex, natural, integers) set out in m rows and n columns. An m × n matrix is rectangular if m 6= n, and square if m = n.     1 2 3 1 i Example 2.1.2. is a 2 × 3 matrix over R, is a 2 × 2 matrix over C. 4 5 6 -i 2 The first matrix is rectangular, the second is square.   x1  x2  A row vector (x1 , x2 ,... , xn ) is a 1 × n matrix. A column vector ..  is an n × 1   .  xn matrix. A general m × n matrix A is written as A = (aij )1≤i≤m,1≤j≤n or simply as A = (aij ) where aij is the entry in row i and column j, i.e.   a11 a12... a1n  a21 a22... a2n  A = .... .  .. ...  am1 am2... amn We call the entry in row position i and column position j of a matrix A the (i, j)-entry of A.   a b If A is a 2 × 2, we often write A =. Here, the (1, 1)-entry of A is a, the c d (1, 2)-entry is b, the (2, 1)-entry is c, and the (2, 2)-entry is d. The m × n zero matrix, denoted0mn or simply 0, is the matrix whose entries are all 0. 0 0 0 For example, 0 = 033 =  0 0 0 . 0 0 0 The n × n identity matrix In (or simply I) is the square n ×n matrix with 1s on the 1 0 0 leading diagonal and 0s everywhere else. For example I = I3 =  0 1 0 . 0 0 1 2 MATRICES 20 2.2 Matrix Operations. 2.2.1 Addition. Addition is only defined for matrices of the same size. To add two matrices, simply add the corresponding entries in each position. More formally, given m × n matrices A = (aij ) and B = (bij ), A + B is the m × n-matrix defined by A + B = (aij + bij ). In other words, the (i, j)-entry of A + B is the sum of the (i, j)-entry of A and the (i, j)-entry of B.       1 2 −1 0 0 2 Example 2.2.1. (i) + =.    3 4  1 1 4 5  a b 0 0 a b 0 0 a b (ii) + = = +. c d 0 0 c d 0 0 c d Subtraction works similarly: If A = (aij ) and B = (bij ) are two m × n matrices then A − B is the m × n matrix defined by A − B = (aij − bij ). 2.2.2 Scalar multiplication If A = (aij ) and λ ∈ R then λA is the m × n -matrix defined by λA = (λaij ). In other words, the (i, j)-entry of λA is λ times the (i, j)-entry of A.     1 2 10 20 Example 2.2.2. 10 3 4  =  30 40  5 6 50 60 2.2.3 Matrix Multiplication. Let A = (aij ) be an m × n matrix and B = (bij ) be a p × q matrix. The product AB is defined if and only if n = p. In other words, AB is defined if and only if the number of columns of the matrix A on the left is equal to the number of rows of the matrix B on the right. 2 MATRICES 21 If n = p, then AB defined, and it is the m × q matrix whose (i, j)-entry is ai1 b1j + ai2 b2j +... + ain bnj , 1 ≤ i ≤ m, 1 ≤ j ≤ q Thus, the (i, j)-entry of AB is obtained as follows: Multiply each entry of the i-th row of A with the corresponding entry of the j-th column of B and take the sum of these numbers. In vector terminology, the (i, j)-entry of AB is the dot product of the i-th row of A with the the j-th column of B. In sigma notation, the (i, j)-entry of AB is equal to n X aik bkj. k=1 In other words, the (i, j)-entry of AB equals n X [(i, k)-entry of A] × [(k, j)-entry of B]. k=1 Here recall the Sigma notation. Sigma Notation. If x1 , x2 , · · · , xn is a list, then we write n X X x1 + x2 + · · · + xn as xk or as xk. k=1 1≤k≤n     3 1 1 0 1 2 1 Example 2.2.3. Let A = ,B = 0 0 4 1 . So A is a 2 × 3-matrix −1 0 4 −1 −1 1 1 and B is a 3 × 4-matrix. (i) A has 3 columns and B has 3 rows. So, |{z} B is defined and is the 2 × 4 matrix: A |{z} 2×3 3×4     2 1 1 0   1 2 1  0 1 0 10 3 AB = 0 4 1  =. −1 0 4 −6 −5 3 4 −1 −1 1 1 (ii) B has 4 columns and A has 2 rows. So |{z} B |{z} A is not defined. 3×4 2×3 Note. When mutiplying matrices, the order of multiplication matters! 2 MATRICES 22   1 1    0 −1  1 2 3 4 5   Example 2.2.4. Let A = ,B =  3 4 . 0 −1 2 1 1   0 0  1 5 (i) Is AB defined? If so, state the size of AB and calculate the (1, 2)-entry of AB. (ii) Is BA defined? If so, state the size of BA and calculate the (4, 4)-entry of BA. Answer: (i) AB is defined. It is a 2 × 2-matrix. The (1, 2)-entry of AB is 1.1 + 2.(−1) + 3.4 + 4.0 + 5.5 = 36. (ii) BA is defined and is a 5 × 5-matrix. The (4, 4)-entry is 0.4 + 0.1 = 0.   1 3    2 1 3  20 20 Example 2.2.5. 3 2 =. 0 2 4 26 20 5 4     1 2 1 3 Example 2.2.6. Let A = and B =. 3 4 2 4      1 2 1 3 5 11 AB = = ; 3 4 2 4 11 25      1 3 1 2 10 14 BA = =. 2 4 3 4 14 20 AB 6= BA !!! The above example shows that matrix multiplication is not a commutative operation in general.           a b 1 0 a b 1 0 a b a b Example 2.2.7. (i) = , =. So c d 0 1 c d 0 1 c d c d for all 2 × 2-matrices A, AI = A and IA = A. (ii) The above generalises to all n. Let A = (aij ) be an n × n-matrix and I = In. Using sigma notation, the (i, j)-entry of IA equals n X [(i, k)-entry of I] × [(k, j)-entry of A]. k=1 2 MATRICES 23 By the definition of I, the (i, k)-entry of I is 0 if k 6= i and is 1 if k = i, hence all terms in the sum disappear except for the termcorresponding to k = i and we obtain n X [(i, k)-entry of I] × [(k, j)-entry of A] = (i, j)-entry of A. k=1 Thus, IA = A. Similarly, AI = A. 2.3 Properties of Matrix Operations. All properties below refer to matrices with real entries but apply also to the case that the entries in any field. 2.3.1. (Properties of matrix addition.) Let m, n be fixed natural numbers. (i) Associativity. For all m × n-matrices A, B, C, (A + B) + C = A + (B + C). (ii) Identity. There is a matrix 0 such that for any m × n-matrix A, A + 0 = A and 0 + A = A. (iii) Inverse. For any m × n-matrix A = (ai j) there is a matrix −A such that A + (−A) = 0 = −A + A, where −A = (−aij ). (iv) Commutativity. For all m × n-matrices A, B, A + B = B + A. 2.3.2. (Properties of multiplication.) Let n be a natural number. The product of two n × n- matrices is defined and is an n × n-matrix. For all n × n-matrices A, B, C we have: (vi) (Associativity) (AB)C = A(BC). (vii) (Distributivity) (A + B)C = AC + BC, and C(A + B) = CA + CB. (viii) (Identity) There exists an n × n-matrix I such that AI = A and IA = A. Note that as shown by Example 2.2.6 matrix multiplication does not satisfy the commutativity property. Remark. The above properties of matrix multiplication also hold for non-square matrices (provided multiplication is defined). 2.3.3. (Properties of Scalar Multiplication) Let A, B be m × n matrices, λ, µ ∈ R. (ix) λ(A + B) = λA + λB. (x) (λ + µ)A = λA + µA. (xi) (λµ)A = λ(µA). 2 MATRICES 24 (xii) 1A = A. The collection of properties (i) -(v) says: The set of m × n-matrices with entries in R form an abelian group under addition. The collection of Properties (i) -(v) and (vi)-(viii) says: The set of n × n-matrices with entries in R form a ring under addition and multiplication. The collection of Properties (i) -(v) and (ix)-(xii) says: The set of m × n-matrices with entries in R form a vector space under addition and scalar multiplication. Remark. The above properties can (and should) be proved. They all follow from the definiton of the matrix addition, scalar multiplication, and matrix multiplication. For instance, Example 2.2.7 gives a proof of Property (viii). 2.4 Inverses. Definition 2.4.1. Let A be an n × n matrix. If there is an n × n matrix B such that AB = I and BA = I, then A is invertible; the matrix B is the inverse of A and we write B = A−1.     3 5 2 −5 Example 2.4.2. Let A = and B =. 1 2 −1 3      3 5 2 −5 1 0 AB = = =I 1 2 −1 3 0 1 and      2 −5 3 5 1 0 BA = = = I. −1 3 1 2 0 1 Thus B = A−1. Not all square matrices are invertible as the next example shows.   1 1 Example 2.4.3. Show that A = is not invertible. 0 0 Answer: Let B = (bij ). Then      1 1 b11 b12 1 0 AB = I ⇐⇒ = 0 0 b21 b22 0 1     b11 + b21 b12 + b22 1 0 ⇐⇒ =. 0 0 0 1 The (2, 2)-entry of the matrix on the left is 0 (irrespective of the chosen values of the bij ’s), and the (2, 2)-entry of the right hand matrix is 1. Hence the last equation cannot hold. This means there is no B such that AB = I, i.e., A is not invertible. 2 MATRICES 25 Proposition 2.4.4. (a) If B and C are both inverses of A, then B = C. (b) If A and B are invertible n × n matrices, then AB is invertible and (AB)−1 = B −1 A−1. (c) If A is invertible, then A−1 is invertible and (A−1 )−1 = A. Proof. (a) Suppose B and C are both inverses of A. Then AB = I and CA = I. Hence, C = CI = C(AB) = (CA)B = IB = B. (b) AB(B −1 A−1 ) = A(BB −1 )A−1 = AIA−1 = AA−1 = I. Similarly, (B −1 A−1 )AB = B −1 (A−1 A)B = B −1 IB = BB −1 = I. Thus, B −1 A−1 is the inverse of AB. (c) Obvious from the definition of inverse. Remark. It can be shown that if AB = I holds then BA = I holds and if BA = I holds then AB = I also holds. So in practice we only need to check one of the two equations. This is what we will do from now on. 2.5 Transpose. Definition 2.5.1. Let A be an m × n-matrix. The transpose of A, denoted AT , is the n × m- matrix whose (j, i)-entry equals the (i, j) entry of A for all i, j, 1 ≤ i ≤ m and all j, 1 ≤ j ≤ n. In other words, AT is the matrix whose columns are the rows of A.   1 2   T 1 3 5 Example 2.5.2. If A =  3 4 , then A =. 2 4 6 5 6 Proposition 2.5.3. Assuming that the matrices have sizes which mean the operations are defined, the following holds. 1. (AT )T = A, 2. (A + B)T = AT + B T , 3. (αA)T = αAT (α ∈ R), 2 MATRICES 26 4. (AB)T = B T AT , 5. I T = I, 6. If A is invertible, then AT is invertible and (AT )−1 = (A−1 )T. Proof. (1) and(5) are obvious from the definition. Proof of (2): Let A = (aij ), B = (bij ) be m × n -matrices. Then (A + B)T and AT + B T are both n × m-matrices. (j, i)-entry of (A + B)T = (i, j)-entry (A + B) = aij + bij = (j, i)-entry of AT + (j, i)-entry ofB T = (j, i)-entry of AT + B T. Proof of (3): Similar to the proof of (ii) and will done as exercise. Proof of (4): Will be done as an exercise. Proof of (5): Will be done as an exercise. 2.6 Determinants To each square matrix can be assigned a number, called the determinant of A. The determinant of A is denoted by |A| or by det(A). Definition 2.6.1. Let A be a square matrix. (i) If A = (a11 ) is a 1 × 1 matrix, the determinant of A is given by |A| = a11. (ii) (Inductive definition) Let n > 1 and let A = (aij ) be an n × n-matrix. For each i, j let the mi,j be the determinant of (n − 1) × (n − 1)-matrix obtained by deleting the i-th row and the j-th column of A. Then n X n+1 |A| = a11 m11 − a12 m12 +... + (−1) a1n m1n = (−1)1+j a1j m1j. j=1 Let us see what the above definition means in the case of 2 × 2 and 3 × 3-matrices.   a11 a12 If A = is a 2 × 2 matrix, then a21 a22 |A| = a11 a22 − a12 a21. In other words, the determinant of a 2 × 2-matrix is the difference between the product of the diagonal entries ofthe matrix  and the product of the non-diagonal entries of the a b matrix. If we write A as , then the formula becomes |A| = ad − bc. c d 2 MATRICES 27   a11 a12 a13 If A =  a21 a22 a23  is a 3 × 3 matrix, then a31 a32 a33 a22 a23 a a a a |A| = a11 − a12 21 23 + a13 21 22 a32 a33 a31 a33 a31 a32 = a11 a22 a33 + a12 a23 a31 + a13 a21 a32 − a13 a22 a31 − a11 a23 a32 − a12 a21 a23. Pictorial visualisation of determinants of 2 × 2 and 3 × 3-matrices: 1 2 1 −2 Example 2.6.2. (i) = 4 − 6 = −2, = 0 − (−6) = 6. 3 4 3 0   1 4 7 (ii) If A =  2 5 8 , then 3 6 9 5 8 2 8 2 5 |A| = 1 −4 +7 6 9 3 9 3 6 = (45 − 48) − 4(18 − 24) + 7(12 − 15) = 0. 2.7 Row and Column expansion. The formula for |A| given in Definition 2.6.1 involves moving along the the first row. Now we record a remarkable fact: the determinant can be computed by moving along any row or any column. 2 MATRICES 28 Theorem 2.7.1. Let A = (aij ) be an n × n-matrix. For any r between 1 and n, we have n X |A| = (−1)r+j arj mrj (r-th row expansion). j=1 For any s between 1and n, we have n X |A| = (−1)i+s ais mis (s-th column expansion). i=1 Proof. Omitted. Pattern of signs for calculating determinants. (−1)i+j = 1 if i + j is even and (−1)i+j = −1 if i + j is odd. So, the signs associated to each matrix position alternate as we move along a row or a column starting with a + on the (1, 1)-position:     + − + −   + − + + −  − + − +  ,  − + − ,   + − + −  and so on.  − + + − + − + − + So, if we calculate determinants using 1st row or 1st column expansion, the pattern of signs will be +, −+, · · ·. If we do 2nd row or 2nd column expansion, the pattern of signs will be −, +, − · · ·. If we do 3rd row or 3rd column expansion, the pattern of signs will be +, −+, · · · Definition 2.7.2. The number mij in Definition 2.6.1(ii) is called the (i, j)-minor of A. The (i, j)-signed minor of A is the number (−1)i+j mij.   1 4 7 Example 2.7.3. Let A =  2 5 8 . Compute |A| expanding along 3 6 9 (i) second column. (ii) third row. Answer: Second column expansion. Since 1 + 2 is odd, the signs are alternating starting with −1 and we get |A| = −a12 m12 + a22 m22 − a32 m32 2 8 1 7 1 7 = −4 +5 −6 3 9 3 9 2 8 = −4(−6) + 5(−12) − 6(−6) = 0. Third row expansion: Since 3 + 1 is even, signs are alternating starting with + and we get 4 7 1 7 1 4 |A| = 3 −6 +9 = 3(−3) − 6(−6) + 9(−3) = 0. 5 8 2 8 2 5 2 MATRICES 29 1 4 7 Example 2.7.4. (i) Calculate 0 0 1. 1 3 −1 2 1 1 3 0 1 4 7 (ii) Calculate. 0 0 0 1 0 1 3 −1 Answer: (i) The second row has many zeros. So we use second row expansion: 1 4 7 4 7 4 7 1 4 1 4 0 0 5 = −0 +0 −1 = −1 − 1.(−1) = 1. 3 −1 1 −1 1 3 1 3 1 3 −1 (ii) Using first column expansion: 2 1 1 3 1 4 7 0 1 4 7 by previous part =2 0 0 5 = 2.1 = 2. 0 0 0 1 1 3 −1 0 1 3 −1 2.8 Upper triangular, lower triangular and diagonal matrices. Definition 2.8.1. A square matrix is called upper triangular if all entries below the main diagonal are are equal to zero, lower triangular if all entries above the main diagonal are equal to zero, and A is called diagonal if all non-diagonal entries are equal to zero.   2 4 0 1    0 −1 5 0  2 0 0 For example,   5 0 0  is a lower  0 0 −3 3  is an upper triangular matrix,  1 1 −3 0 0 0 4  2 0 0 triangular matrix and  0 −1 0  is a diagonal matrix. 0 0 −3 If one computes the determinant of an upper (lower) triangular matrix by successive first column (row) expansion one obtains: Proposition 2.8.2. The determinant of any upper triangular, lower triangular or diagonal matrix is simply the product of the diagonal entries of the matrix. 2 4 0 1 2 0 0 2 0 0 0 −1 5 0 So for example, = 24, 5 0 0 = 0 and 0 −1 0 = 6. 0 0 −3 3 1 1 −3 0 0 −3 0 0 0 4 2 MATRICES 30 2.9 Properties of determinants. Proposition 2.9.1. The following hold. |I| = 1. |0| = 0. |AT | = |A|. If A is an n × n-matrix, then |αA| = αn |A|, (α ∈ R). |AB| = |A||B| The first four properties follow easily from the row and column expansion formula for deter- minant and by induction-if you think about them you will see why they hold. The proof of |AB| = |A||B| is harder. The determinant is a powerful concept. We will finish this section by listing one further im- portant property of determinants (to be memorised). Theorem 2.9.2. Let A be a square matrix. Then A is invertible if and only if |A| = 6 0. 2.10 Trace. Definition 2.10.1. The trace of an n × n square matrix A = (aij ) denoted Tr(A), is the sum of the diagonal entries of A, that is i=n X Tr(A) = a11 + a22 +... + ann = aii. i=1   1 2 3 Example 2.10.2. If A =  4 5 6 , then Tr(A) = 1 + 5 + 9 = 15. 7 8 9 Proposition 2.10.3. Let A and B be square matrices of the same size. Tr(A + B) = Tr(A) + Tr(B), Tr(αA) = αTr(A) (α ∈ R). Tr(AT ) = Tr(A). Proof. Exercise. 2 MATRICES 31 2.11 Elementary row operations There are three types of elementary row operations that can be performed on matrices: (i) multiply one row by a non-zero constant; (ii) swap two rows; (iii) replace one row with the sum of itself and the multiple (positive or negative) of another row. If a matrix B is obtained from a matrix A by successively performing elementary row operations, then we say that B is row equivalent to A. Every elementary row operation is reversible, so if B is row equivalent to A, then A is row equivalent to B.   2 3 4 Example 2.11.1.. 2 2 2   2 3 4 Multiplying Row 2 by 1/2 gives:. 1 1 1   1 1 1 Swapping Row 1 and Row 2 of the new matrix:. 2 3 4  1 1 1 Replacing Row 2 with Row 2 -2 Row 1 of the new matrix:. All the above matrices 0 1 2 are row equivalent to one another. Definition 2.11.2. A matrix A is in echelon form if the following hold: all rows containing only zeros are at the bottom of the matrix; the first non-zero entry in any row is to the right of the first non-zero entry of the row above it. A matrix A is in reduced echelon form the following holds: A is in echelon form; The first non-zero entry in any row is equal to 1; and The first non-zero entry in any row is the only non-zero entry in its column.     1 2 3 4 1 0 0 2 Example 2.11.3.  0 5 6 7  is in echelon form but not in reduced echelon form;  0 1 0 3  0 0  8 9  0 0 1 4 1 0 0 2 is in reduced echelon form;  0 1 0 3  is not in echelon form. 1 0 1 4 2 MATRICES 32 2.12 Row reduction Row reduction is the process of applying elementary row operations to a matrix to obtain a matrix in echelon form.   1 1 2 9 Example 2.12.1. Reduce  2 4 −3 1  to echelon form and to reduced echelon form. 3 6 −5 0 Solution. The first step is to make all other entries (below the leading diagonal) in the first column 0.     1 1 2 9 1 1 2 9 ρ2 −2ρ1 ρ3 −3ρ1 −→  0 2 −7 −17  −→  0 2 −7 −17 . 3 6 −5 0 0 3 −11 −27 The next step is to place a 1 in the second position on the diagonal:   1 1 2 9 ρ2 /2 −→  0 1 −7/2 −17/2 . 0 3 −11 −27 Reduce all entries in the third column below the diagonal to zero:   1 1 2 9 ρ3 −3ρ2 −→  0 1 −7/2 −17/2 . 0 0 −1/2 −3/2 The matrix is now in echelon form. Now the next step is to place a 1 on the third diagonal position:   1 1 2 9 −2ρ3 −→  0 1 −7/2 −17/2 . 0 0 1 3 We take the process further to remove the entries above the diagonal -this is the backward part of the process.This part of the process is called Gauss-Jordan elimination. Reduce the entries above the third diagonal to zero:     1 1 2 9 1 1 0 3 ρ2 +7ρ3 /2 ρ1 −2ρ3 −→  0 1 0 2  −→  0 1 0 2  0 0 1 3 0 0 1 3 Reduce the entry above the second diagonal to zero:   1 0 0 1 ρ1 −ρ2 −→  0 1 0 2 . 0 0 1 3 This is the reduced echelon form. 2 MATRICES 33 Note that all matrices in the backward part of the process are all in echelon form. So the echelon form of a matrix is not unique. However, the reduced echelon form is unique. We have the following: Every matrix is row equivalent to a unique matrix in reduced echelon form. Remarks. The sequence of row operations for changing a matrix to its reduced echelon form is not unique. For example, it can be convenient to change the first non-zero entry in a row to a 1 by multiplying that row by the reciprocal of the first non-zero entry before making all entries below the leading entry equal to 0. We may equally speak of elementary column operations but we will stick to row operations in this course. 2.13 Row reduction and systems of linear equations Consider a system of linear equations 2x + 6y = 4 3x − y = 16 The augmented matrix of the system is the matrix   2 6 4. 3 −1 16 The vertical line indicates the separation of the coefficients of the variables on the left hand side of the system from the values of the right hand side. The matrix on the left hand side of the vertical line is called the coefficient matrix of the system. To solve the system: (i) multiply first equation by 1/2 to give: x + 3y = 2 3x − y = 16 (ii) Subtract 3 times the first equation from the second equation to give: x + 3y = 2 −10y = 10 2 MATRICES 34 (iii) Multiply second equation by −1/10: x + 3y = 2 y = −1 (iv) Subtract 3 times the second equation from the first (or use back substitution) x = 5 y = −1 This can be carried out on the augmented matrix.       1     2 6 4 ρ1 /2 1 3 2 ρ2 −3ρ1 1 3 2 − 10 ρ2 1 3 2 ρ1 −3ρ2 1 0 5 −→ −→ −→ −→ 3 −1 16 3 −1 16 0 −10 10 0 1 −1 0 1 −1 giving the solution x = 5, y = −1. Note that we could also have stopped the row reduction process when we reached an echelon form and could have then proceeded to solve by backward substitution. General method for solving systems of linear equations: Do row operations on the augmented matrix to bring the coefficient matrix into (reduced) echelon form. This method is sometimes called Gauss Jordan elimination. Example 2.13.1. x+y−z =1 2x + 2y = 2 x+z =0 Solution.       1 1 −1 1 1 1 −1 1 1 1 −1 1  2 2 0 2  ρ2 −2ρ 1 ,ρ3 −ρ1 −→  0 0 2 0 2 ↔ρ3  ρ−→  0 −1 2 −1  1 0 1 0 0 −1 2 −1 0 0 2 0     1 1 −1 1 1 1 −1 1 (−1)ρ2 ρ3 /2 −→  0 1 −2 1  −→  0 1 −2 1 . 0 0 2 0 0 0 1 0 Could use back substitution to solve or can proceed:     1 1 0 1 1 0 0 0 ρ2 +2ρ3 ,ρ1 +ρ3 ρ1 −ρ2 −→  0 1 0 1  −→  0 1 0 1 . 0 0 1 0 0 0 1 0 Solution: x = 0, y = 1, z = 0. 2 MATRICES 35 2.14 Types of solution sets. Example 2.14.1. Solve x+y−z =1 2x + 2y = 2 x+y+z =0 Solution.         1 1 −1 1 1 1 −1 1 1 1 −1 1 1 1 −1 1  2 2 0 2  ρ2 −2ρ1 ,ρ3 −ρ1 −→  0 0 2 ρ2 /2 0  −→  0 0 1 ρ3 −2ρ2 0  −→  0 0 1 0 . 1 1 1 0 0 0 2 −1 0 0 2 −1 0 0 0 −1 The last row corresponds to the equation 0x + 0y + 0z = −1 which clearly has no solution. So, the system does not have a solution. Example 2.14.2. x+y−z =1 2x + 2y = 2 x + y + z = 1. Solution. The coefficient matrix is the same as the previous example so doing the same operations but on the new augemented matrix we obtain     1 1 −1 1 1 1 −1 1  2 2 0 2  −→  0 0 1 0 . 1 1 1 1 0 0 0 0 Proceeding further   1 1 0 1 ρ1 +ρ2 −→  0 0 1 0 . 0 0 0 0 This corresponds to z = 0, x + y = 1. There are infinitely many solutions: z = 0, y = λ, x = 1 − λ, where λ is any real number. Summary of types of solutions. (No solution) Reduced echelon form contains a row with all entries zero on the left of the line but the entry in this row after the vertical line is a non zero number. In this case the system has no solution. Such a system is called inconsistent. 2 MATRICES 36 Suppose that the system is not inconsistent. Then the number of non-zero rows of the reduced echelon form is less than or equal to the number of variables in the system. – (Unique solution) If the number of non-zero rows of the reduced echelon form is equal to the number of variables in the system, then the system has exactly one solution. – (Infinitely many solutions) If the number of non-zero rows of the reduced echelon form is less than the number of variables in the system, then the system has infinitely many solutions. Remark. If the number of variables is equal to the number of equations, i.e. the coefficient matrix is a square matrix, then there is a unique solution if and only if the reduced echelon form of the coefficient matrix is I. 2.15 Row reduction and Inverse Gauss-Jordan elimination method: Row reduce an n × n matrix A to reduced echelon form: If the reduced echelon form is is I, then A is invertible and we can obtain A−1 by applying the same operations to I. If the reduced echelon form is not I, then A is not invertible.   1 2 Example 2.15.1. Let A =. 1 4       1 2 1 0 ρ2 −ρ1 1 2 1 0 1 2 ρ2 /2 1 0 −→ −→ 1 4 0 1 0 2 −1 1 0 1 −1/2 1/2       ρ1 −2ρ2 1 0 2 −1 ρ2 /2 1 2 1 0 ρ1 −2ρ2 1 0 2 −1 −→ −→ −→. 0 1 −1/2 1/2 0 1 −1/2 1/2 0 1 −1/2 1/2   2 −1 A−1 =. −1/2 1/2 Checking our result we have      −1 1 2 2 −1 1 0 AA = =. 1 4 −1/2 1/2 0 1 Note that our answer for A−1 coincides with the formula in Proposition 2.17.1. 2 MATRICES 37   1 2 Example 2.15.2. Let A =. 2 4     1 2 1 0 ρ2 −2ρ1 1 2 1 0 −→. 2 4 0 1 0 0 −2 1 The row of zeros before the vertical line means that the reduced echelon form of A is not I. Hence A is not invertible.   1 2 3 Example 2.15.3. Let A =  2 5 6 . Find A−1 if it exists. 3 6 8 Solution.       1 2 3 1 0 0 1 2 3 1 0 0 1 2 3 1 0 0 2 −2ρ1 3 −3ρ1  2 5 6 0 1 0  ρ−→  0 1 0 −2 1 0  ρ−→  0 1 0 −2 1 0  3 6 8 0 0 1 3 6 8 0 0 1 0 0 −1 −3 0 1   1 2 3 1 0 0 −ρ3 −→  0 1 0 −2 1 0  (At this stage we may already conclude that A is invertible.) 0 0 1 3 0 −1   1 0 0 −4 −2 3 ρ1 −3ρ3 ,ρ1 −2ρ2 −→  0 1 0 −2 1 0 . 0 0 1 3 0 −1   −4 −2 3 A−1 =  −2 1 0 . 3 0 −1 Check the answer:      −4 −2 3 1 2 3 1 0 0  −2 1 0   2 5 6  =  0 1 0 . 3 0 −1 3 6 8 0 0 1   1 2 3 Example 2.15.4. Let A =  0 5 6 . Find A−1 if it exists. 3 6 9 Solution.     1 2 3 1 0 0 1 2 3 1 0 0 3 −3R1  0 5 6 0 1 0  R−→  0 5 6 −2 1 0 . 3 6 9 0 0 1 0 0 0 −3 0 1 Reduced echelon form 6= I, hence A is not invertible. 2 MATRICES 38   1 2 −4 Example 2.15.5. Let A =  0 2 6 . Find A−1 if it exists. 0 0 1 Solution.       1 2 −4 1 0 0 1 2 −4 1 0 0 1 2 −4 1 0 0 1/2ρ2 ρ2 −3ρ3 A =  0 2 6 0 1 0  −→  0 1 3 0 1/2 0  −→  0 1 0 0 1/2 −3  0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1   1 2 0 1 0 4 ρ1 +4ρ3 −→  0 1 0 0 1/2 −3  0 0 1 0 0 1   1 0 0 1 −1 10 ρ1 −2ρ2 −→  0 1 0 0 1/2 −3 . 0 0 1 0 0 1   1 −1 10 −1 A =  0 1/2 −3 . 0 0 1 Check the answer:      1 −1 10 1 2 −4 1 0 0  0 1/2 −3   0 2 6  =  0 1 0 . 0 0 1 0 0 1 0 0 1 Note that since A is upper triangular and all diagonal entries of A are non-zero, |A| 6= 0 and we could have concluded right at the beginning that A is invertible. 2.16 Systems of linear equations and invertible matrices. Let a11 x1 + · · · + a1n xn = b1... am1 x1 + · · · + amn xn = bm be a general system of m-linear equations in n variables x1 , x2 , · · · , xm. This system can be written as a matrix equation (where on the left we have the product of an m × n matrix with an n × 1 matrix and on the right we have an m × 1-matrix.)      a11... a1n x1 b1 ....  ..  = ..  or .. .  .  am1... amn xn bm 2 MATRICES 39 Ax = b where     a11... a1n x1 A = .....  is the coefficient matrix of the system, x = ..  is the n × 1 .  .  am1... amn xn   b1 ..  matrix of variables and b = .  is the n × 1 matrix of values. bm Now suppose that m = n and suppose that the matrix A is invertible. Multiplying both sides of the equation Ax = b by A−1 gives A−1 Ax = A−1 b; Ix = A−1 b; x = A−1 b. This gives us a method to solve equations (only when the coefficient matrix is invertible): If A is invertible, then the system has a unique solution given by x = A−1 b. Example 2.16.1. Solve the systems. (i) x + 2y + 3z = 1 2x + 5y + 6z = 1 3x + 6y + 8z = −1. (ii) x + 2y + 3z = 0 2x + 5y + 6z = 0 3x + 6y + 8z = 2.     1 2 3 x Solution. For both systems the coefficient matrix is A =  2 5 6 , x =  y .   3 6 8 z −4 −2 3 −1 A =  −2 1 0  (see Example 2.15.3). 3 0 −1 2 MATRICES 40 (i)     x 1 A y  =  1 . z −1          x 1 −4 −2 3 1 −9  y  = A−1  1  =  −2 1 0   1  =  −1 . z −1 3 0 −1 −1 4 x = −9, y = −1, z = 4. (ii)    x 0 A y  =  0 . z 2          x 0 −4 −2 3 0 6  y  = A−1  0  =  −2 1 0  0  =  0 . z 2 3 0 −1 2 −2 x = 6, y = 0, z = −2. Advantage of the method: If A is invertible, once A−1 has been calculated, can be used to solve for any system of values. Disadvantage: Only works if the number of variables is equal to the number of equations and A is invertible. Let us summarize some of what we have learnt in the last few sections. Theorem 2.16.2. Let A be an n × n matrix, b an n × 1 column matrix with real entries and x an n × 1 column matrix of variables. The following statements are equivalent. 1. |A| = 6 0. 2. A is an invertible matrix. 3. The reduced echelon form of A is I. 4. The system of equations Ax = b has a unique solution. 2 MATRICES 41 2.17 Finding the inverse of a matrix using minors. Determinants can be used to even find the inverse of a matrix. We first treat the case of 2 × 2-matrices.   a b Proposition 2.17.1. Let A =. If |A| = 6 0, c d     −1 1 d −b 1 d −b A = =. |A| −c a ad − bc −c a If |A| = 0, then A is not invertible. Proof.   1 d −b a b −c a ad − bc c d =     1 da − bc db − bd 1 ad − bc 0 = = I. ad − bc −ca + ac −cb + ad ad − bc 0 ad − bc       2 3 5 −3 5/13 −3/13 Example 2.17.2. If A = , then A−1 = 1 =. −1 5 13 1 2 1/13 2/13 Let A be an n × n matrix. Recall from the section on determinants (section 2.7) that the (i, j)-minor mij of A is the determinant of the (n − 1) × (n − 1) matrix obtained by deleting the i-th row and the j-th column of A and that the signed (i, j) minor of A is the number (−1)i+j mij. The signed (i, j) minor of A is called the (i, j)-cofactor of A. Definition 2.17.3. Let A be an n × n matrix, and let mij be the (i, j)-minor of A. The cofactor matrix also called the matrix of signed minors of A is the n×n-matrix whose (i, j)-entry is (−1)i+j mij. The adjoint matrix of A is the transpose of the cofactor matrix of A. Theorem 2.17.4. Let A be an n×n matrix, and let C be the matrix of signed minors (cofactor matrix) of A. Then 1. C T A = |A|I. 6 0, A−1 = 2. Provided that |A| = 1 |A| CT. 2 MATRICES 42   2 4 3 Example 2.17.5. Let A =  0 1 −1  3 5 7 (i) Find all the minors of A. (ii) Find the the matrix of cofactors of A. (ii) Calculate A−1 (if it exists). Solution. (i) Minors: 1 −1 0 −1 0 1 m11 = = 12, m12 = = 3, m13 = = −3, 5 7 3 7 3 5 4 3 m21 = = 13, m22 = 5, m23 = −2 5 7 m31 = −7, m32 = −2, m33 = 2. (ii) Cofactor matrix     m11 −m12 m13 12 −3 −3 C =  −m21 m22 −m23  =  −13 5 2 . m31 −m32 m33 −7 2 2 (iii) Expanding along the second row (it’s got a 0) gives 2 3 2 4 det(A) = 0 + − (−1) = (14 − 9) + (10 − 12) = 3 6= 0 3 7 3 5 so A is invertible.  T   12 −3 −3 12 −13 −7 1 T 1 1 A−1 = C = −13 5 2  =  −3 5 2  |A| 3 3 −7 2 2 −3 2 2     12 −13 −7 4 −13/3 −7/3 1 =  −3 5 2 = −1 5/3 2/3 . 3 −3 2 2 −1 2/3 2/3   a b Now suppose that A = is 2 × 2-matrix. The minors are: m11 = d, m12 = c, m21 = b c d   d −c m22 = a. The cofactor matrix is C =. If |A| = 6 0, −b a     −1 1 T d −b 1 d −b A = C = = |A| −c a ad − bc −c a as in Proposition 2.17.1. 3 SETS, FUNCTIONS AND RELATIONS 43 3 Sets, Functions and Relations 3.1 Sets. A set is any well-defined collection of distinct objects called the elements of the set. There are several ways to describe sets. One could give a verbal rule describing which elements are members of the set, or one could explicitly list each member of the set and enclose the list in curly brackets., or one could use a combination of the two methods. Example 3.1.1. The following are four different descriptions of the same set. The set of all integers n such that n2 is less than 10. {−1, −2, −3, 0, 1, 2, 3}. {n ∈ Z : n2 < 10}. {n ∈ Z : −3 ≤ n ≤ 3}. The notation : in the above is used to denote “such that”; a single vertical line may also be used, e.g. in the above example we could also write {n ∈ Z | n2 < 10}. In general, X = {x : x has certain properties } or X = {x|x has certain properties }. If a set is described as an explicit list within curly brackets, then the list can be written in any order, and repeats do not count. For example, {3, 1, 4, 2} = {1, 2, 3, 4} = {1, 3, 3, 2, 4, 4}. Some standard notation. If A is a set, then a ∈ A means that “a is an element of A” or “a is in A” and a ∈ / A means that “a is not an element of A”. The symbol ⇔ means “if and only if ”. The symbol ⇒ means “implies”. ∃ means “there exists”. ∀ means “for all”. Some standard sets. The empty set is the set containing no elements and is denoted by ∅ or {}. 3 SETS, FUNCTIONS AND RELATIONS 44 N: the set of all positive integers: N = {1, 2, 3, 4...}. N is also called the set of natural numbers. Z: the set of all integers: Z = {... , −2, −1, 0, 1, 2,...}. Q: the set of all rational numbers Q = {p/q : p, q ∈ Z and q 6= 0}. R: the set of all real numbers (rational and irrational). C: the set of all complex numbers. For n a natural number, Matn (R): the set of all n × n-matrices with real number entries. √ Example 3.1.2. (i) 1 + i ∈ {z ∈ C : |z| = 2}. √ (ii) 1 + i 6∈ {z ∈ R : |z| = 2}. This is because 1 + i is not a real number. (iii) 1 + i ∈ {z ∈ C : |z| ≤ 3}. (iv) 1 + i 6∈ {z ∈ C : |z| < 2}.   1 −2 (v) ∈ {A ∈ Mat2 (R) : AT = A}. −2 3   3 1 (vi) 6∈ {A ∈ Mat2 (R) : Tr(A) = 0}. −2 3   3 1 (vii) ∈ {A ∈ Mat2 (R) : Tr(A) = 6, |A| = 11}. −2 3 Note that as in the last example, if several conditions are placed, they are separated by a comma. They may also be separated by “and”. 3.2 Subsets Two sets are equal if they have the same elements. If every element of a set A is also an element of the set B, then we say that A is a subs