20070206, 02:35  #1 
May 2005
Argentina
272_{8} 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. 
20070206, 17:03  #2 
∂^{2}ω=0
Sep 2002
República de California
26053_{8} Posts 
Could you please define "geometric multiplicity?" (It sounds like something relating to the eigenspace, but I want to be sure.)

20070206, 22:04  #3 
May 2005
Argentina
2·3·31 Posts 

20070206, 22:18  #4 
∂^{2}ω=0
Sep 2002
República de California
10110000101011_{2} 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 rankdeficient. In that case the best one can do is to find a set of pseudoeigenvectors (real eigenvectors plus some noneigenvectors to "fill in" the rankdeficient elements of the eigenspace corresponding to the particular problematic repeated eigenvalues) which "nearly" diagonalize the matrix. 
20070208, 16:28  #5 
May 2005
Argentina
2×3×31 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. 
20070208, 17:13  #6 
∂^{2}ω=0
Sep 2002
República de California
3·3,769 Posts 
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.

20070212, 19:31  #7  
∂^{2}ω=0
Sep 2002
República de California
3×3,769 Posts 
Quote:


20070212, 20:27  #8  
Bamboozled!
May 2003
Down not across
2764_{16} Posts 
Quote:
Paul Last fiddled with by xilman on 20070212 at 20:28 Reason: Fix tag 

20070212, 22:25  #9 
∂^{2}ω=0
Sep 2002
República de California
3·3,769 Posts 
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 L^{p}'s, but failing that, simply to achieve closure.
p.s.: Norm's is bestknown 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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