# Math Help - How to find the companion form of a matrix from its eigenvalues?

1. ## How to find the companion form of a matrix from its eigenvalues?

In the literature, there are a lot of explanations about finding the jordan canonical form of a matrix. However, there is almost nothing about finding a matrix from its eigenvalues, especially multiple eigenvalues. I need to obtain the transpose companion form (Bush form) of a matrix whose eigen values are given. For example how can I find the companion form of a matrix A whose eigen values are -1,-1, and -1 (i.e. three repeated eigenvalues.)

We know that S^-1*A*S=J. Since we know the eigen values, we can write J. And A=S*J*S^-1. There are some formulas to find S when the eigen values are distinct (e.g. S=[1 1 1; e1 e2 e3; e1^2 e2^2 e3^2] for a 3x3 matrix). However, when the eigen values are multiple this formulas fail. For such a case, how can we obtain S?

To find a general formulation for our case I tried to make some calculations. I found a formula for two repeated and one distinct eigen values case. i.e. When eigen values are e1 e1 e3, S=[1 0 1; e1 1 e3; e1^2 2*e1 e3^2]. However, for triple eigen values case (e1 e1 e1) I couldn't find any solution.

If there is some having an idea, please share it

2. We have had basically this same question a number of times now.

You can't recover a matrix (or its "companion" form) just from its eigenvalues. You would also have to know its eigenvectors.

For example,
$\begin{bmatrix}-1 & 0 & 0 \\ 0 & -1 & 0\\ 0 & 0 & -1\end{bmatrix}$,

$\begin{bmatrix}-1 & 1 & 0 \\ 0 & -1 & 0\\ 0 & 0 & -1\end{bmatrix}$,

and $\begin{bmatrix}-1 & 1 & 0 \\ 0 & -1 & 1\\ 0 & 0 & -1\end{bmatrix}$,

are all Jordan Normal forms for matrices having -1 as a triple eigenvalue.

3. ## Thank you HallsofIvy

Thanks for your remark, HallsofIvy. You are right but our problem is not the type of the jordan form. For example, let's assume the last jordan form is the case. In such a condition how can we find the required matrix of S (or indirectly A)?