December 15th, 2017, 10:16 AM  #1 
Member Joined: Sep 2013 Posts: 84 Thanks: 2  For every A matrix exists a B diagonal matrix that A is similar to B?
Hello I feel that this statement is not true, but I don't know how to prove it. We know that A is similar to B if there exists an invertible P matrix so that B=inv(P)*A*P Any idea how to do this? 
December 15th, 2017, 01:34 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,457 Thanks: 1338 
this won't be true for matrices $A$ who's eigenvectors form a rank deficient matrix. In this case $P^{1}$ won't exist. 
December 16th, 2017, 02:42 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
What definition of "similar matrices" are you using?

December 16th, 2017, 04:17 AM  #4 
Member Joined: Sep 2013 Posts: 84 Thanks: 2  
December 16th, 2017, 05:00 AM  #5 
Member Joined: Sep 2013 Posts: 84 Thanks: 2 
What if A is not a square matrix? Does it have any similar matrix? On this site: http://www.maths.manchester.ac.uk/~p...TH10212/notes9 it's talking about n*n matrices ( page 12) 
December 16th, 2017, 05:25 AM  #6 
Member Joined: Sep 2013 Posts: 84 Thanks: 2 
after all, I came up with this solution: A has to have n linearly independent eigenvector to be diagonalizable, and this doesn't stand for example for this matrix: A=[1 1; 0 1] this has only 1 eigenvector instead of 2 linearly independent vectors Am I right? Or is it considered to have 2 eigenvector which are equal so not linearly independent? Last edited by ricsi046; December 16th, 2017 at 05:55 AM. 
December 16th, 2017, 06:04 AM  #7  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2642 Math Focus: Mainly analysis and algebra  Quote:
 
December 16th, 2017, 07:59 AM  #8  
Senior Member Joined: Sep 2015 From: USA Posts: 2,457 Thanks: 1338  Quote:
Two equal vectors are not linearly dependent so yes, the eigenvectors of $A$ form a rank deficient matrix and thus A is not diagonalizable.  
December 17th, 2017, 12:31 AM  #9 
Member Joined: Sep 2013 Posts: 84 Thanks: 2 
Thanks

December 17th, 2017, 05:38 AM  #10  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  Quote:
 

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