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From the definition of eigenvectors of a (square) matrix $A$, I can obtain the result: $$ A=P\Omega P^{-1} $$ where $P$ is unitary matrix (columns are eigenvectors), and $\Omega $ is a diagonal matrix of eigenvalues.

But I am wondering a very different situation: is it possible to decompose a given matrix $A$ as: $$ A=M \Omega M^{-1} $$ where $M$ is NOT unitary (the columns are not orthogonal). In other words, how to find a non-orthogonal matrix $M$ making $A=M \Omega M^{-1}$ holds ($\Omega$ is diagonal), given an arbitrary square matrix $A$?

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  • $\begingroup$ What does the star mean here in terms of matrices? Because I normally use that as the conjugate of a complex valued matrix $\endgroup$
    – Triatticus
    Jan 20, 2017 at 5:18
  • $\begingroup$ The star should be $T$, transpose. $\endgroup$
    – whitegreen
    Jan 20, 2017 at 5:26
  • $\begingroup$ Note that you can only get your first display if $A$ is symmetric (or hermitian, in the complex case). $\endgroup$ Jan 20, 2017 at 6:34
  • $\begingroup$ oh, yes. I should not assume $P^{-1}=P^T$ from the outset. $\endgroup$
    – whitegreen
    Jan 20, 2017 at 7:27

2 Answers 2

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That $P$ is unitary is a relativly rare case! Always happens, when $A$ is "normal", which means $AA^H = A^H A$. It can occure more often, but I#m not sure about that.

One example to make a difference is by using $$ A = \begin{pmatrix}1 &2 \\ 0 & 1\end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 &5\end{pmatrix}\begin{pmatrix}1 &2 \\ 0 & 1\end{pmatrix}^{-1} = \begin{pmatrix}2&6\\0&5 \end{pmatrix} $$ The collumns of $P$ are not orthogonal.

Keep in mind, that it always depends on $A$ and it's eigenvalues and eigenvectors. If the vectors are orthogonal, you can find an orthonormal basis. If not, you can not. However you can just judge this by calculationg the eigenvalues.

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  • $\begingroup$ Thank you very much! I should read the textbook much more carefully. I often worked with Hermitian operators and thus always saw orthogonal eigenvectors, which made me misunderstand the general idea of eigendecomposition. $\endgroup$
    – whitegreen
    Jan 20, 2017 at 8:11
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For a square matrix $A$ of order $n$, $A=M\Omega M^{-1}$ holds iff $A$ has $n$ linearly independent eigenvectors, where $M$ contains the $n$ eigenvectors as its columns and obviously $det(M)\ne 0$. Moreover, to satisfy the restriction for $M$ (non-unitary), matrix $A$ should not be normal .

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