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Consider a squared non-symmetric complex valued matrix $A \in \mathbb{C}^{m\times m}$. Because the matrix is squared, I can use SVD or Eigendecomposition to decompose this matrix as:

$ A = Q \Lambda Q^{-1} \quad \text{or} \quad A=U\Sigma V^* $

Being in general, $\Lambda$ the matrix of complex-valued eigenvalues, and $\Sigma$ the matrix of singular values that are real-valued, by convention.

I understand that if the matrix $A$ would be symmetric and positive-definite, both SVD and eigendecomposition become equivalent, being the eigenvalues real-valued and identical to singular values. However, for the case of non-symmetric positive-definite matrices, this obviously not the case, as eigvals are complex and singular values are real.

I am interested in understanding the relationships between left/right singular vectors with eigenvectors, and the relationship between singular values and eigenvalues for this scenario. Any help is more than appreciated.

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    $\begingroup$ If $A$ is symmetric with non-negative eigenvalues, SVD and eigendecomposition are equivalent. More generally, this is not the case $\endgroup$ May 8, 2023 at 14:51
  • $\begingroup$ Edited the question for precision as suggested by @BenGrossmann $\endgroup$
    – Danfoa
    May 9, 2023 at 14:41
  • $\begingroup$ See my answer regarding singular values and eigenvalues. I'm not aware of any result comparing eigenvectors to singular values, but I suspect that something can be said in the case that $A$ has real eigenvalues and $Q$ has a small condition number. $\endgroup$ May 9, 2023 at 15:48
  • $\begingroup$ Also, to correct my earlier statement, $A$ would need to be Hermitian with non-negative eigenvalues in order for it to be positive definite; I hadn't noticed that we were considering complex matrices earlier. A symmetric matrix with non-real entries such as $$ \pmatrix{1&1+i\\1+i&2} $$ cannot be "positive definite" under the usual definition of that word for complex matrices. $\endgroup$ May 9, 2023 at 15:50

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One powerful result describing the relationship between singular values and eigenvalues is the set of inequalities $$ \prod_{j=1}^k |\lambda_j| \leq \prod_{j=1}^k \sigma_j, \quad 1 \leq k \leq n, $$ where $n$ is the size of the matrix $A$ and the indices are chosen such that $\sigma_1 \geq \cdots \geq \sigma_n$ and $|\lambda_1| \geq \cdots \geq |\lambda_n|$. In fact, the two products are necessarily equal for $k = n$ (with both sides equal to $|\det(A)|$. This result implies a more general result known as "Weyl's Majorant Theorem", which leads to a lot of inequalities of this kind. For instance, we have $$ \sum_{j=1}^k |\lambda_j|^p \leq \sum_{j=1}^k \sigma_j^p, \quad 1 \leq k \leq n $$ for all exponents $p \geq 0$.

In a sense, there is no result that yields a stronger relationship in general than this. In particular, Weyl's Majorant theorem has the following converse: if $\lambda_i, \sigma_i$ are complex and non-negative numbers (respectively) ordered with non-increasing magnitude such that we have $$ \prod_{j=1}^k |\lambda_j| \leq \prod_{j=1}^k \sigma_j, \quad 1 \leq k \leq n-1,\\ \prod_{j=1}^n |\lambda_j| = \prod_{j=1}^n \sigma_j, $$ then there necessarily exists a matrix $A$ whose eigenvalues are equal to the $\lambda_i$ and whose singular values are equal to the $\sigma_i$.

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    $\begingroup$ For more information, see chapter 2 of Bhatia's Matrix Analysis. This answer is mostly pasted from my earlier post here $\endgroup$ May 9, 2023 at 15:30
  • $\begingroup$ Thanks for the reference and the detailed answer Ben. One doubt, $k$ is above an index value on the set $[1, n]$. Thus the phrase "In fact, the two products are necessarily equal for $k=n$ (with both sides equal to $|det(A)|$" is I belive imprecise, maybe due to a trivial typo. Do you mean if the rank of the matrix is $n$? $\endgroup$
    – Danfoa
    May 18, 2023 at 14:33
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    $\begingroup$ @Danfoa I mean that this is true regardless of the rank of the matrix. If the rank of an $n \times n$ matrix is less than $n$, then it will have a zero singular value and a zero eigenvalue, which means that both products will be equal to $0$, which is also the determinant of the matrix. $\endgroup$ May 18, 2023 at 16:15

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