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I saw a question some time ago, asking about the eigenvalues of the matrix $$A=\begin{pmatrix}5&-3&0\\-3&5&0\\0&0&2\end{pmatrix}$$ which were then shown to be $\lambda=\left\{8,2,2\right\}$; and I thought, what a nice feature, we have a symmetric matrix and the eigenvalues seem to be $\lambda=\left\{5-(-3),5+(-3),2\right\}$. So I tried to show that it was indeed a property of the more general matrix $$M=\begin{pmatrix}a&b&0\\b&a&0\\0&0&c\end{pmatrix}$$ and sure enough, I found $\lambda=\left\{a+b,a-b,c\right\}$.

And I remembered that in fact, I had seen a few of these matrices in a Quantum Mechanics course, in the context of perturbation theory, where an Hamiltonian $\hat{H_0}=\text{diag}(a,a,c)$ was said to be degenerate as its eigenvalues were $\lambda=\left\{a,a,c\right\}$, but it was possible to lift the degeneracy by applying a perturbation $\hat{H_p}=\begin{pmatrix}0&b&0\\b&0&0\\0&0&0\end{pmatrix}$, leading to an overall Hamiltonian described by the matrix $M$ above, with distinct eigenvalues.

But I digress. My question is: knowing the eigenvalues, can we build a corresponding matrix? This question (or something close to it) has probably already been asked, but I'm not only interested in building one matrix from a set of eigenvalues, but all possible matrices.

Going back to perturbation theory, my question is directly motivated by trying to find perturbation matrices $\hat{H_p}$ that do not modify the eigenvalues of the unperturbed system $\hat{H_0}$. An (almost trivial) example is $$\hat{H}=\begin{pmatrix}a&0&b\\0&a&0\\0&0&c\end{pmatrix},\quad\hat{H_0}=\begin{pmatrix}a&0&0\\0&a&0\\0&0&c\end{pmatrix},\quad\hat{H_p}=\begin{pmatrix}0&0&b\\0&0&0\\0&0&0\end{pmatrix}$$ where $\hat{H}$ and $\hat{H_0}$ obviously have the same eigenvalues.

Thus, given a set of eigenvalues, is there a way of determining the set of matrices to which they belong?

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Demosthene
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  • Yes, we can at least determine the similarity classes using canonical forms. – Daniel May 25 '15 at 16:06
  • @SolidSnake How? – Demosthene May 25 '15 at 16:07
  • It's not clear what you mean with 'determine'. One could say the set you're looking for is the set of all matrices with such and such eigenvalues. There, it's determined. I think the best you can hope for is what the Legendary Hero said. – Git Gud May 25 '15 at 16:09
  • @Demonsthene Now that think about it, we need the multiplicity of each eigenvalue too (which is equivalent to knowing the characteristic polynomial). The idea is the following: assume you have the characteristic polynomial of the matrix, then you can all possible minimal polynomials for a matrix with that char and therefore all matrices with those eigenvalues (all similarity classes I mean). – Daniel May 25 '15 at 16:13
  • @SolidSnake How do you go about building these similarity classes? And is there an easy way to decide, given a random matrix, whether or not it belongs to a similarity class (of the matrix I started with)? – Demosthene May 25 '15 at 16:16
  • @SolidSnake That was already implied in your first comment. – Git Gud May 25 '15 at 16:16
  • @Demosthene Given a similarity class and a random matrix $A$, you can always find (for instance) the rational canonical form of both (a representative of the class and $A$) and deduce if the matrix belongs to $A$ (two matrices are similar iff they have the same rational canonical form). The process of going from the characteristic polynomial to the similarity classes is quite complicated to explain and it uses some theory of canonical forms (for example, it uses results such as the minimal polynomials divides the characteristic polynomial and they have the same roots) – Daniel May 25 '15 at 16:20
  • @SolidSnake So in the end, isn't it easier to just compute the eigenvalues directly? – Demosthene May 25 '15 at 16:27
  • @Demosthene No. There are non-similar matrices with the same characteristic polynomial (only for $3\times 3$ matrices you conclude they're similar if they have the same c. polynomial, for $4\times 4$ you need the characteristic and minimal polynomial to be equal). – Daniel May 25 '15 at 16:29
  • @SolidSnake Thanks for all the comments, I need some time to put my thoughts in order now :) – Demosthene May 25 '15 at 16:37
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    @Demosthene That's a fascinating subject! http://www.math.niu.edu/~beachy/courses/423/canon07.pdf – Daniel May 25 '15 at 16:40
  • @SolidSnake Thanks for the notes, I'll have a read :) – Demosthene May 25 '15 at 16:42

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Depending what you mean by "find", this is straightforward if all eigenvalues are distinct, because then any matrix having those eigenvalues is diagonalizable http://en.wikipedia.org/wiki/Diagonalizable_matrix . In that case every invertible matrix P gives rise to a matrix 

Matrix_with_those_eigenvalues = P * diag([eigenvalues]) * inv(P)

using MATLAB notation (i.e., middle matrix is diagonal with eigenvalues on the diagonal). Matrix_with_those_eigenvalues has those eigenvalues. If you do this for all invertiibe matrices P, you will have all possible matrices having those eigenvalues.

The case when eigenvalues are not distinct is left to you as an exercise.

Mark L. Stone
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