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I want to find eigenvalues and eigenvectors of this matrix: $$ \begin{bmatrix} 2 & 3 & 0 \\ 3 & 6 & 1 \\ 0 & 1 & 6 \\ \end{bmatrix} $$

So eigenvalues:

$$ \begin{vmatrix} 2-\lambda_1 & 3 & 0 \\ 3 & 6-\lambda_2 & 1 \\ 0 & 1 & 6-\lambda_3 \\ \end{vmatrix} $$

then $$-\lambda^3+14\lambda^2-50\lambda+16=-(8-\lambda)(\lambda^2-6\lambda+2)$$

and the eigen values are: $\lambda_1=3-\sqrt{7}, \lambda_2=3+\sqrt{7}, \lambda_3=8$

So then I want to find eigenvectors for eigenvalue $\lambda_1=3-\sqrt{7}$

I know that:

$$$$ \begin{bmatrix} 2-(3-\sqrt{7}) & 3 & 0 \\ 3 & 6-(3-\sqrt{7}) & 1 \\ 0 & 1 & 6-(3-\sqrt{7}) \\ \end{bmatrix} $$$$ and that matrix I multiply with matrix:

$$$$ \begin{bmatrix} x \\ y \\ z\\ \end{bmatrix} $$$$

and this gives me:

$$$$ \begin{bmatrix} (\sqrt{7}-1)x+3y\\ 3x+(\sqrt{7}+3)y+z \\ y+(\sqrt{7}+3)z \\ \end{bmatrix} $$$$

and then I get three equation:

$$\begin{cases} (\sqrt{7}-1)x+3y=0\\[2ex] 3x+(\sqrt{7}+3)y+z =0\\[2ex] y+(\sqrt{7}+3)z=0 \end{cases}$$

Now, every such system will have infinitely many solutions, because if e is an eigenvector, so is any multiple of e. So our strategy will be to try to find the eigenvector with $x=?$

How I decide the $x$ value when the correct answer is:

$x=0.87, y=-0.479, z=0.085$.

Yes I know that when I set the $x=0.87$ then the equalitons gives me the correct $y$ and $z$ values.

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  • $\begingroup$ could you clarify the question? $\endgroup$
    – J. W. Tanner
    Nov 4, 2019 at 18:01
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    $\begingroup$ As you said, if $v$ is an eigenvector, then so is $cv$ for any nonzero scalar $c$. So, you can assign $x$ some arbitrary convenient value such as $x=1$, then compute $y$ and $z$. Looks like the vector you called the "correct answer" is simply normalized so its length $\sqrt{x^2 + y^2 + z^2}$ is $1$. But any nonzero scalar multiple of this "correct answer" is also correct. $\endgroup$
    – user169852
    Nov 4, 2019 at 18:02
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    $\begingroup$ @Bungo Even when normalized, there’s still a sign ambiguity. I would guess that it’s chosen so that the first nonzero element is positive. $\endgroup$
    – amd
    Nov 4, 2019 at 18:10

1 Answer 1

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The strategy is that you define one of the variables $x$, $y$, or $z$ as independent variable. Then two other variables will be dependent on it. Let's choose $z$ to be the independent. In this case your eigenvector corresponding to $\lambda_1=3-\sqrt{7}$ is $(\dfrac{(2+\sqrt{7})(4+\sqrt{7})}{3}z,-(3+\sqrt{7})z,z)$. There is no one correct answer. Once you fix $z$ you obtain another vector.

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