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 2007-02-06, 02:35 #1 Damian     May 2005 Argentina 101110102 Posts Linear algebra proof Can someone point me to the proof that if the geometric multiplicity of each eigenvalue is equal to the corresponding algebraic multiplicity, then the matrix is diagonalizable. Thanks in advance.
 2007-02-06, 17:03 #2 ewmayer ∂2ω=0     Sep 2002 Repรบblica de California 5×2,351 Posts Could you please define "geometric multiplicity?" (It sounds like something relating to the eigenspace, but I want to be sure.)
2007-02-06, 22:04   #3
Damian

May 2005
Argentina

2×3×31 Posts

Quote:
 Originally Posted by ewmayer Could you please define "geometric multiplicity?" (It sounds like something relating to the eigenspace, but I want to be sure.)
The geometric multiplicity is the dimenion of the eigenspace.

 2007-02-06, 22:18 #4 ewmayer ∂2ω=0     Sep 2002 Repรบblica de California 5·2,351 Posts Then it's quite simple: for a repeated eigenvalue (the only case one need be concerned about w.r.to possible nondiagonalizability), if the geometric multiplicity of the corresponding eigenspace is equal to the algebraic multiplicity of the eigenvalue (call that K), that means that one can find K linearly independent eigenvectors, hence the matrix is diagonalizable. Put another way, one only winds up with a Jordan form (nondiagonalizability) if the eigenspace is rank-deficient. In that case the best one can do is to find a set of pseudo-eigenvectors (real eigenvectors plus some non-eigenvectors to "fill in" the rank-deficient elements of the eigenspace corresponding to the particular problematic repeated eigenvalues) which "nearly" diagonalize the matrix.
 2007-02-08, 16:28 #5 Damian     May 2005 Argentina BA16 Posts Thanks, and other question Thank you very much. I've got a new question: is there a proof that for distinct eigenvalues, there correspond linear independent eigenvectors, that does not use mathematical induction? Thanks in advance, Damian.
2007-02-08, 17:13   #6
ewmayer
2ω=0

Sep 2002
Repรบblica de California

267538 Posts

Quote:
 Originally Posted by Damian Thank you very much. I've got a new question: is there a proof that for distinct eigenvalues, there correspond linear independent eigenvectors, that does not use mathematical induction?
This would appear to follow directly from the definition of an eigenvector. Try this: assuming that for 3 distinct eigenvalues l1,l2,l3 with corresponding eigenvectors x,yz, one of the eigenvectors is a linear combination of the other 2. e.g. z = a*x+b*y. Multiply by the matrix, and you should pretty quickly get a contradiction.

2007-02-12, 19:31   #7
ewmayer
2ω=0

Sep 2002
Repรบblica de California

2DEB16 Posts

Quote:
 Originally Posted by ewmayer This would appear to follow directly from the definition of an eigenvector. Try this: assuming that for 3 distinct eigenvalues l1,l2,l3 with corresponding eigenvectors x,yz, one of the eigenvectors is a linear combination of the other 2. e.g. z = a*x+b*y. Multiply by the matrix, and you should pretty quickly get a contradiction.
OK, I verified that this does lead to a simple proof, but one still needs to also show that given a starting point of one {eigenvalue,eigenvector} pair, the eigenvector for the *second* distinct eigenvalue must be LI of the first, which in this case reduces to "not a multiple" of. Again easy to show, but in the end it does amount to proof by induction.

2007-02-12, 20:27   #8
xilman
Bamboozled!

"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across

2DC916 Posts

Quote:
 Way Out in Hilbert Space
An infinite-dimensional exit? If so, where does it lead to?

Paul

Last fiddled with by xilman on 2007-02-12 at 20:28 Reason: Fix tag

2007-02-12, 22:25   #9
ewmayer
2ω=0

Sep 2002
Repรบblica de California

5·2,351 Posts

Quote:
 Originally Posted by xilman An infinite-dimensional exit? If so, where does it lead to?
Like the sign (you know, the one that pops up almost everywhere) says: In Hilbert Space, all roads converge (if they converge) to a li'l place called Norm's Functional Rest Stop. I'm hoping to get there at some point so I can start to unload some of my collection of old vinyl Lp's, but failing that, simply to achieve closure.

p.s.: Norm's is best-known for its "eat a burger, drink a beer and smoke adjoint" special, but interestingly, they also offer a nice lineup of Cauchy foods.

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