# Matrix diagonalization: what if the (basis) vectors are not orthogonal?

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 $A$?

• What does the star mean here in terms of matrices? Because I normally use that as the conjugate of a complex valued matrix – Triatticus Jan 20 '17 at 5:18
• The star should be $T$, transpose. – whitegreen Jan 20 '17 at 5:26
• Note that you can only get your first display if $A$ is symmetric (or hermitian, in the complex case). – Gerry Myerson Jan 20 '17 at 6:34
• oh, yes. I should not assume $P^{-1}=P^T$ from the outset. – whitegreen Jan 20 '17 at 7:27

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.
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 .