# Eigenvalues and Eigenvectors of a 3 by 3 matrix

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.

• could you clarify the question? – J. W. Tanner Nov 4 '19 at 18:01
• 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. – Bungo Nov 4 '19 at 18:02
• @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. – amd Nov 4 '19 at 18:10

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.