A matrix A is not invertible if and only if 0 is an eigenvalue of A which is if and only if if and only if
I've been browsing over some Linear Algebra this summer, and came across this problem in a practice book. Any help on this? It should be solvable using only elementary methods.
Let A be an nxn matrix with characteristic polynomial
Prove that A is invertible if and only if is not equal to 0.
if a0=0, then 0 is an eigenvalue. So there exist a nonzero vector X such that AX=0X=0. If A is invertible, then A^(-1)AX=A^(-1)0=0=X, which will make contradiction. So A is not invertible. If A is not invertible, there exists nontrivial solution for the equation groups represented by A. That is, there is a nonzero X such that AX=0=0X. So 0 is an eigenvalue. Since f(t)=(t-r1)(t-r2)...(t-rk),where ri is eigenvalue, a0=(-1)^(k)r1...rk. So a0=0