Given eigenvectors $v_1, v_2, \dots, v_n$ and eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_n$, how do I construct a matrix whose eigenvectors and eigenvalues are $v$ and $\lambda$?
The straightforward way of doing this is to encapsulate all $n^2$ constraints into a linear system and solve for each element of the matrix $M_i$. I.e.,
$$
\begin{bmatrix}
v_{11} & v_{12} & \dots & v_{1n} & 0 & 0 &\dots & 0 & \dots & 0 & 0 &\dots & 0 \\
0 & 0 &\dots & 0 & v_{11} & v_{12} & \dots & v_{1n} & \dots & 0 & 0 &\dots & 0\\
& & & & & & \vdots \\
0 & 0 &\dots & 0 & 0 & 0 &\dots & 0 & \dots & v_{n1} & v_{n2} & \dots & v_{nn}\\
\end{bmatrix}
\begin{bmatrix}
M_1 \\ M_2 \\ \vdots \\ M_{n^2}
\end{bmatrix}
=
\begin{bmatrix}
\lambda_1 v_{11} \\ \lambda_1 v_{12} \\ \vdots \\\lambda_n v_{nn}
\end{bmatrix}
$$
Is there a more elegant way?
Best Answer
Your system of equations is $Mv_1=\lambda_1v_1,\ldots,Mv_n=\lambda_nv_n$. Or equivalently, $M(v_1,\ldots,v_n)=(\lambda_1 v_1,\ldots,\lambda_n v_n)$, where $V:=(v_1,\ldots,v_n)$ is the $n\times n$-matrix with columns $v_1,\ldots,v_n$. You can write this as $MV=VD$ where $D$ is the diagonal matrix with diagonal entries $\lambda_1,\ldots,\lambda_n$. So, assuming $V$ is invertible, that is, that your given eigenvectors are linearly independent, you get $M=VDV^{-1}$. Thus to calculate $M$ this way, all you need to do is to find the inverse of the matrix of eigenvectors, and multiply three matrices together.