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 December 15th, 2017, 10:16 AM #1 Member   Joined: Sep 2013 Posts: 93 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: 2,584 Thanks: 1430 this won't be true for matrices $A$ who's eigenvectors form a rank deficient matrix. In this case $P^{-1}$ won't exist. Thanks from ricsi046 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
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 Originally Posted by Country Boy What definition of "similar matrices" are you using?
A is similar to B if there exists an invertible P matrix so that B=inv(P)*A*P December 16th, 2017, 05:00 AM #5 Member   Joined: Sep 2013 Posts: 93 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 1-2) December 16th, 2017, 05:25 AM #6 Member   Joined: Sep 2013 Posts: 93 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
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 Originally Posted by ricsi046 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 1-2)
You can't have a diagonal matrix that isn't square. December 16th, 2017, 07:59 AM   #8
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 Originally Posted by ricsi046 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?
sounds remarkably like what I said in post 2

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: 93 Thanks: 2 Thanks December 17th, 2017, 05:38 AM   #10
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 Originally Posted by ricsi046 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?
If a matrix has one eigenvector, it has an infinite number of them since any scalar multiple is also an eigenvector. Tags diagonal, exists, matrix, similar Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post silviatodorof Linear Algebra 2 March 22nd, 2015 05:28 AM annakar Linear Algebra 0 January 10th, 2013 09:58 AM lamhmh Applied Math 2 July 11th, 2011 12:12 PM

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