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.
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 $\sigma_i$.