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Problem 3 on page 882 here, translated:

The matrix $A=\begin{bmatrix} 1 & 1 & 0 \\ 0 & 2 & 0 \\ 2 & 1 & -1 \\ \end{bmatrix}$ has eigen-vectors $\begin{bmatrix} 1 & 1 & 1\end{bmatrix}^T$, $\begin{bmatrix} 1 & 0 & 1\end{bmatrix}^T$ and $\begin{bmatrix} 0 & 0 & 1\end{bmatrix}^T$. What is $A^{11} \bar{x}$, when $\bar{x}=\begin{bmatrix} 2 & 1 & 1 \end{bmatrix}^T$?

Now I think the key point is to realize

$$\begin{bmatrix} 2 & 1 & 1 \end{bmatrix}^T=\begin{bmatrix} 1 & 1 & 1 \end{bmatrix}^T+\begin{bmatrix} 1 & 0 & 1 \end{bmatrix}^T-\begin{bmatrix} 0 & 0 & 1 \end{bmatrix}^T$$

then we get

$$A^{11}\bar{x}= A^{11}\bar{x}_1+A^{11}\bar{x}_2-A^{11}\bar{x}_3$$

and since they are eigenvectors, we have $A\bar{x}=\lambda\bar{x}$ but here I mess up with things. I can calculate the eigenvalues from the determinant but I cannot understand which eigenvalue correspond to which eigenvector. According to my draft, the eigenvalues are $\lambda_h=1$, $\lambda_y=2$ and $\lambda_z=-1$ but how can I know which is $\lambda_1, \lambda_2$ and $\lambda_3$?

Draft

This is my work but it is not essential to understand the problem, it probably show if I have misunderstood something.

enter image description here

hhh
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    You have the eigenvectors $x_1$, $x_2$ and $x_3$. Why not just calculate $A\cdot x_1$ etc. to see which eigenvalues they have? – example Apr 04 '12 at 09:40

1 Answers1

2

You can check easily that $A\bar x_1=2\cdot \bar x_1$: $$\left(\begin{array}{rrr}1 & 1 & 0 \\ 0& 2 & 0 \\2 &1 &-1 \\ \end{array} \right) \; \left( \begin{array}{c}1 \\ 1\\ 1 \\ \end{array} \right)=\left( \begin{array}{c}2 \\ 2\\ 2 \\ \end{array} \right)=2\left( \begin{array}{c}1 \\ 1\\ 1 \\ \end{array} \right) $$ So, $\bar x_1$ is an eigenvector with eigenvalue 2. In a similar way you'll find that $\left( \begin{array}{c}1 \\0\\ 1 \\ \end{array} \right)$ has eigenvalue $1$ and $\left( \begin{array}{c}0 \\0\\ 1 \\ \end{array} \right)$ has eigenvalue $-1.$

Xabier Domínguez
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