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I was trying to answer the question of whether the diagonalization of a matrix is unique and found that it is "unique up to a permutation of the entries." (see: Is a diagonalization of a matrix unique?)

However, I found a slightly different way to diagonalize A, which I think does not depend on eigenvalues.

Conjugacy (A-orthogonality) is defined such that for a matrix A the directions $p_i$ for i=1,...,n satisfy $$p_i^TAp_j = 0 \; \text{if} \; i\neq j $$

Let $$ P = [p_1 \; p_2 \; ... \; p_n] $$

So, $ P^TAP $ is a diagonal matrix.

Since we have found a way to diagonalize A without using eigenvectors and the diagagonal matrix does not depend on eigevalues, I was wondering why we can say that matrix diagonalization is "unique up to a permutation of the entries."

Doesn't this also contradict this answer: Is it possible to diagonalize a matrix without eigenvalues?

Nick Hill
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    Your definition of a diagonalizable matrix is not correct. A matrix $A$ is diagonalizable iff there exists $P$ such that $P^{-1}AP$ is diagonal (not $P^TAP$). If all eigenvalues are distinct, then $P$ can be chosen to be orthogonal, so $P^{-1}=P^T$. I guess this got you confused. – SMM Jul 14 '20 at 20:40

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In a diagonalization, the matrix that is multiplied from the left must be the inverse of the matrix that is multiplied from the right. In your case, $P^T=P^{-1}$ does not necessarily hold. If it does, then the diagonal matrix will contain the eigenvalues.

Reinhard Meier
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  • So, are you saying that since $ P^T \neq P^{-1} $, my formulation is not considered a diagonalization? Is it simply a matter of definition? – Nick Hill Jul 14 '20 at 21:05
  • Yes, it is a matter of definition. I have never seen the term "matrix diagonalization" being used for something that cannot be expressed as $P^{-1}AP.$ – Reinhard Meier Jul 14 '20 at 21:10