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I'm having trouble starting to solve this. Do I solve it with an augmented matrix? Can someone walk me through the first one so I know how to solve the rest?

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    $\begingroup$ Hint: $A[0,-1,1]^T=-2[0,-1,1]^T$ which implies $-2$ is an eigenvalue. $\endgroup$ Nov 15, 2016 at 4:49

2 Answers 2


One usual method for calculating the eigenvalue $\lambda$ of a matrix $A$ from one of its eigenvectors $v$ is the Rayleigh-quotient \begin{align*} \lambda = \frac{v^* A v}{v^* v} \end{align*} If the eigenvectors are real as in your example this is equivalent to \begin{align*} \lambda = \frac{v^T A v}{v^T v} \end{align*} The formula works since with $Av=\lambda v$ you get \begin{align*} \frac{v^* A v}{v^* v} = \frac{v^* \lambda v}{v^* v} = \lambda \frac{v^* v}{{v^* v}}= \lambda. \end{align*} This method also works for approximative eigenvalues corresponding to approximative eigenvectors where it is not so simple to recognize per inspection that $Av$ is a multiple of $v$.


Eigenvalues and eigenvectors have the property that $Av=\lambda v$ where $A$ is your matrix, $v$ is an eigenvector, and $\lambda$ is an eigenvector. Then, for any of your eigenvectors, we can find the corresponding eigenvalue by solving the equation $Av=\lambda v$. For example, using the first eigenvector, we get:

$\begin{pmatrix}3&0&0\\0&-2&0\\2&3&1 \end{pmatrix}*\begin{pmatrix}0\\-1\\1 \end{pmatrix}=\lambda \begin{pmatrix}0\\-1\\1 \end{pmatrix} \implies \begin{pmatrix}0\\2\\-2 \end{pmatrix}=\lambda \begin{pmatrix}0\\-1\\1 \end{pmatrix} \implies \lambda=-2$

Thus, for the first eigenvector, we see that the corresponding eigenvalue is $-2$. Now follow this process for the other two to get the other eigenvalues.


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