# Eigenvectors of the companion matrix [duplicate]

Suppose one has an Hermitian square matrix $A$ with $p$ is the characteristic polynomial $$p(x)= a_0 + a_1 x + \cdots + a_{n-1}x^{n-1} + x^n ~,$$ and define the companion matrix of $p$ as $$\tilde{A}=\begin{bmatrix} 0 & 0 & \dots & 0 & -a_0 \\ 1 & 0 & \dots & 0 & -a_1 \\ 0 & 1 & \dots & 0 & -a_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \dots & 1 & -a_{n-1} \end{bmatrix}.$$ For the definition of companion matrix, the eigenvalues of a the hermitian matrix $A$ coincide with the eigenvalues of the non-hermitian matrix $\tilde{A}$ (and with the roots of the polynomial $p$).

The question is: What about eigenvectors? Is there a relationship between the eigenvectors of the original hermitian matrix $A$ and the eigenvectors of the companion matrix $\tilde{A}$?

Note that there is no unitary transformation between the two matrices (one is hermitian and the other is not). Therefore one cannot find a unitary matrix $U$ such that $\tilde{A}=U A U^\dagger$ so that the eigenvectors transforms simply as $\tilde{v}=U v$. Maybe in this case one can still define a non-unitary transformation which does the job.

This is a follow up of this question.

## marked as duplicate by user99914, Daniel W. Farlow, user91500, JonMark Perry, Joey ZouAug 24 '16 at 4:45

• The left eigenvectors ($e\tilde{A}=\lambda e$) are just $[1,\lambda,\lambda^2,...,\lambda^{n-1}]$ – Empy2 Aug 1 '15 at 17:41
• @Michael do you have any reference or can you show me a proof of this (also sketched)? – sintetico Aug 1 '15 at 17:44
• It is a very simple calculation because of the form of $\tilde{A}$. – Empy2 Aug 1 '15 at 17:45
• @Michael And what about right eigenvectors? I am sorry for the naive question, but I find it difficult to deal with non-hermitian matrices. – sintetico Aug 1 '15 at 18:05
• Yes, sorry, I don't know what the right eigenvectors are. – Empy2 Aug 1 '15 at 18:48

This is more of an exploratory than definitive answer...but hopefully still helpful.

I suspect there is probably not much that can be said in terms of actually relating the eigenvectors. But what could be the most useful insight regarding these matrices (Hermitian vs Companion) are that they represent "extreme" cases of matrices with the same characteristic polynomial, specifically in terms of the amount of eigenvectors associated with each eigenvalue.

On the one hand is the Hermitian matrix which is always diagonalizable - it has a basis of eigenvectors (which is the maximum possible) - and even more can be said, they form a unitary basis. On the other hand the companion matrix is nonderogatory - its minimum polynomial and characteristic polynomial coincide. What this implies is that there is only one eigenvector per eigenvalue, which is the minimum possible (we are of course assuming the characteristic polynomial splits).

Here is a simple example with repeated roots: $I_2$. It's of course Hermitian, and already in diagonal form, find the companion matrix of its characteristic polynomial and find the eigenvector - there is only one.

What about if we have a characteristic polynomial with no repeated roots? Unfortunately, even though in this case both matrices will have a basis of eigenvectors, not much more can be said - again, here is a simple example to explore: $$\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$$

• If there is only one eigenvector per eigenvalue, how about a scalar times the eigenvector, isn't it still an eigenvector? – Linfeng Li Nov 26 '16 at 18:56
• Yes, you are right, it should read: 'one linearly independent eigenvector...' – Christiaan Hattingh Nov 26 '16 at 19:04

I was experimenting and found that you can form the right eigenvectors quite easily as well.

If $r$ is one of the roots of the characteristic polynomial $p(x)$ then listing the coefficients of $q(x)=p(x)/(x-r)$ upward starting with the $1$ at the bottom gives a right eigenvector corresponding to $\lambda=r$. For instance, consider $p(x)=(x-1)(x-2)(x-3)=x^3-6x^2+11x-6$. Then we can construct the $\lambda=2$ eigenvector $[3,-4,1]^t$ for the companion matrix using the coefficients from $q(x)=p(x)/(x-2)=x^2-4x+3$.

The proof is an easy exercise (in the sense that I was able to do it in 15 minutes) and left to the reader.

• As rule of thumb, I never trust proofs which are "an easy exercise left to the reader" ;) – sintetico Jan 28 '16 at 14:27