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From Bretscher's Linear Algebra with Applications:

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where $A$ is a real matrix in $ \mathbb{R}^{n \times m}$ and the singular values of $A$ are the square roots of the eigenvalues of the symmetric $A^TA$, numbered in descending order and repeated up to algebraic multiplicity. Note that positive semi-definite $A^TA \in \mathbb{R}^{m \times m}$ has $m$ complex (possibly nondistinct) eigenvalues $\lambda_j$ from the characteristic polynomial and fundamental theorem of algebra, and it can be proven that all $m$ of these will be real, so the singular values will be $\sigma_i=\sqrt{\lambda_j} \in \mathbb{R}$.

My question: suppose a non-symmetric $A\in \mathbb{R}^{n \times n}$ is square, so that $A=U\Sigma V^T$ for diagonal $\Sigma$ and orthogonal $U$ and $V^T$, which are all square matrices in $\mathbb{R}^{n \times n}$. Thus, non-symmetric $A$ is similar to diagonal matrix $\Sigma$ via orthogonal matrix $U$ ("orthogonally diagonalizable"), so this contradicts one direction of the spectral theorem for a real matrix $A$:

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What went wrong?

jskattt797
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1 Answers1

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I think the error I made is this:

$AV=U \Sigma$ is different from $AU=U \Sigma$, so $A$ is not necessarily similar to $\Sigma$. Thus $A$ is not necessarily orthogonally diagonalizable, so there is no contradiction.

Even though the SVD sandwiches a diagonal matrix between two invertible matrices, this decomposition is not a diagonalization unless $U=V$. This makes sense because all matrices have an SVD, but not all matrices are diagonalizable.

Then I wondered: if square $A$ is symmetric, we can orthogonally diagonalize $A$ as $A=SDS^T$ by the spectral theorem - when does this decomposition coincide with the singular value decomposition? What is the relationship between the singular value decomposition and the eigenvalue decomposition? This has discussion here https://en.wikipedia.org/wiki/Singular_value_decomposition#Relation_to_eigenvalue_decomposition and here Is $U=V$ in the SVD of a symmetric positive semidefinite matrix?

jskattt797
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