# Eigenvectors and eigenvalues- finding the matrix

I am aware of the relation between eigenvectors and eigenvalues, and that the following holds: If A is an n×n matrix, the nonzero n-component column vector x is an eigenvector for eigenvalue λ if Ax=λx.

I was wondering, if I was given some eigenvectors and some eigenvalues of a matrix A, and it's not specified which eigenvectors correspond to which eigenvalues, is there a way to determine that, and hence find matrix A?

Thank you!

• If you are given $n$ eigenvalues, then you know what the matrix is; it is the diagonal matrix with the given eigenvalues as its diagonal. The eigenvectors are in a sense irrelevant, in this case. If you are given fewer eigenvalues, I don't see in general how you could determine the form of the matrix. – Chris Mar 24 '18 at 3:29
• @Chris In general, $A$ not need be diagonal. $A$ can be written in the form: $A = P \cdot J \cdot P^{-1}$ where $J$ is the corresponding Jordan matrix, while $P$ has linearly independent eigenvectors as its columns. – thanasissdr Mar 24 '18 at 3:35
• If you know that $v$ is an eigenvector, then $v \neq 0$ and $Av$ is a multiple of $v$. So, find some $[v]_k \neq 0$ and then $\lambda = {[Av]_k \over [x]_v}$. – copper.hat Mar 24 '18 at 3:39
• @thanasissdr oops my mistake, you're totally right of course. Thanks for the correction. – Chris Mar 24 '18 at 3:50

## 1 Answer

I think what you are asking is given that if you are given that $\{v_1,\cdots,v_n\}$ are linearly independent eigenvectors for a linear transformation and that $\{\lambda_1,\cdots,\lambda_n\}$ are the eigenvalues for the same transformation (counting multiplicity), is this enough to recover the matrix (specifically, by determining which eigenvalues are associated with which Eigen vectors).

The answer is no. There is no way to determine which eigenvectors are associated with which eigenvalues with that information. Suppose that we were given that $\begin{bmatrix} 1 \\ 0\end{bmatrix}$ and $\begin{bmatrix} 0 \\ 1\end{bmatrix}$ are eigenvectors and that $0$ and $1$ are the eigenvalues for a $2\times2$ matrix. Then both $\begin{bmatrix} 1 & 0\\ 0 & 0\end{bmatrix}$ and $\begin{bmatrix} 0 & 0\\ 0 & 1\end{bmatrix}$ are distinct matrices that give rise to this, with a different eigenvector-eigenvalue pairing.

In general, if $A$ is a matrix, $\{v_1,\cdots,v_n\}$ are linearly independent eigenvectors, and $\{\lambda_1,\cdots,\lambda_n\}$ are your eigenvalues such that for each $i$, $\lambda_i$ is associated with $v_i$, then if $V$ is the matrix where the $i^{th}$ column is the $v_i$ and $\Lambda$ is the matrix with the eigenvalues on the diagonal such that $\Lambda_{i,i} = \lambda_i$, then $A=V\Lambda V^{-1}$.