# $\lambda$ is an eigenvalue of $A$ if and only if $\lambda^2$ is an eigenvalue of $A^2$

Is this statement true? It's easy to prove the forward direction: $$Av=\lambda v\Rightarrow\lambda (Av)=\lambda(\lambda v)\Rightarrow A(\lambda v)=\lambda^2 v\Rightarrow A(Av)=\lambda^2 v\Rightarrow A^2v=\lambda^2 v$$ But the backward direction is eluding me. I figured the contrapositive statement, $Av\neq\lambda v\Rightarrow A^2v\neq\lambda^2 v$, might help, but I don't see how. Thanks in advance.

• What if $A=-I$ ? Commented Jan 1, 2018 at 23:42
• The reverse direction needs some interpretation. For example, $(-1)^2$ is an eigenvalue of $I^2$ but $-1$ is not an eigenvalue of $I$. Commented Jan 1, 2018 at 23:43

The backward direction is false. Consider that $A^2v=\lambda^2 v$. Then $(A^2-\lambda^2I)v=0$, so $(A-\lambda I)(A+\lambda I)v=0$, implying that at least one of $\lambda$ or $-\lambda$ are eigenvalues of $A$, but not necessarily $\lambda$.
In the real plane rotation through a quarter of a circle has no (real) eigenvalues but its square is $-I$, which has eigenvalue $-1$ with multiplicity $2$.
• I don't see what this has to do with the question. The phrasing suggests that you intend it to be a counterexample, but it doesn't appear to contradict anything, since there is no $\lambda \in \mathbb R$ for which $\lambda^2 = -1$. Commented Jan 2, 2018 at 0:52