Prove/disprove: if $X$ is an eigenvector of $T$ then X is a singular matrix

Question

I have this question:

Let $A$ be a non-scalar matrix of an order $(N\times N)$. And let $T:M_{n\times n}^R \rightarrow M_{n \times n}^R$ such that: $T(X) = AX$ for every $ X \in M_{n\times n}^R$

Prove/disprove: if $X$ is an eigenvector of $T$ then X is a singular matrix.

Now I think that this is true but I don't know how to prove this,

I first said that if $X$ is an eigenvector and $T(X) = AX$, then $A$ must be its eigenvalue. and if it's an eigenvalue, it must somehow zero $T$'s determinant (I'm not sure about this), anyway, I'm having difficulties correlating this info to $X$,

Some help? Thanks!

Answer 1

Suppose that $X$ is an eigenvector of $T$ with the eigenvalue $\lambda$. Then, $AX=T(X)=\lambda X$, or $(A-\lambda I)X=0$. If $X$ is nonsingular, then $X^{-1}$ exists and $A-\lambda I=\big((A-\lambda I)X\big)X^{-1}=0\,X^{-1}=0$, which violates the assumption that $A$ is not a scalar matrix.

I think it is a good exercise for you to show that $T$ is diagonalizable if and only if $A$ is so.