# Finding eigenvalues and eigenvectors of a real matrix

• Aug 17th 2009, 05:30 PM
chella182
Finding eigenvalues and eigenvectors of a real matrix
Okay, so the matrix is...

$\left(\begin{array}{cc}6 & 5 \\ -5 & 0\end{array}\right)$

I solved $det(A-\lambda I)=0$ to find $\lambda$ (which is the method I was taught), and I ended up with $\lambda=3\pm4i$, but now I'm a bit stuck. I've written down...

$\left(\begin{array}{cc}6 & 5 \\ -5 & 0\end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right)=(3\pm4i)\left(\begin{array}{c} x \\ y\end{array}\right)$

... but I can't seem to manipulate it to get the eigenvectors.
• Aug 17th 2009, 05:38 PM
ynj
$AX=\lambda X\Longleftrightarrow (A-\lambda I)X=0$..
you just have to solve the right side equation, which is a linear equation groups,for a specific $\lambda$, say $\lambda=3+4i$
$A-\lambda I=\left(\begin{array}{cc}3-4i& 5 \\ -5 & -3-4i\end{array}\right)$
• Aug 17th 2009, 05:39 PM
chella182
Yeah I know that... I'm not stuck on that (Worried) I'm stuck on what to do once I've found the eigenvalues.
• Aug 17th 2009, 07:42 PM
mr fantastic
Quote:

Originally Posted by chella182
Okay, so the matrix is...

$\left(\begin{array}{cc}6 & 5 \\ -5 & 0\end{array}\right)$

I solved $det(A-\lambda I)=0$ to find $\lambda$ (which is the method I was taught), and I ended up with $\lambda=3\pm4i$, but now I'm a bit stuck. I've written down...

$\left(\begin{array}{cc}6 & 5 \\ -5 & 0\end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right)=(3\pm4i)\left(\begin{array}{c} x \\ y\end{array}\right)$

... but I can't seem to manipulate it to get the eigenvectors.

Read this: The Case of Complex Eigenvalues
• Aug 17th 2009, 11:03 PM
chella182
Uhm, sort of thanks, but I'm still not coming up with it for my example. Exam's in an hour and a half anyway. Oh well.
• Aug 17th 2009, 11:19 PM
mr fantastic
Quote:

Originally Posted by chella182
Uhm, sort of thanks, but I'm still not coming up with it for my example. Exam's in an hour and a half anyway. Oh well.

Where exactly is your problem? For $\lambda = 3 + 4i$ the two equations reduce to the single equation $(3 - 4i) x + 5y = 0$. The link I gave you tells you what to do with this.

Ideally you would have met and resolved this sort of question earlier than a few hours before an exam.
• Aug 18th 2009, 05:08 AM
chella182
Sorry, but I really don't think it's your place to tell me how to revise for my exams. I have been revising for this exam for MONTHS now, and the threads I posted last night were just a last-minute attempt to iron out a few creases.

I do understand the theory of it, I just have problems reducing it to one equation, which is what I wanted explaining.

I know you won't care, but the exam went well and this didn't matter in the end 'cause the question I was given had real eigenvalues (which I find easier to get my head around), so hooray!
• Aug 18th 2009, 06:05 AM
mr fantastic
Quote:

Originally Posted by chella182
Sorry, but I really don't think it's your place to tell me how to revise for my exams. Mr F says: When someone flys the woe-is-me flag with remarks like "Exam's in an hour and a half anyway. Oh well." then I think the observation I made is justified.

I have been revising for this exam for MONTHS now, and the threads I posted last night were just a last-minute attempt to iron out a few creases. Mr F says: It's a pretty major crease.

I do understand the theory of it, I just have problems reducing it to one equation, which is what I wanted explaining. Mr F says: It would have been helpful to say that in the first place. Then you might have got precisely the explanation you apparently needed much sooner.

I know you won't care, Mr F says: To coin a quote, "I really don't think it's your place to tell me" whether or not I care ....

but the exam went well and this didn't matter in the end 'cause the question I was given had real eigenvalues (which I find easier to get my head around), so hooray! Mr F says: Good for you. I hope your result reflects that. (Not that I care, of course .... (Wink))