Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Quick questions:

  • if you have a diagonal matrix $A$ and a unitary matrix $B$. Do $A$ and $B$ commute?

  • if $A$ and $B$ are positive definite matrices. if $a$ is an eigenvalue of $A$ and $b$ is an eigenvalue of $B$, does it follow that $a+b$ is an eigenvalue of $A+B$?

share|cite|improve this question
No and no (that is, not always). Have you tried any examples? – Qiaochu Yuan Apr 30 '12 at 20:25
up vote 5 down vote accepted

For the first question, the answer is no, an explicit example is given by $A:=\pmatrix{1&0\\ 0&2}$ and $B=\pmatrix{1&1\\ -1&1}$. An other way to see it's not true is the following: take $S$ a symmetric matrix, then you can find $D$ diagonal and $U$ orthogonal (hence unitary) such that $S=U^tDU$ and if $U$ and $D$ commute then $S$ is diagonal.

For the second question, the answer is "not necessarly", because the set $$\{a+b\mid a\mbox{ eigenvalue of }A,b\mbox{ eigenvalue of }B\}$$ may contain more elements than the dimension of the space we are working with.

share|cite|improve this answer
(Just as a little resource, \mid is probably the best markup for the vertical bar in set notation.) – rschwieb Apr 30 '12 at 22:49
@rschwieb I agree, especially when we don't have division. – Davide Giraudo May 1 '12 at 8:25

Perhaps you know of so-called normal matrices, those complex matrices $M$ that commute with their transpose conjugate $M^{\dagger}$? If the answer to your first question were yes, there would be only "trivial", i.e. diagonal, normal matrices. Indeed, any normal matrix is unitarily equivalent to a diagonal one (all normal endomorphisms of a hermitian vector space are diagonalisable in an orthonormal basis).

share|cite|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.