# Do $A$ and $p(A)$ have the same eigenvectors?

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!

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

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.

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 $A$.

• 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. – Friedrich Philipp Mar 5 '16 at 3:27
• what means for you the sentence every vector is not an eigenvector of $A$? – Tsemo Aristide Mar 5 '16 at 3:29
• That's what you wrote: "Every vector is [...] not an eigenvector of $A$". – Friedrich Philipp Mar 5 '16 at 3:31
• So what means for you the sentence every vector is an eigenvector of $A^n$ but not an eigenvector of $A$? – Tsemo Aristide Mar 5 '16 at 3:32
• 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 $A$. – Friedrich Philipp Mar 5 '16 at 3:33

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$~A$.

• Beautiful examples! Thanks a lot! – syeh_106 Mar 9 '16 at 6:17

What happens if the polynomial is constant?

• That was in essentially already mentioned in the first comment (and in the first answer) – Friedrich Philipp Mar 5 '16 at 3:28
• 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". – Mariano Suárez-Álvarez Mar 5 '16 at 3:39
• (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) – Mariano Suárez-Álvarez Mar 5 '16 at 3:40
• As to your last comment (at least for comments this is ok to say ;o)): I meant the answer of Ian. – Friedrich Philipp Mar 5 '16 at 3:47