0
$\begingroup$

It is easy to show that if $A$ is a square matrix, and $p(x)=c_nx^n+...+c_1x+c_0$ is a polynomial, then any eigenvector of $A$ (corresponding to eigenvalue $\lambda$) must also be an eigenvector of $p(A)$ (corresponding to eigenvalue $p(\lambda)$). Is the converse also true, i.e. is any eigenvector of $p(A)$ also an eigenvector of $A$? If so, how do we prove it? Thank you!

$\endgroup$
  • 1
    $\begingroup$ What if $c_i = 0$? $\endgroup$ – Michael Biro Mar 5 '16 at 3:03
5
$\begingroup$

No, basically because it can happen that the geometric multiplicity of $p(\lambda)$ is higher than that of $\lambda$. In turn this could happen because $A$ was defective, or because $p$ maps multiple distinct eigenvalues of $A$ to the same number. Here is an example of each of these phenomena:

  • $A=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ and $p(x)=x^2$. Then everything is an eigenvector of $p(A)$ but not vice versa.
  • Any square $A$ not a multiple of the identity, with $p$ constant. Again, then everything is an eigenvector of $p(A)$ but not vice versa.
$\endgroup$
3
$\begingroup$

No, take $A$ a non trivial nilpotent matrix of order $n$ and $p(X)=X^n$. Every vector is an eigenvector of $A^n=0$ but not an eigenvector of $A$.

$\endgroup$
  • $\begingroup$ This is only an example for special matrices. In fact, you can always choose the minimal or characteristic polynomial. Then $p(A) = 0$. Also your statement that every vector is not an eigenvector of $A$ is false since a nilpotent matrix has at least one eigenvector. $\endgroup$ – Friedrich Philipp Mar 5 '16 at 3:27
  • $\begingroup$ what means for you the sentence every vector is not an eigenvector of $A$? $\endgroup$ – Tsemo Aristide Mar 5 '16 at 3:29
  • $\begingroup$ That's what you wrote: "Every vector is [...] not an eigenvector of $A$". $\endgroup$ – Friedrich Philipp Mar 5 '16 at 3:31
  • $\begingroup$ So what means for you the sentence every vector is an eigenvector of $A^n$ but not an eigenvector of $A$? $\endgroup$ – Tsemo Aristide Mar 5 '16 at 3:32
  • $\begingroup$ This is not misunderstandable in my opinion. It means that every vector you choose is an eigenvector of $A^n$ but not an eigenvector of $A$. $\endgroup$ – Friedrich Philipp Mar 5 '16 at 3:33
2
$\begingroup$

No, the converse is false. Worse, for every matrix $A$ that is not a multiple of the identity (so that not all vectors are eigenvectors) there exist polynomials where this fails. Take for instance for $p$ a polynomial that annihilates $A$ (which always exists, for instance take the minimal or the characteristic polynomial of$~A$), then $p[A]=0$ and every vector is an eigenvector for$~p[A]$, while this is not so for$~A$.

One can also make this fail in smaller ways. For instance,whenever $A$ has at least two distinct eigenvalues $\lambda,\mu$, taking any polynomial $p$ with $p[\lambda]=p[\mu]$ (for instance $p=(X-\lambda)(X-\mu)$) will ensure that the eigenspaces $V_\lambda,V_\mu$ of$~A$ for $\lambda$ and $\mu$ are contained in the same eigenspace $W$ of$~p[A]$ (for the eigenvalue $p[\lambda]=p[\mu]$); then $V_\lambda\oplus V_\mu\subseteq W$ consists entirely of eigenvectors for$~p[A]$, but it contains many non-eigenvectors for$~A$.

$\endgroup$
  • $\begingroup$ Beautiful examples! Thanks a lot! $\endgroup$ – syeh_106 Mar 9 '16 at 6:17
1
$\begingroup$

What happens if the polynomial is constant?

$\endgroup$
  • $\begingroup$ That was in essentially already mentioned in the first comment (and in the first answer) $\endgroup$ – Friedrich Philipp Mar 5 '16 at 3:28
  • $\begingroup$ If the comment in the question intended to say that «$c_i=0$ for all i» or for positive i, then that intention did not realize. I read the other questions and it is not quite apparent that the trivial case of constant polynomials is considered there. "Essentially mentioned" is, often, quite different from "mentioned". $\endgroup$ – Mariano Suárez-Álvarez Mar 5 '16 at 3:39
  • $\begingroup$ (Saying "the first answer" is usually not very helpful, as the order of answers depends on one's settings: referring to answers by mentioning the author works much better) $\endgroup$ – Mariano Suárez-Álvarez Mar 5 '16 at 3:40
  • $\begingroup$ As to your last comment (at least for comments this is ok to say ;o)): I meant the answer of Ian. $\endgroup$ – Friedrich Philipp Mar 5 '16 at 3:47

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.