Let $A$ be a (real or complex) square matrix, let $alpha neq 0$.

Is it true that $lambda$ is an eigenvalue of $A$ if and only if and only if $alpha lambda$ is an eigenvalue of $alpha A$?

I think yes, here is why I suppose so: $lambda$ is an eigenvalue of $A ifflambda I- A$ is non injective $iff alpha lambda I – alpha A$ is non injective $iff alpha lambda$ is an eigenvalue of $alpha A$.

Is this correct?