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.

  • $\begingroup$ could you clarify the question? $\endgroup$ Nov 4, 2019 at 18:01
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ @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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.