6 1. Preliminaries equations in n unknowns, Ax = b, we can find the general solution by using Gaussian elimination to reduce A to row echelon form. Since the elementary row operations of Gaussian elimination clearly preserve the row space, the row rank of A is the number of leading ones (equivalently, number of nonzero rows) in a row echelon form of A. One can then show that the set of columns of A corresponding to the positions of the leading ones in an echelon form is a basis for the column space of A hence, the row rank equals the column rank. The matrix A will be invertible if and only if the transformation T is invertible. Hence, A is invertible if and only if m = n and A has rank n, or equivalently, nullity 0. We also use the word nonsingular. The reader has no doubt seen some version of the following list of equivalent conditions. Theorem 1.4. Let A be an n × n matrix. Then the following are equivalent. (1) The rank of A is n. (2) The columns of A are linearly independent. (3) The columns of A span Fn. (4) The rows of A are linearly independent. (5) The rows of A span Fn. (6) For any b in Fn, there is exactly one solution to Ax = b. (7) The only solution to Ax = 0 is the trivial solution x = 0. (8) The matrix A is invertible. (9) The determinant of A is nonzero. (10) All of the eigenvalues of A are nonzero. 1.6. Change of Basis and Similarity Let V be an n-dimensional vector space, and suppose we have two bases for V, labeled B = {v1,..., vn} and R = {r1,..., rn}. Let x = ∑n j=1 xjvj. Then [x]B = x1 x2 . . xn and [x]R = ∑n j=1 xj[vj]R. Let P be the n × n matrix with [vj]R in column j. We then have [x]R = P [x]B. We call P the change of basis matrix. The matrix P is invertible, and thus [x]B = P −1 [x]R. Now let T : V W be a linear transformation from an n-dimensional vector space V to an m-dimensional vector space W. Suppose B and R are two bases for the space V, and suppose C and S are two bases for the space W. Consider the following two matrix representations for T : A = [T ]B,C and B = [T ]R,S (see Figure 1.1). Then for any x V, we have [T (x)]C = A[x]B, (1.1) [T (x)]S = B[x]R. (1.2)
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