If $A$ is similar to $cA$ then $A $ is nilpotent

Question

If $A$ is a $n\times n $matrix such that $A$ is similar to $cA$, for complex scalar $c$ and $|c| \neq 1$, show that $A$ is nilpotent.

I tried to prove that if $A$ has an eigenvalue $k$ then $ck$ is an eigenvalue for $cA$ But we can't say that $ck=k$ (if this is true then the statement is obviously true), we don't ensure the order of eigenvalues for each matrix, that is, $ck =l$ for another eigenvalue $l$ of $A$. Any help?

Answer 1

Let $\lambda$ be an eigenvalue of $A$ with the largest magnitude. The eigenvalue of $cA$ with largest magnitude is then $c \lambda$, but since $A,cA$ are similar, we must have $|\lambda |= |c| | \lambda | $ from which it follows that $\lambda = 0$. Hence all eigenvalues are zero.

Answer 2

If $p(X)$ is the minimal polynomial of $cA$, then $p(cX)$ is a polynomial which vanishes on $A$, so it is divisible by the minimal polynomial $q(X)$ of $A$.

Now, if $A$ and $cA$ are similar, then the two matrices have the same minimal polynomial so this tells us that $p(cX)$ is divisible by $p(X)$. Show that this implies that $p$ is a power of $X$.

Answer 3

As you observed, if $\lambda$ is an eigenvalue of $A$, then $c\lambda$ is an eigenvalue of $cA$. Since you are assuming that $A$ and $cA$ are similar, this implies that

if $\lambda$ is an eigenvalue of $A$, then so is $c\lambda$.

Now of $\lambda$ is an eigenvalue of $A$ and it is not zero, the infinitely many complex numbers $$\lambda,c\lambda,c^2\lambda,c^3\lambda,\dots$$ are pairwise different if $|c|\neq0$, and this tells us that $A$ has infinitely many different eigenvalues. Of course, this is impossible.

It follows that $0$ is the only eigenvalue of $A$, and then that $A$ is nilpotent.