7. (25 points) Let A be an nx n matrix. Justify each of the following statements

Question

7. (25 points) Let A be an nx n matrix. Justify each of the following statements. If it is correct, give a brief proof. Otherwise give an example to show that it it not correct (a) If matrix A is invertible, then its determinant cannot be 0. (b) If matrix A is invertible, then its columns are linearly independent. c)If matrix A is invertible, then its rows are linearly independent. (lint: consider the transpose matrix and use part (b) d) if matrix A is invertible, then there is an unique solution to the system of linear equation Ax -0. (e) If matrix A is invertible, then it cannot have 0 as eigenvalue

Explanation / Answer

(a). If the matrix A is invertible, then there exists a matrix B ( or A-1) such rthat AB = BA = In. Then det(A)*det(B) = det(In) =1. If det(A)= 0, then det(In) = 0,det(B) = 0 which is a contradiction.Hence, if A is invertible, then det(A) cannot be 0.

(b) If the columns of A are linearly dependent, then the rows of AT are linearly dependent so that by elementary row operations, AT can be reduced to a matrix B with a zero row. Thus, det(A) =det(AT) = k*det(B), where k is a scalar. Further, since det(B) = 0, hence det(A) = 0 so that A cannot be invertible (party (a) above), which is a contradsiction. Hence the columns of A are linearly independent.

( c). If the rows of A are linearly dependent, then A can be reduced to a matrix B with a zero row. Thus, det(A) = k*det(B), where k is a scalar. Further, since det(B) = 0, hence det(A) = 0 so that A cannot be invertible (party (a) above), which is a contradsiction. Hence the rows of A are linearly independent.

(d). If A is invertible and AX = 0, then, on multiplying to the left by A-1, we get A-1(AX) = A-1. 0 or, X = 0. Thus, X = 0 is the only(unique) solution to AX = 0.

(e). If A has 0 as an eigenvalue with a corresponding eigenvector v, then Av = 0.v = 0 so that on multiplying to the left by A-1, we get v = 0, a contradiction as an eigenvector is always non-zero. Hence 0 cannot be an eigenvalue of an invertible matrix A.

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