Prove that Det(A-E)=0 if and only if AC=C

Question

We have some $n \times n$ matrix $A$ and $n \times 1$ vector C. Let $E$ be the identity matrix. $$Det(A-E)=0 \iff AC=C.$$

Me and a few friends have been trying to prove it, but none of us could. Thank you!

Answer 1

you need to know that $det(A)$ if and only if the columns of $A$ are linearly dependent. once you have that, applying it to the matrix $A_I$ tells you that there is a nonzero vector $c$ such that $(A_I)c = 0$. that means $$Ac = c, c \neq 0.$$

Answer 2

note as $\det(A-I)=0$ therefore $1$ is an eigenvalue of $A$ so there has to be an eigenvector corresponding to it which is $c$