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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: $$\lambda_1 = -1 \qquad \lambda_2 = 1$$

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There are two ways to solve for the eigenvalues.
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 $\lambda_2$.

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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.

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  • $\begingroup$ Of course! Thank you! :) $\endgroup$ – Ani Jun 6 '16 at 23:23

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