# Given a set of Eigenvectors - find the Eigenvalues

This is a question on my practice exam for Linear Algebra, however the solution that I was given provided insufficient information as to how the answer came about.

Given a matrix: $$A = \frac15\begin{bmatrix}-3 & 4\\ 4 & 3\end{bmatrix}$$ and eigenvectors $v_1$ and $v_2$: $$v_1 = \begin{bmatrix} 2\\ -1\end{bmatrix} \qquad v_2 = \begin{bmatrix} 1\\ 2\end{bmatrix}$$ Find the corresponding eigenvalues.

The expected eigenvalues: $$\lambda_1 = -1 \qquad \lambda_2 = 1$$

One way is to directly solve for eigenvalues from the matrix using $\det(A-\lambda I)=0$, you would only have at most two eigenvalues and two sets of eigenvectors, so just solve it straight and you will find $\lambda_1$ and $\lambda_2$
The second way is to use the basic definition of eigenvalues: $$Ax=\lambda x$$ $A=\frac15 \begin{bmatrix}-3&4\\4&3\end{bmatrix}$, $x_1=\begin{bmatrix}2\\-1\end{bmatrix}$
So$$Ax_1=\frac15 \begin{bmatrix}-3&4\\4&3\end{bmatrix}\begin{bmatrix}2\\-1\end{bmatrix}=\begin{bmatrix}-2\\1\end{bmatrix}=\lambda_1 x_1$$ Therefore $\lambda_1=-1$.
Similarily plug in $x_2$ you can get $\lambda_2$.
Compute $Av_i$, you should get $\lambda_iv_i$.
For example for $v_1=\begin{pmatrix} 2 \\ -1 \end{pmatrix}$, $Av_1=\begin{pmatrix} -2 \\ 1 \end{pmatrix}=(-1)\begin{pmatrix} 2 \\ -1 \end{pmatrix}=(-1)v_1$ and you can conclude that $-1$ is the corresponding eigenvalue.