# Symmetric matrix with eigenvalues

1. Apr 29, 2009

### Ylle

1. The problem statement, all variables and given/known data
Let {u1, u2,...,un} be an orthonormal basis for Rn and let A be a linear combination of the rank 1 matrices u1u1T, u2u2T,...,ununT. If

A = c1u1u1T + c2u2u2T + ... + cnununT

show that A is a symmetric matrix with eigenvalues c1, c2,..., cn and that ui is an eigenvector belonging to ci for each i.

2. Relevant equations

No clue...

3. The attempt at a solution

I'm really stuck with this problem. So I'm really just hoping for a little hint or something.
It SOUNDS easy, but as I said, I have no clue where to start.

Hope you can help.

Regards

2. Apr 29, 2009

### tiny-tim

Hi Ylle!
(i assume you can prove that A is symmetric)

You just need to prove that Aui = ciui

(and remember the basis is orthonormal )

3. Apr 29, 2009

### Ylle

Re: Eigenvalues

Hi :)

I'm not totally sure about the first, but I have this (See the picture added). I don't know if it is proved there ?
For it to be symmetric it's it has to be: A = AT, right ?

And the next I can't figure out what you mean :?
If you add ui to A I guess you can remove the u1-n and u1-nT because they become the identitymatrix for all of them. And then you have c1-nui left. So how do I make the c1-n to ci ? :)

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4. Apr 30, 2009

### tiny-tim

Hi Ylle!

Yes, A = AT.

But your proof doesn't work, because transpose is like inverse … it alters the order of things … so (AB)T = BTAT
Just write Aui in full (using the orthonormality of the u's) …

what do you get?

5. Apr 30, 2009

### Ylle

Re: Eigenvalues

Well, my guess is you have:

A = ciuiuiT

If you then add ui on both sides you get:

Aui = ciuiuiTui

And because uiuiT = <ui,<ui> = ||ui|| = 1 since it's a basis, you have:

Aui = ciui

Don't know if that is correct ?

6. Apr 30, 2009

### tiny-tim

Hi Ylle!

(btw, please don't say "add" … say "multiply", or if you prefer a more neutral word, "apply" )
Yup, that's fine …

that's what "orthonormal" is all about!

but … you still have to deal with all the u's in the ∑ that aren't that particular ui

use the orthonormal property again.