Properties of Determinants PDF

# 3.1-3.2 Properties of determinants Previously we used expansion of a determinant with respect to the first row. Turns out, one can use any row (or even a column) and get exactly the same result. But the formula needs a small modification. The following theorem is hard to prove, so we will just st...

# 3.1-3.2 Properties of determinants Previously we used expansion of a determinant with respect to the first row. Turns out, one can use any row (or even a column) and get exactly the same result. But the formula needs a small modification. The following theorem is hard to prove, so we will just state it and use it whenever we need to. Suppose *A* is an (*n* x *n*) matrix, and submatrix *A_i,j* is obtained from *A* by removing row *i* and column *j*. The number *C_i,j* = (-1)^i+j *det(A_i,j)* is called the (*i,j*) **cofactor** of *A*. ## Theorem The determinant of *A* can be computed by a cofactor expansion with respect to any row or column. ### Expansion w.r. to row *i*: *det(A) = a_i,1C_i,1 + a_i,2C_i,2 + ... + a_i,nC_i,n* ### Expansion w.r. to column *j*: *det(A) = a_1,jC_1,j + a_2,jC_2,j + ... + a_n,jC_n,j* Amazingly, the result is always the same! ## Properties of determinants ### Theorem If matrix *A* has a zero row, or a zero column, *det(A) = 0*. **Proof:** Suppose row *i* has all zeros, i.e. a_i,1=a_i,2=... a_i,n=0. Use this formula: *Det(A) = 0.C_1,1 + 0.C_1,2 + ... + 0.C_1,n= 0.* Same story with a zero column. ## Theorem 2 If *A* is an upper- or lower-triangular matrix, *det(A)* is equal to the product of diagonal entries: *det(A) = a_1,1a_2,2...a_n,n*. **Proof:** Let's prove for a lower triangular (4x4) matrix: *A = * a_1,1 0 0 0 * a_2,1 a_2,2 0 0 * a_3,1 a_3,2 a_3,3 0 * a_4,1 a_4,2 a_4,3 a_4,4 det (A) = a_1,1C_1,1 + 0.C_1,2 + 0 C_1,3+ 0.C_1,4 = a_1,1 det * a_2,2 0 0 * a_3,2 a_3,3 0 * a_4,2 a_4,3 a_4,4 = a_1,1 a_2,2 det * a_3,3 0 * a_4,3 a_4,4 = a_1,1 a_2,2 a_3,3 a_4,4 A similar proof holds in the general case. ## Theorem 3 I would like to study this theorem in more detail, than is given in the book. **(a)** Suppose matrix *A* is made of *B* by interchanging two adjacent rous. Same holds for a column interchange. Then, det(B)= -det(A) (determinant changes sign). Let's check for (2x2) matrices. Suppose *A = (a_b^c_d), det(A) = ad - bc*. Interchange rows: *B=(_a^c_d), det(B)=cb-ad = - (ad-bc) = -det(A)*. What about a larger matrix? Suppose we interchange row #2 and row #3 in a (3x3) matrix *A*. *det(A) = a_1,1 det(A_1,1) - a_1,2 det(A_1,2) + a_1,3 det(A_1,3)* Suppose *B* is *A* with interchanged 2-nd and 3-rd rows. Then *det(B)= a_1,1 det(B_1,1) - a_1,2 det(B_1,2) + a_1,3 det(B_1,3)* but *det(B_i,j)=-det(A_i,j)* because these submatrices have their rows interchanged. Therefore, *det(B) = - det(A)* Larger sizes are similar. **Same result holds if two non-adjacent rows (or columns are interchanged: the determinant changes sign.** **(b)** If two rows in a matrix are the same, the determinant equals to 0. (Same goes for colums) **Proof:** Suppose two rows in matrix *A* are the same, and matrix *B* is *A* with the two rows interchanged. On one hand, *B = A*, so det(B)=det(A). On the other hand, det(B) = -det(A). Combining these two equations we get det(A) = - det(A). Solve for det(A)! Obtain det(A) = 0. **(c)** Linearity with respect to a sum of rows Suppose the first row of a matrix is a sum of two row vectors: *(a_1,1 a_1,2 ... a_1,n) = (b₁ b₂ ... b_n) + (d₁ d₂ ... d_n)* Then *det(A) = a_1,1C_1,1 + a_1,2C_1,2 + ... a_1,nC_1,n = (b₁ + d₁)C_1,1 + (b₂ + d₂)C_1,2 + ... + (b_n + d_n)C_1,n = (b₁C_1,1 + b₂C_1,2 + ... b_nC_1,n) + (d₁C_1,1 + d₂C_1,2 + ... d_nC_1,n) = det * b₁ b₂ ... b_n + det * d₁ d₂ ... d_n *where yellow shading indicates the same entries as in matrix A. Same is true for any row, not just the first one.* **(d)** Adding a row from a matrix to another row does not change the determinant. **Sketch of a proof:** Suppose *A = (a_1,1 ... a_1,n) a_2,1 ... a_2,n ... a_n,1 ... a_n,n and B = (a_1,1 ... a_1,n a_2,1 ... a_2,n ... a_n,1 ... a_n,n) Same as in A* Then, *det(B) = det * a_1,1 ... a_1,n * a_2,1 ... a_2,n * ... * a_n,1 ... a_n,n + det * a_2,1 ... a_2,n * a_2,1 ... a_2,n * ... * a_n,1 ... a_n,n = det(a_1,1 ... a_1,n) + 0 = det(A)* These two rows are the same=>det(...)=0 So *det(B) = det(A)*, the determinant did not change. This holds for any rows or columns, not just the first two **(e)** If a column (or a row) is multiplied by a scalar λ, the determinant is multiplied by λ **Proof:** Let’s say *A = (a_1,1 ... a_1,n) B = (λa_1,1 λa_1,2 ... λa_1,n) same as A* Then, *det(B) = (λa_1,1)C_1,1 + (λa_1,2)C_1,2 + ... + (λa_1,n)C_1,n = λ(a_1,1C_1,1 + a_1,2C_1,2 + ... + a_1,nC_1,n) = λ det(A), as we wanted Same holds for any row or column to show* **Corollary:** det(λA) = λⁿdet(A). Prove it yourself. ## Why any of these is important? Recall that we can modify a matrix to an echelon form, by row interchanges, and by adding multiples of rows to other rows. The former changes the sign and the latter does not change the determinant. Therefore, if * A ~ U*, and *U* is an echelon form obtained without scaling the individual rows, then *det(A) = ± det(U)*. If *U* has all pivots (≠0), then *det(U) = the product of the pivots ≠ 0*. Therefore, *det(A) ≠ 0* and both *U* and *A* are **invertible**. If *U* does not have all pivots, *U* and *A* are **singular**. But *det(U) = 0* (since 0's on the diagonal), and then *det(A) = ±det(U) = ± 0 = 0*. We have proved that, if *det(A)=0*, *A* is **singular** (not invertible). And that, if *det(A) ≠0*, *A* is **invertible** (not singular). So, we have proved the following theorem: ## Theorem 4 A Square matrix *A* is **invertible** if and only if *det(A) ≠ 0* ## In addition Reduction to echelon form can be used to compute a determinant of a matrix (instead of the cofactor expansion). For large matrices this is much faster. This method is incorporated into Matlab, Python, and other commonly used computer systems.