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The matrix: $$ A=\begin{pmatrix} 1 & -1 & 1 \\ -1 & 1 & -1 \\ -1 & 1 & -1 \\ \end{pmatrix} $$ has two real eigenvalues, one of multiplicity $1$ and one of multiplicity $2$. Find the eigenvalues and a basis of each eigenspace.

$\lambda_1$ has multiplicity $1$, with basis $\begin{bmatrix} ? \\ ? \\ ? \end{bmatrix}$, and $\lambda_2$ has multiplicity $2$, with basis $\begin{bmatrix} ? \\ ? \\ ? \end{bmatrix}$, $\begin{bmatrix} ? \\ ? \\ ? \end{bmatrix}$.

The polynomial I get upon doing $(\lambda I-A)$ is: $$ (\lambda^3-\lambda^2) = (\lambda-1)\lambda^2.$$

I'm not entirely sure what to do after this point... Any input?

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  • $\begingroup$ First step: what are the roots of that polynomial? What is the multiplicity of each of them? You should be able to get that far, at least, since you have the factored form, and then we can worry about the rest. $\endgroup$ – Brian M. Scott Mar 30 '13 at 0:12
  • $\begingroup$ The problem I'm facing here is with the word multiplicity. I'm not sure how this comes into play with what is asked here. $\endgroup$ – Dimitri Mar 30 '13 at 0:16
  • $\begingroup$ @Amzoti: How can you say that Dimitri’s eigenvalues are correct? He hasn’t actually given any! $\endgroup$ – Brian M. Scott Mar 30 '13 at 0:18
  • $\begingroup$ Haha, I had gotten to that part, I just never mentioned it above; I should have. $\endgroup$ – Dimitri Mar 30 '13 at 0:20
  • $\begingroup$ This matrix is singular (two rows equal, the other one just their negative)... $\endgroup$ – vonbrand Mar 30 '13 at 1:29
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The multiplicity of a root $r$ of a polynomial $p(x)$ is the number of times that $x-r$ appears as a factor when $p(x)$ is fully factored into linear factors. You have the equation $\lambda^2(\lambda-1)=0$, which is fully factored into the linear factors $\lambda$, $\lambda$, and $\lambda-1$. Thus, $0$ is the root of multiplicity $2$, and $1$ is the root of multiplicity $1$.

Now you want to find the eigenvectors. For a given eigenvalue $\lambda$, these are the vectors $v$ such that $Av=\lambda v$ or, equivalently, $(A-\lambda I)v=\vec 0$. For $\lambda=0$ this equation becomes

$$\begin{bmatrix}1&-1&1\\-1&1&-1\\-1&1&-1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}\;,$$

a matrix equation that you solve by your favorite method. I’d just row-reduce the augmented matrix to

$$\left[\begin{array}{rrr|r}1&-1&1&0\\0&0&0&0\\0&0&0&0\end{array}\right]$$

and choose a basis for the two-dimensional solution space.

Then do the same thing for $\lambda=1$; your starting point should be the equation

$$\begin{bmatrix}0&-1&1\\-1&0&-1\\-1&1&-2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}\;.$$

This will actually be a little easier, since the solution space will be one-dimensional.

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  • $\begingroup$ Now I understand. Thanks so much, I thought of doing the same thing previously but wasn't sure if it made any sense. I'll give this a try $\endgroup$ – Dimitri Mar 30 '13 at 0:36
  • $\begingroup$ @Dimitri: You’re welcome; if you get stuck, give me a ping. $\endgroup$ – Brian M. Scott Mar 30 '13 at 0:37
  • $\begingroup$ And it's perfect, I found the solution space for both eigenvalues and it worked out nicely. Thanks again. $\endgroup$ – Dimitri Mar 30 '13 at 0:44
  • $\begingroup$ @Dimitri: Excellent! Glad to help. $\endgroup$ – Brian M. Scott Mar 30 '13 at 0:44

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