# Do singular values dominate eigenvalues?

Suppose $A$ is an $n \times n$ complex matrix with singular values $s_1 \ge s_2 \ge \cdots \ge s_n$ and eigenvalues $(\lambda_i)_{i=1}^{n}$ arranged so that $|\lambda_1| \ge |\lambda_2| \ge \cdots \ge |\lambda_n|$. Is it true that $s_1 \ge |\lambda_1|,\ s_1+s_2 \ge |\lambda_1| + |\lambda_2|,$ and so on?

• For normal matrices, it seems clear that we have the even stronger statement $s_j \ge \lambda_j$ for each $j$. On the other hand, for non-normal matrices this stronger statement fails. – David Ketcheson Oct 26 '14 at 11:05
• @David: How does this correlate with $\prod s_i=\prod |\lambda_i|=|\det A|$? – Ilya Bogdanov Oct 26 '14 at 16:14
• Am I missing something, or for normal matrices $s_i=|\lambda_i|$, since the eigendecomposition is an SVD up to phase factors? – Federico Poloni Oct 26 '14 at 18:00
• @Federico: Surely;) – Ilya Bogdanov Oct 26 '14 at 19:18
• A relevant reference here is the classic paper of Horn, ams.org/mathscinet-getitem?mr=61573 , showing that the Weyl inequalities in Yanqi's answer in fact generate the complete set of relations between singular values and eigenvalues. Also, one can prove the inequalities by applying Gram-Schmidt to the eigenvectors to conjugate $A$ by a unitary matrices to upper-triangular form (so that the eigenvalues become diagonal entries), and then applying the SVD and orthogonality relations for the singular vectors. – Terry Tao Oct 26 '14 at 20:31

Let $\lambda(A)$ denote the vector of eigenvalues and $s(A)$ the vector of singular values (arranged in decreasing order). The claim of the question is whether $|\lambda(A)|^{\downarrow} \prec_w s(A)$. The answer to this question is yes (the proof follows using the tensor power trick Yanqi Qui mentioned above).

Theorem (Weyl's Majorant theorem). Let $f: (0,\infty)\to(0,\infty)$ be such that $f(e^t)$ is convex and monotone increasing in $t$. Then $$[f(\vert\lambda_1\vert),\ldots, f(|\lambda_n|)] \prec_w [f(s_1),\ldots,f(s_n)].$$

As a corollary, using the function $t \mapsto t^p$ for any $p \ge 0$, we obtain an affirmative answer to the OP's question (actually, by merely using the special case $p=1$). Actually, Weyl's Majorant theorem provides the following log-majorization (which implies the above result): \begin{equation*} \log |\lambda(A)| \prec \log s(A), \end{equation*} and this version also answers Ilya's comment, namely $\prod_i^n |\lambda_i| = \prod_i^n s_i$.

Another closely related result is that of absolute values of the Hermitian part of a complex matrix. Let $\Re(A) := (A+A^*)/2$. Then, we have

Theorem (Fan-Hoffman). For every matrix $A$, we have \begin{eqnarray*} \lambda_j^{\downarrow}(\Re(A)) &\le& s_j(A),\qquad 1\le j \le n\\ |\lambda(\Re(A))| &\prec_w& s(A). \end{eqnarray*}

Reference

R. Bhatia, Matrix Analysis (Springer GTM, 169).

The singular values of $A$ are the eigenvalues of $|A|$. Since $$\lambda_1(A) \lambda_2(A) \cdots \lambda_k(A)$$ is an eigenvalue of $\underbrace{A\wedge A\wedge \dots \wedge A }_{\text{$k$times}}$, we have \begin{align*} |\lambda_1(A) \lambda_2(A) \cdots \lambda_k(A) | & \le \| A\wedge A\wedge \dots \wedge A \| \\ &= \Big\| \left|A\wedge A\wedge \dots \wedge A\right| \Big\| \\ & = \Big\| |A|\wedge |A|\wedge \dots \wedge |A| \Big\| \\ & = \lambda_1(|A|) \lambda_2(|A|) \cdots \lambda_k(|A|) .\end{align*} It follows that $$\sum_{k = 1}^n \varphi(|\lambda_k(A)|) \le \sum_{k=1}^n\varphi(\lambda_k(|A|)) ,$$ for any $\varphi$ such that $t \rightarrow \varphi(e^t)$ is convex non-decreasing.

• I realize I didn't really answer the original question. Sorry. – Yanqi QIU Oct 26 '14 at 16:25
• Actually your answer does answer the original question, by creating weak majorization inequalities for the entire range, rather than just the $k=1,..,n$ case. – Suvrit Oct 26 '14 at 23:58