# Determine All Matrices Satisfying Some Conditions on Eigenvalues and Eigenvectors ## Problem 423

Determine all $2\times 2$ matrices $A$ such that $A$ has eigenvalues $2$ and $-1$ with corresponding eigenvectors
$\begin{bmatrix} 1 \\ 0 \end{bmatrix} \text{ and } \begin{bmatrix} 2 \\ 1 \end{bmatrix},$ respectively. Add to solve later

## Solution.

Suppose that $A$ is a $2\times 2$ matrix having eigenvalues $2$ and $-1$ with corresponding eigenvectors
$\begin{bmatrix} 1 \\ 0 \end{bmatrix} \text{ and } \begin{bmatrix} 2 \\ 1 \end{bmatrix},$ respectively.
Then since $A$ has two distinct eigenvalues, the matrix $A$ is diagonalizable.
As we know eigenvectors, we can diagonalize $A$ by the matrix
$S:=\begin{bmatrix} 1 & 2\\ 0& 1 \end{bmatrix}.$ That is, we have
$S^{-1}AS=\begin{bmatrix} 2 & 0\\ 0& -1 \end{bmatrix}.$ The inverse matrix of $S$ is given by
$S^{-1}=\begin{bmatrix} 1 & -2\\ 0& 1 \end{bmatrix}.$ It follows that we have
\begin{align*}
A&=S\begin{bmatrix}
2 & 0\\
0& -1
\end{bmatrix}S^{-1}\6pt] &= \begin{bmatrix} 1 & 2\\ 0& 1 \end{bmatrix} \begin{bmatrix} 2 & 0\\ 0& -1 \end{bmatrix} \begin{bmatrix} 1 & -2\\ 0& 1 \end{bmatrix}\\[6pt] &=\begin{bmatrix} 2 & -6\\ 0& -1 \end{bmatrix}. \end{align*} Therefore, the only matrix satisfying the given conditions is \[A=\begin{bmatrix} 2 & -6\\ 0& -1 \end{bmatrix}. Add to solve later

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