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I have a question where I do not understand the answer, nor how it was derived. The problem asks to find the eigenvalues and corresponding eigenvector of the matrix:

$$ \begin{pmatrix} 3 & 2 & 4 \\ 2 & 0 & 2 \\ 4 & 2 & 3 \\ \end{pmatrix} $$

Solving the characteristic equation allowed me to find the eignevalues λ = -1 and λ = 8.

Finding the associated eigenvector for λ = -1 is where I have problems, as I end up (with gaussian elimination) with a matrix:

$$ \begin{pmatrix} 2 & 1 & 2 & | & 0\\ 0 & 0 & 0 & | & 0\\ 0 & 0 & 0 & | &0\\ \end{pmatrix} $$

Leaving (using variables of $ x_1, x_2, x_3$): $ 2x_1+x_2+2x_3 = 0 $, giving $ x_1 = 1/2 x_2 - x_3 $, which is as far as I get. *


The answers I have to this question provided details that it gives $$ v_1 = -\frac{1}{2} v_2 - v_3 $$ for the eigenvector components, and that because there are two degrees of freedom we can construct two eigenvectors.

It says that we can write $ \begin{bmatrix} v_1 & v_2 & v_3 \end{bmatrix} $ as $ -\frac{1}{2} v_2 \begin{bmatrix} -1 & 2 & 0 \end{bmatrix} + v_3 \begin{bmatrix} -1 & 0 & 1 \end{bmatrix} $ which allows us to identify the eigenvectors $ \begin{bmatrix} -1 & 2 & 0 \end{bmatrix}^T $ and $ \begin{bmatrix} -1 & 0 & 1 \end{bmatrix}^T $, as this is one of the cases where one eigenvalue has two linearly independent associated eigenvectors.


I'm looking to understand how they arrived at those numbers as I can't see any obvious explanation

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Where you wrote $2x_1+x_2+x_3=0$, you meant to write $2x_1+x_2+2x_3=0$. Solving for $x_1$ gives $x_1=-(1/2)x_2-x_3$, which is exactly what's in the given answers, except for using $x$ instead of $v$.

Now you can take $x_2=2$, $x_3=0$, giving $x_1=-1$, so $(x_1,x_2,x_3)=(-1,2,0)$ and that's one eigenvector. You can take $x_2=0$, $x_3=1$, giving $x_1=-1$, so $(x_1,x_2,x_3)=(-1,0,1)$, your other eigenvector.

You could take other values of $x_2,x_3$ --- there's nothing special about the values in the answers --- all you need is two linearly independent results.

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  • $\begingroup$ Ok, I think I get it. So you just use any two values for $ x_2, x_3 $ that produces something easy to work with, and make sure they are linearly independent. Would $ x_2 = 2, x_3 = 1 $ therefore $ e_1 $ = (-2, 2, 1) and $ x_2 = 0, x_3 = 1 $ therefore $ e_2 $ = (-1, 0, 1) be a valid answer? I've had the same problem in another question, so using this would (1,1,0) and (0,1,0) be ok for {{1,1/2,-1/2},{1/2,1,1/2},{-1/2,1/2,1}}? $\endgroup$ – Adam M-W Oct 13 '12 at 12:18
  • $\begingroup$ The easiest way to check whether those are OK is to check whether they are eigenvectors and whether they are linearly independent. If they pass both tests, you win. $\endgroup$ – Gerry Myerson Oct 13 '12 at 21:54

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