Is there a theorem that allows one to decompose any real $(n \times n)$ matrix $A$ into $P \Lambda P^{\mathsf{T}}$ where $P$ is symmetric and idempotent and $\Lambda$ is a diagonal matrix of eigenvalues of $A$? If so, where can I find its proof?
Factorization of real square matrix into eigenvalues and symmetric and idempotent matrix
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linear-algebra
eigenvalues-eigenvectors
matrix-decomposition
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0Do you mean $P^2 = I$ instead of "$P$ is idempotent"? – 2017-02-15
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0No, I meant $P^{2} = P$ and $P^{\mathsf{T}} = P$. – 2017-02-15
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0That's a very strange. Did anything in particular motivate this question? – 2017-02-15
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0I just realized where my mistake was. I needed the $P$'s to disappear as I raise $A^{j}$, but of course this will also happen in a decomposition $A = P \Lambda P^{-1}$, because then $A^{2} = P \Lambda P^{-1} P \Lambda P^{-1} = P \Lambda^{2} P^{-1}$. Actually a pretty silly mistake. But thanks for response. – 2017-02-15
1 Answers
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No, there is no such theorem. It is impossible to do so.
In particular: take $A$ is invertible and non-diagonal. $P$ must be invertible in order for the product $P\Lambda P^T$ to be invertible. However, the only invertible idempotent matrix is $I$. So, we would need to have $A = I\Lambda I^T = \Lambda$. But $\Lambda$ is diagonal, and $A$ is not.
So, there is no such $P$ and $\Lambda$ for an invertible, non-diagonal $A$.
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0Thank you. Just for the record though, a solution to $A = P \Lambda P^{\mathsf{T}}$ exists if $A$ is a _symmetric_ matrix, by [Theorem 5.8 in Basilevsky (1983)](https://books.google.com/books?id=ScssAwAAQBAJ&pg=PA200). But I was looking for a more general result, which as you've shown cannot be. – 2017-02-15
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1@Durder that's for an **orthogonal** matrix $P$ (and that result known as the spectral theorem for symmetric matrices). Your result doesn't seem "more general"; it just seems completely different. – 2017-02-15