# How to relate SVD and Eigendecomposition of squared complex non-symmetrical matrix

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.

• If $A$ is symmetric with non-negative eigenvalues, SVD and eigendecomposition are equivalent. More generally, this is not the case May 8, 2023 at 14:51
• Edited the question for precision as suggested by @BenGrossmann May 9, 2023 at 14:41
• 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. May 9, 2023 at 15:48
• 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. May 9, 2023 at 15:50

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 $$\sigma_i$$.
• 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 $n$? May 18, 2023 at 14:33
• @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. May 18, 2023 at 16:15