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? 
this won't be true for matrices $A$ who's eigenvectors form a rank deficient matrix. In this case $P^{1}$ won't exist. 
What definition of "similar matrices" are you using? 
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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 12) 
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? 
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Two equal vectors are not linearly dependent so yes, the eigenvectors of $A$ form a rank deficient matrix and thus A is not diagonalizable. 
Thanks 
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