# Thread: Eigenvector (easy)question

1. ## Eigenvector (easy)question

Hi guys,

im given this matrix here:

|3 2|
|3 8|

and the eigenvalues were 2 and 9.

I found the eigenvectors where 'lamda' = 2

but when i use the eigenvalue for 9 my answer were (1, 1/3) but the book gave me the answer as being (1, 3).

Could someone explain to me why they have (1,3)?

Thanks. Your contribution is much appreciated.

2. Well you get $3x_1=x_2$ so if you set $x_1=t$ then $x_2=3t$

So we have $\left(\begin{array}{cc}x_1\\x_2\end{array}\right) =t\left(\begin{array}{cc}1\\3\end{array}\right)
$

3. thanks!

If its not too much to ask, could you show me your working too? Still a bit confused as to why u let 3x1 = x2

Would my original answer be correct too?

4. Hi Redeemer_Pie,

Remember that if $\binom{s_1}{t_1}$ is an eigenvector corresponding to an eigenvalue of $\lambda_1$ then $r\binom{s_1}{t_1}$ where $r$ is some constant is also an eigenvector corresponding to an eigenvalue of $\lambda_1$ .

We have $M\:=\:\begin{bmatrix}3&2 \\ 3&8 \end{bmatrix}$

Note that for any eigenvalue $\lambda$ and the corrsponding eigenvector $s$, $(M-\lambda I)s=0$ so for $\lambda = 9$ we have

$\:\begin{bmatrix}\text{-}6&2 \\ 3&\text{-}1 \end{bmatrix}\binom{x}{y}=0$ hence

$3x-y=0 \implies y=3x$ . If we let $x=1$ then $y=3$ so

$s=r\binom{1}{3}$ where r is some constant.

Hope this helps.

5. OH!!!!!

Got it!!! thanks Sean12345

Forgot to mention.

The answer i had originally, would that also be correct?

6. Any nonzero multiple of an eigenvector is correct.
That point is it forms an eigenspace and you need to know what part of the whole space it spans.