MHB Finding eigenvectors for a double root

[brunette15]

I have the following matrix: [-1 3 -3(2)^0.5 ; 3 -1 -3(2)^0.5 ; -3(2)^0.5 -3(2)^0.5 2]

I was able to find the eigenvalues as -4 and 8.

I am now trying to find the corresponding eigenvectors how since -4 is a double root i am unsure how to go about this.

I have tried using gaussian elimination however i always end up with too many unknown variables.

Can anyone please work me through this?

Thanks in advance! :)

 

[Dethrone]

Hi brunette15,

Your eigenvalues are correct. Let's start with finding the eigenspace corresponding to $\lambda=-4$. Let $E_{\lambda=-4}$ be the eigenspace corresponding to $\lambda=-4$, then:

$E_{\lambda=-4}=\text{null}(-4I-A)=\text{null}\left[\begin{array}{c}-3 & -3 & 3 \sqrt{2} \\ -3&-3 & 3 \sqrt{2} \\ 3 \sqrt{2} & 3 \sqrt{2} & -6 \end{array}\right]$

Now, instead of using GE, you can notice that the 2nd and 3rd row are both scalar multiplies of the first one, so:
$E_{\lambda=-4}=\text{null}\left[\begin{array}{c}-3 & -3 & 3 \sqrt{2} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$

Solving $(-4I-A)X=0$, where $X=\left[\begin{array}{c}x_1 \\ x_2 \\ x_3 \end{array}\right]$, we get $-3x_1-3x_2+3\sqrt{2}x_3=0$.
We can say $x_1=-1x_2+\sqrt{2}x_3$, $x_2=x_2$, $x_3=x_3$, and:

$X=\left[\begin{array}{c}-x_2+\sqrt{2} x_3 \\ x_2 \\ x_3 \end{array}\right]=\left[\begin{array}{c}-1 \\ 1 \\ 0 \end{array}\right]x_2+\left[\begin{array}{c}\sqrt{2} \\ 0 \\ 1 \end{array}\right]x_3$
Finally,
$E_{\lambda=-4}=\text{span}\left[\begin{array}{c}-1 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{c}\sqrt{2} \\ 0 \\ 1 \end{array}\right]$

Let me know if you have any questions, or if there is a calculation error somewhere. I leave you with $\lambda=8$.

 

[brunette15]

Hi brunette15,

Your eigenvalues are correct. Let's start with finding the eigenspace corresponding to $\lambda=-4$. Let $E_{\lambda=-4}$ be the eigenspace corresponding to $\lambda=-4$, then:

$E_{\lambda=-4}=\text{null}(-4I-A)=\text{null}\left[\begin{array}{c}-3 & -3 & 3 \sqrt{2} \\ -3&-3 & 3 \sqrt{2} \\ 3 \sqrt{2} & 3 \sqrt{2} & -6 \end{array}\right]$

Now, instead of using GE, you can notice that the 2nd and 3rd row are both scalar multiplies of the first one, so:
$E_{\lambda=-4}=\text{null}\left[\begin{array}{c}-3 & -3 & 3 \sqrt{2} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$

Solving $(-4I-A)X=0$, where $X=\left[\begin{array}{c}x_1 \\ x_2 \\ x_3 \end{array}\right]$, we get $-3x_1-3x_2+3\sqrt{2}x_3=0$.
We can say $x_1=-1x_2+\sqrt{2}x_3$, $x_2=x_2$, $x_3=x_3$, and:

$X=\left[\begin{array}{c}-x_2+\sqrt{2} x_3 \\ x_2 \\ x_3 \end{array}\right]=\left[\begin{array}{c}-1 \\ 1 \\ 0 \end{array}\right]x_2+\left[\begin{array}{c}\sqrt{2} \\ 0 \\ 1 \end{array}\right]x_3$
Finally,
$E_{\lambda=-4}=\text{span}\left[\begin{array}{c}-1 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{c}\sqrt{2} \\ 0 \\ 1 \end{array}\right]$

Let me know if you have any questions, or if there is a calculation error somewhere. I leave you with $\lambda=8$.

Thankyou!:D