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 $\displaystyle 3x_1=x_2$ so if you set $\displaystyle x_1=t$ then $\displaystyle x_2=3t$

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

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

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

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

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

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

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

Hope this helps.

5. OH!!!!!

Got it!!! thanks Sean12345

Forgot to mention.