Eigenvector of complex Eigenvalues

[bowlbase]

Homework Statement

$A=\begin{bmatrix} 16 &{-6}\\39 &{-14} \end{bmatrix}$

Homework Equations

The Attempt at a Solution

I did $A=\begin{bmatrix} 16-\lambda &{-6}\\39 &{-14-\lambda} \end{bmatrix}$

and got that $\lambda_1=1+3i$ and $\lambda_2=1-3i$

The solution is partially given for both the vectors:

$x_1=\begin{bmatrix} 1+i \\ ?+?i\end{bmatrix}$
I should fill in the "?". I tried placing 1+3i into replace lambda:

$\begin{bmatrix} 16-(1+3i) &{-6}\\39 &{-14-(1+3i) } \end{bmatrix}$
Which I then get:
$\begin{bmatrix} 15-3i &{-6}\\39 &{-15-3i } \end{bmatrix}$


I don't know what to do from here. I've tried all sorts of ways to get this figured out but I just keep getting the wrong answers.

 

[Mark44]

Homework Statement

$A=\begin{bmatrix} 16 &{-6}\\39 &{-14} \end{bmatrix}$

Homework Equations

The Attempt at a Solution

I did $A=\begin{bmatrix} 16-\lambda &{-6}\\39 &{-14-\lambda} \end{bmatrix}$

and got that $\lambda_1=1+3i$ and $\lambda_2=1-3i$

The solution is partially given for both the vectors:

$x_1=\begin{bmatrix} 1+i \\ ?+?i\end{bmatrix}$
I should fill in the "?". I tried placing 1+3i into replace lambda:

$\begin{bmatrix} 16-(1+3i) &{-6}\\39 &{-14-(1+3i) } \end{bmatrix}$
Which I then get:
$\begin{bmatrix} 15-3i &{-6}\\39 &{-15-3i } \end{bmatrix}$I don't know what to do from here. I've tried all sorts of ways to get this figured out but I just keep getting the wrong answers.

Add -39 times the first row to 15 - 3i times the second row. That should result in the second row being all zeros. This seems to be a pretty messy problem to do by hand.

 

[bowlbase]

For whatever reason all of the problems for this section are like this. I just hope our test tomorrow is simpler...

Got ahead of myself on posting. 1 sec while I do what you said.

They did cancel so I'm left with

$(-585+117i)x_1=-234x_2$

$x_1=1+i=\frac{-234x_2}{-585+117i}$

which is $x_2(\frac{135}{338}+\frac{27i}{338})=1+i$

so $x_2=\frac{26}{9}+\frac{52i}{27}$

I got it wrong somewhere.

 

[bowlbase]

I figure it out:

$(-585+117i)x_1+234x_2=0$ where $x_1=1+i$

$\frac{(-585+117i)(1+i)}{-234}=x_2$

$x_2=3+2i$

This is really exactly the same so I'm not sure what happened up there. Probably just calculations. I used my calculator once I had it in the second equation form.