Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A = \left( {a_{ij} } \right)$ be a matrix over $\mathbb R$, of size $n \times n$. Let $\left\{ \lambda _k \right\}_{k = 1}^n$ be the $n$ eigenvalues of the matrix. Prove the following inequality:

$$ \sum_{k = 1}^n \left| {\lambda _k } \right|^2 \leqslant \sum_{i = 1}^n \sum\limits_{j = 1}^n \left| {a_{ij} } \right|^2 . $$

I can't prove it, and I'm not sure if in the problem one assumes that there exist $n$ real eigenvalues, or it's also true for complex eigenvalues. D:

share|improve this question
I'm fairly certain this follows from Jordan form, for example, and that one has an equality iff $A$ is diagonal over $\mathbf{C}$. Will come back to it if I can... –  tkr Nov 14 '11 at 22:13
Jordan form won't quite work because RHS isn't invariant under general conjugations. –  p.s. Nov 14 '11 at 22:34
Yeah, that is why I hesitated writing an answer. Glad someone wrote it out. –  tkr Nov 14 '11 at 22:44
Right hand side is equal to $\operatorname{Tr}(AA^*)$ and trace is equal to the sum of the eigenvalues of the matrix. You might want to continue from there. –  user13838 Nov 14 '11 at 23:03

2 Answers 2

This is in fact true even for matrices over $\mathbb{C}$. Both the left and right side are invariant under conjugating the matrix $A$ by a unitary matrix $Q$. Therefore by taking the Schur decomposition we can assume without loss of generality that $A$ is upper triangular (but potentially complex even if the original $A$ was real). Since the eigenvalues of an upper triangular matrix are just the diagonal entries, the inequality follows.

share|improve this answer

This are Schur's Inequalities. There are many proofs that can be found. It basically directly follows from the Schur decomposition.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.