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 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: 2,039 Thanks: 1063 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: 3,261 Thanks: 894 What definition of "similar matrices" are you using?
December 16th, 2017, 04:17 AM   #4
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Quote:
 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: 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 1-2)
 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
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Quote:
 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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Quote:
 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: 84 Thanks: 2 Thanks
December 17th, 2017, 05:38 AM   #10
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Quote:
 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.

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