Suppose $A^3-3A^2+2A$ is not invertible. Show that $A$ has a real eigenvalue

Question

I do not know how to start with this question. Could you give me some inspiration? Thank you in advance.

Answer 1

Let $B=A^3-3A^2+2A$. Since $B$ is not invertible there is $v\ne0$ such that $Bv=0$.

Now $B=A (A - I) (A - 2I)$.

Let $v_0=(A - I) (A - 2I)v$. If $v_0\ne0$, then $Av_0=0$ and so $0$ is an eigenvalue of $A$ with eigenvector $v_0$.

Otherwise, let $v_1=(A - 2I)v$. If $v_1\ne0$, then $(A-I)v_1=0$ and so $1$ is an eigenvalue of $A$ with eigenvector $v_1$.

Otherwise, $(A-2I)v=0$ and so $2$ is an eigenvalue of $A$ with eigenvector $v$.

Therefore, at least one of $0,1,2 \in \mathbb R$ is an eigenvalue of $A$.