Mathematics Lecture Notes PDF

Mathematics Abhishek Kaushik  Linear algebra Linear: having to do with lines, planes, etc. Algebra: solving equations involving unknowns. What is system of linear equations ? Write few linear as well as non linear equations ?  It will be very important to...

Mathematics Abhishek Kaushik  Linear algebra Linear: having to do with lines, planes, etc. Algebra: solving equations involving unknowns. What is system of linear equations ? Write few linear as well as non linear equations ?  It will be very important to us to understand systems of linear equations both algebraically and geometrically. Algebraically (writing equations for their solutions). Geometrically (drawing pictures and visualizing).  Solve this system of equations: A system of linear equations need not have a solution. For example, there do not exist numbers x and y making the following two equations true simultaneously: Definition. A system of equations is called inconsistent if it has no solutions. It is called consistent otherwise.  Pictures of Solution Sets Before discussing how to solve a system of linear equations below, it is helpful to see some pictures of what these solution sets look like geometrically One Equation in Two Variables. Consider the linear equation x + y = 1. We can rewrite this as y = 1 - x, which defines a line in the plane: the slope is -1, and the x-intercept is 1. A solution to the system of both equations is a pair of numbers (x, y) that makes both equations true at once. In other words, it as a point that lies on both lines simultaneously. We can see in the picture above that there is only one point where the lines intersect: therefore, this system has exactly one solution. (This solution is (3,2)) Usually, two lines in the plane will intersect in one point, but of course this is not always the case. Consider now the system of equations So the system of linear equations can have: Only one solution (unique solution) No solution Infinite many solutions The planes defined by the equations x + y + z = 1 and x - z = 0 intersect in the red line, which is the solution set of the system of both equations. Row Reduction The Elimination Method Solve (1.1.1) using the elimination method. All three operation will still valid  Row Echelon Form We want to reduce the augmented matrix to Echelon form, to get it solved.  Reduced Row Echelon Form A matrix is in reduced row echelon form if it is in row echelon form, and in addition: If an augmented matrix is in reduced row echelon form, the corresponding linear system is viewed as solved. row echelon form of the matrix is What happens when one of the non-augmented columns lacks a pivot? : Multiple solutions Summary Systems of Linear Equations: Geometry A vector in R3, and its coordinates. Add these vectors geometrically Subtract these vectors geometrically Linear Combinations of the vector Demo to show w is not in Span of v1 and v2 https://textbooks.math.gatech.edu/ila/demos/spans.html?v1=2,-1,1&v2=1,0,-1&target=0,2,2&range=5 The vector w is not contained in Span {v1, v2 }, so the equation Ax = b is inconsistent. Demo to show w is in Span of v1 and v2 https://textbooks.math.gatech.edu/ila/demos/spans.html?v1=2,-1,1&v2=1,0,-1&target=1,-1,2&range=5 The vector w is contained in Span {v1, v2 }, so the equation Ax = b is consistent. How to solve AX = 0 ? Since RHS is 0. so we can ignore Construction of augmented matrix. (Compare to this example, where we solved the corresponding homogeneous equation.) (Compare this example.) According to the key observation, this is supposed to be a translate of a span by p. Indeed, we saw in the first example that the only solution of Ax = 0 is the trivial solution, i.e., that the solution set is the one-point set {0}. The solution set of the inhomogeneous equation Ax = b is has a nontrivial solution. We solve this by forming a matrix and row reducing  A set of two noncollinear vectors v, w is linearly independent: Neither is in the span of the other, so we can apply the first criterion. The span got bigger when we added w, so we can apply the increasing span criterion. The set of three vectors {v, w, u} below is linearly dependent: u is in Span {v, w}, so we can apply the first criterion. The span did not increase when we added u, so we can apply the increasing span criterion In the picture below, note that v is in Span {u, w}, and w is in Span {u, v}, so we can remove any of the three vectors without shrinking the span. These three vectors {v, w,u} are linearly dependent: indeed, {v, w} is already linearly dependent, so we can use the third fact.