# Thread: Eigenvectors correspondig to eigenvalues

1. ## Eigenvectors correspondig to eigenvalues

Hello,
Let's assume that vector $\bar v$ is eigenvector corresponding to eigenvalue $\lambda_{A}$ of matrix $A$. Also $\bar v$ corresponds to eigenvalue $\lambda_{B}$ of matrix $B$. Now the questions:

I Is the $\bar v$ eigenvector of matrix $A+B$?
II Is the $\bar v$ eigenvector of matrix $AB$?

If answer is yes to atleast one question, it must be shown, to which eigenvalue vector $\bar v$ corresponds.

I think I should use following theorem:
"If $\bar v$ is eigenvector corresponding to eigenvalue of matrix $A$, then $\bar v$ is is eigenvector corresponding to $A^k$'s $\lambda^k$".

But somehow, I can't get anywhere, though I inspected eigenvalues and -vectors definitions closely. Also it's no problem to me calculate those vectors and values in normal cases. So, any help is appreciated. Thank you!

2. Try computing the quantities

$(A+B)\bar{v}$ and $AB\bar{v}.$ What do you get?

3. Hmm. I got:
$(A+B)\bar{v}$
$=\bar{v}A+\bar{v}B$
$=\bar{v}\lambda_{A}+\bar{v}\lambda_{B}$

So, I think $\bar v$ corresponds to $\lambda_{A}$ and $\lambda_{B}$, in this case. If I calculated it right. Second:

$\bar{v}(AB)$
$=(\bar{v}A)B$
$=A(\bar{v}B)$
$=(\bar{v}\lambda_{A})\lambda_{B}$
$=\lambda_{A}(\bar{v}\lambda_{B})$

Which means that, it isn't eigenvector of $AB$. Does this make any sense or Am I completely lost? Thanks for your help again! It's been very useful.

4. I am so not used to seeing leftward-looking operators. I can see that you're on the right track, though. Try this on for size:

$(A+B)\bar{v}=A\bar{v}+B\bar{v}=\lambda_{A}\bar{v}+ \lambda_{B}\bar{v}=(\lambda_{A}+\lambda_{B})\bar{v }.$ What does that tell you?

And, we also have

$AB\bar{v}=A(\lambda_{B}\bar{v})=\lambda_{B}A\bar{v }=\lambda_{B}\lambda_{A}\bar{v}=(\lambda_{A}\lambd a_{B})\bar{v}.$

What can you conclude?

5. Revisiting following basic definition:
"If $A\bar{v}=\lambda\bar{v}$, then $\lambda$ is eigenvalue of $A$ and $\bar{v}$ eigenvector corresponding to $\lambda$",

I would like to conclude, that $\bar{v}$ is eigenvector of both (which is obviously wrong). This also should say something to me, but...

6. Right. You need to have the nonzero condition on the eigenvector in there as well. Why can't $\bar{v}$ be an eigenvector of both $A+B$ and $AB?$ Haven't I basically just proven that it is?

7. Yeah, you have it correct. You've been very kind again, so word of thanks for you. I tend to make wrong guesses quite often, so I wasn't serious. $\bar{v}$ can be eigenvector of both, Q.E.D.

8. You're welcome. Have a good one!