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December 15th, 2017, 11:16 AM   #1
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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?
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December 15th, 2017, 02:34 PM   #2
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this won't be true for matrices $A$ who's eigenvectors form a rank deficient matrix.

In this case $P^{-1}$ won't exist.
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December 16th, 2017, 03:42 AM   #3
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What definition of "similar matrices" are you using?
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December 16th, 2017, 05:17 AM   #4
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Quote:
Originally Posted by Country Boy View Post
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
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December 16th, 2017, 06:00 AM   #5
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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)
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December 16th, 2017, 06:25 AM   #6
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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 06:55 AM.
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December 16th, 2017, 07:04 AM   #7
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Quote:
Originally Posted by ricsi046 View Post
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.
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December 16th, 2017, 08:59 AM   #8
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Quote:
Originally Posted by ricsi046 View Post
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.
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December 17th, 2017, 01:31 AM   #9
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December 17th, 2017, 06:38 AM   #10
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Quote:
Originally Posted by ricsi046 View Post
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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