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For instance let $$A=\begin{pmatrix} 3 & -1 & -1 \\ 2 & 1 &-2 \\ 0 & -1 & 2 \\ \end{pmatrix}$$ be a matrix and $$u_1=\begin{pmatrix} 1 \\ 1 \\ 1 \\ \end{pmatrix},$$ $$u_2=\begin{pmatrix} 1 \\ 0 \\ 1\\ \end{pmatrix},$$ $$u_3=\begin{pmatrix} 0 \\ -1 \\ 1 \\ \end{pmatrix}.$$ its eigenvectors. What are its eigenvalues?

Is there anything more simple than doing $A-λI$?

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    $\begingroup$ Just solve $Ax=\lambda x$ for $\lambda$. $\endgroup$ – David Mitra May 13 '15 at 10:04
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    $\begingroup$ The easiest way is to do apply the matrix to each vector... $\endgroup$ – Martigan May 13 '15 at 10:04
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If $u$ is an eigenvector of $A$ and $\lambda$ is the corresponding eigenvalue, you know the following: $$Au = \lambda u$$ So in your example, you can do the following (I’ll take the second one): $$Au_2 = \begin{pmatrix}3 & -1 & -1 \\ 2 & 1 & -2 \\ 0 & -1 & 2\end{pmatrix}\begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix} = \begin{pmatrix}2 \\ 0 \\ 2\end{pmatrix} = 2 u_2$$ So $\lambda_2 = 2$. Do the same for $u_1$ and $u_3$.

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