Eigenvector (easy)question

**Redeemer_Pie** · June 7th 2009, 06:28 AM

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.

**Spec** · June 7th 2009, 07:01 AM

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

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

**Redeemer_Pie** · June 7th 2009, 07:06 AM

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?

**Sean12345** · June 7th 2009, 07:11 AM

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.

**Redeemer_Pie** · June 7th 2009, 07:18 AM

OH!!!!!

Got it!!! thanks Sean12345

Forgot to mention.

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

**matheagle** · June 7th 2009, 07:51 AM

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.

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