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The given matrix is:

$$ \begin{pmatrix}3 & 1 & 6 \\ 2 & 1 & 0 \\ -1 & 0 & -3\end{pmatrix}\qquad $$

I got the characteristic polynomial of $$x^3 - x^2 - 5x - 3 = 0$$

which factors down to $$(x+1)^2 * (x-3) = 0$$

I see that it has eigenvalues of -1 and 3.

I know I'm almost there, I plugged in the eigenvalues to $A-\lambda I$ but completely forgot how to find the eigenvectors after this.

When $\lambda$ = 3, I got:

$$\begin{pmatrix}0 & 1 & 6 \\ 2 & -2 & 0 \\ -1 & 0 & 0\end{pmatrix} \begin{pmatrix}x _1 \\ x_2 \\ x_3\end{pmatrix}=0\qquad$$

and when $\lambda$ = -1, I got:

$$\begin{pmatrix}4 & 1 & 6 \\ 2 & 0 & 0 \\ -1 & 0 & 4\end{pmatrix}\begin{pmatrix}x _1 \\ x_2 \\ x_3\end{pmatrix}=0\qquad$$

Where do I go from here? Row-reduce the $3x3$ matrices to solve?

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  • $\begingroup$ Now let $x=\begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix}$ and find all $x\neq 0$ such that $(A-\lambda I)x=0$. $\endgroup$ – Git Gud Jun 4 '13 at 15:18
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I am going to use the approach you are using so you can see your issues.

We are given: $A = \begin{bmatrix}3 & 1 & 6 \\ 2 & 1 & 0 \\ -1 & 0 & -3\end{bmatrix}$

We set up and and solve: $|A - \lambda I| = 0$, which yields:

$$\left|\begin{matrix}3-\lambda & 1 & 6 \\ 2 & 1-\lambda & 0 \\ -1 & 0 & -3-\lambda\end{matrix}\right| = 0$$

This yields a characteristic polynomial and eigenvalues as:

$$-\lambda^3+\lambda^2+5 \lambda+3 = -(\lambda-3) (\lambda+1)^2 = 0 ~~~\rightarrow ~~~ \lambda_1 = 3, \lambda_{2,3} = -1$$

We have multiplicities of $1$ and $2$ for those eigenvalues.

To find the eigenvectors, we generally solve $[ A - \lambda_i I]v_i = 0$, but since we have a repeated eigenvalue, we may need to change that strategy and find a generalized eigenvalue (I'll let you deal with the details of this and geometric multiplicities).

So, for $\lambda_1 = 3$, we have:

$[A- 3I]v_1 = \begin{bmatrix}0 & 1 & 6 \\ 2 & -2 & 0 \\ -1 & 0 & -6\end{bmatrix}v_1 = 0$

Doing row-reduced-echelon-form (RREF), yields:

$\begin{bmatrix}1 & 0 & 6 \\ 0 & 1 & 6 \\ 0 & 0 & 0\end{bmatrix}v_1 = 0$

Thus, $b = -6c, a = -6c \rightarrow ~~\text{let}~~ c = 1 \rightarrow a = b = -6, v_1 = (-6,-6,1)$.

Repeating this same process for the second eigenvalue, we have as RREF:

$\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & -2 \\ 0 & 0 & 0\end{bmatrix}v_1 = 0$

This yields an eigenvector of $v_2 = (-2, 2, 1)$.

Unfortunately, we cannot get another linearly independent eigenvector, so need to get a generalized one, by doing $[A - \lambda I]v_3 = v_2$, so we have:

$\begin{bmatrix}4 & 1 & 6 \\ 2 & 2 & 0 \\-1 & 0 & 2\end{bmatrix}v_3 = \begin{bmatrix} -2 \\ 2 \\ 1 \end{bmatrix}$

After RREF, we arrive at:

$\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & -2 \\ 0 & 0 & 0\end{bmatrix}v_3 = \begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix}$

So, we have: $a = -1 -2c, b = 2 + 2c \rightarrow ~~ \text{let} ~~ c = 0 \rightarrow a = -1, b = 2$, thus $v_3 = (-1,2,0)$

Putting all of this together, we have the eigenvalue/eigenvector pairs:

  • $\lambda_1 = 3, v_1 = (-6, -6, 1)$
  • $\lambda_2 = -1, v_2 = (-2, 2, 1)$
  • $\lambda_3 = -1, v_3 = (-1,2,0)$
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  • $\begingroup$ Nice work here, Amzoti! Really nice +1 $\endgroup$ – Namaste Jun 5 '13 at 0:09
  • $\begingroup$ This was extremely helpful, you have no idea. I got stuck while finding the third vector and thought I was doing everything wrong. $\endgroup$ – Jesus Jun 5 '13 at 14:55
  • $\begingroup$ Nice to know when you've helped, indeed! So-so day for me...a bit slow...but I guess we'll need to get accustomed to that! $\endgroup$ – Namaste Jun 6 '13 at 0:23
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it is enough you find: $$\left(\begin{bmatrix}3 & 1 & 6 \\ 2 & 1 & 0 \\ -1 & 0 & -3\end{bmatrix}- 3I\right)\begin{bmatrix}x _1 \\ x_2 \\ x_3\end{bmatrix}=0\qquad \text{and} \qquad \left(\begin{bmatrix}3 & 1 & 6 \\ 2 & 1 & 0 \\ -1 & 0 & -3\end{bmatrix}+ I\right)\begin{bmatrix}x _1 \\ x_2 \\ x_3\end{bmatrix}=0 \ .$$

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  • $\begingroup$ Aren't some parentheses missing? $\endgroup$ – egreg Jun 4 '13 at 16:51
  • $\begingroup$ Yes you are right Dear egreg :) $\endgroup$ – Somaye Jun 4 '13 at 18:46
  • $\begingroup$ nice answer $\Large\color{green}{✔}^\color{red}{❁}$ @somaye $\endgroup$ – Software Jun 6 '13 at 18:00
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Remember an eigenvector $v_\lambda$ of the matrix $M$ associated to the eigenvalue $\lambda$ is a vector such $Mv_\lambda=\lambda v_\lambda$ or equivalently $(M-\lambda I)v_\lambda=0$. So, you're trying to find a vector (or vectors depending on multiplicity) $v_\lambda$ which spans the nullspace of the matrix $M-\lambda I$. Do you know how to find the nullspace of a matrix?

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