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-   -   For every A matrix exists a B diagonal matrix that A is similar to B? (http://mymathforum.com/linear-algebra/343083-every-matrix-exists-b-diagonal-matrix-similar-b.html)

 ricsi046 December 15th, 2017 10:16 AM

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?

 romsek December 15th, 2017 01:34 PM

this won't be true for matrices \$A\$ who's eigenvectors form a rank deficient matrix.

In this case \$P^{-1}\$ won't exist.

 Country Boy December 16th, 2017 02:42 AM

What definition of "similar matrices" are you using?

 ricsi046 December 16th, 2017 04:17 AM

Quote:
 Originally Posted by Country Boy (Post 585905) 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

 ricsi046 December 16th, 2017 05:00 AM

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)

 ricsi046 December 16th, 2017 05:25 AM

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?

 v8archie December 16th, 2017 06:04 AM

Quote:
 Originally Posted by ricsi046 (Post 585913) 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.

 romsek December 16th, 2017 07:59 AM

Quote:
 Originally Posted by ricsi046 (Post 585915) 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.

 ricsi046 December 17th, 2017 12:31 AM

Thanks

 Country Boy December 17th, 2017 05:38 AM

Quote:
 Originally Posted by ricsi046 (Post 585915) 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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