5
$\begingroup$

Let $V$ be a (finite-dimensional) vector space and $A \colon V \to V$ a linear map.

Is it true that, if $v$ is an eigenvector of $A\circ A$, then $v$ is an eigenvector of $A$?

I know the converse statement is true.

$\endgroup$
6
$\begingroup$

Hint: Consider $$A=\begin{pmatrix}1&0\\0&-1\end{pmatrix}. $$ Or $$A=\begin{pmatrix}0&1\\0&0\end{pmatrix}. $$

$\endgroup$
3
$\begingroup$

A rotation by 90° as represented by a matrix e.g. $$A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$$ has no real eigenvalue ($\chi_A(\lambda)=\lambda^2+1$ is the characteristic polynomial) and so no eigenvectors. The rotation by 180° $$A^2=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$$ has the whole space as a space of eigenvectors and zero vector .

$\endgroup$
  • 1
    $\begingroup$ I love this answer. It's very intuitive, since $A\cdot v$ is orthogonal to $v$, yet $A^2 \cdot v$ is anti parallel. $\endgroup$ – infinitezero Sep 15 at 20:56
2
$\begingroup$

No, it is false: consider in a real vector space $$A=\begin{bmatrix} 0&1\\1&0\end{bmatrix}\implies A\circ A=\begin{bmatrix} 1&0\\0&1\end{bmatrix}$$ Then the vector $\begin{bmatrix} 1\\0\end{bmatrix}$ is an eigenvector of $A^2$ and not of $A$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.