Symmetric matrix with eigenvalues

  • Thread starter Ylle
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  • #1
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Homework Statement


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.



Homework Equations



No clue...

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
 

Answers and Replies

  • #2
tiny-tim
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Hi Ylle! :smile:
… show that A is a symmetric matrix with eigenvalues c1, c2,..., cn and that ui is an eigenvector belonging to ci for each i.
(i assume you can prove that A is symmetric)

You just need to prove that Aui = ciui

(and remember the basis is orthonormal :wink:)
 
  • #3
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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
tiny-tim
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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 ?
Hi Ylle! :smile:

Yes, A = AT.

But your proof doesn't work, because transpose is like inverse … it alters the order of things … so (AB)T = BTAT :wink:
And the next I can't figure out what you mean :?
Just write Aui in full (using the orthonormality of the u's) …

what do you get? :smile:
 
  • #5
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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
tiny-tim
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Hi Ylle! :smile:

(btw, please don't say "add" :eek: … say "multiply", or if you prefer a more neutral word, "apply" :wink:)
… And because uiuiT = <ui,<ui> = ||ui|| = 1 since it's a basis, you have:

Aui = ciui

Don't know if that is correct ?
Yup, that's fine …

that's what "orthonormal" is all about! :biggrin:

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

use the orthonormal property again. :smile:
 

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