I'm trying to solve the following problem and hope for some helpful insights on how to approach this:
In the 3-dimensional case, for and a given set of eigenvalues and eigenvectors and chosen values $a$ and $b$, $c$, find the set of 3 by 3 matrices that have the corresponding eigenvalues and eigenvectors (if it exists). Some elements of the matrix and eigenvectors are fixed values.
For example in the 3-dimensional case:
Eigenvalues:
$ \lambda_1=3 $, $\lambda_2 = -1$ and $\lambda_3 = 2$
Eigenvectors:
$v_1=\begin{bmatrix} 1 \\ 2 \\a \end{bmatrix} $, $ v_2=\begin{bmatrix} 2 \\ b \\1 \end{bmatrix} $ and $ v_3=\begin{bmatrix} c \\ 0 \\-1 \end{bmatrix} $
The matrix has the following fixed elements:
$$ \begin{bmatrix} x_{11} & 2 & x_{13} \\ x_{21} & x_{22} & 1 \\ x_{31} & 3 & x_{33} \end{bmatrix} $$
In other words: Let's say I set $a=2$ and $b=-1$ and $c=1$. What are the possible matrices with fixed values $x_{12}=2$,$x_{23}=1$ and $x_{32}=3$ with eigenvalues $ \lambda_1=3 $, $\lambda_2 = -1$ and $\lambda_3 = 2$ and eigenvectors $v_1=\begin{bmatrix} 1 \\ 2 \\2 \end{bmatrix} $, $ v_2=\begin{bmatrix} 2 \\ -1 \\1 \end{bmatrix} $ and $ v_3=\begin{bmatrix} 1 \\ 0 \\-1 \end{bmatrix} $.
Also, it would be interesting to know under what circumstances solutions exist and how many possible solutions there are?
My first idea was to solve the eigenvalue equations for each eigenvalue and eigenvector:
$$ (A-\lambda I)v=0 $$
for example for $\lambda_1$ and $v_1 $:
$$ \begin{bmatrix} x_{11}-3 & 2 & x_{13} \\ x_{21} & x_{22}-3 & 1 \\ x_{31} & 3 & x_{33}-3 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\2 \end{bmatrix} =\begin{bmatrix} 0 \\ 0 \\0 \end{bmatrix} $$
However, the values of $x$ must be true for all eigenvectors and not just for one. So I got stuck (or I'm missing something).
My second idea was to use the fact that a matrix $M$ can be obtained when knowing the eigenvectors and eigenvalues by $M=PDP^{-1}$ where $P$ is the matrix with the eigenvectors as columns and $D$ the diagonal matrix of eigenvalues. However, this does not take care of the constraints. Stuck again.
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