MathOverflow is a question and answer site for professional mathematicians. 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

Probably a well-know question, but I haven't solved it, so I'll ask.

I can show that every matrix in $M_2(\mathbb{R})$ is the sum of two squares of matrices in $M_2(\mathbb{R})$. If $n>2$, I can also show that every matrix in $M_n(\mathbb{R})$ is the sum of three squares of matrices in $M_n(\mathbb{R})$.

So my question is : Is every matrix in $M_n(\mathbb{R})$ is the sum of two squares of matrices in $M_n(\mathbb{R})$ (n>2)?

share|cite|improve this question
According to Wikipedia, every matrix is the average of two unitaries ( – Steve Huntsman Apr 9 '10 at 23:56
That sounds wrong to me. The set of unitary n by n matrices is compact, so the set of averages of two unitary matrices is compact too. So the latter set cannot equal the space of all n by n matrices. – Bjorn Poonen Apr 10 '10 at 0:28
Wikipedia missed the requirement that the norm of the matrix should be ≤ 1. – François G. Dorais Apr 10 '10 at 0:46
It's fixed now. – Steve Huntsman Apr 10 '10 at 1:22
One remark: Waring's problem usually takes place in the context of integers, not real numbers. So one might ask: what can one say about sums of squares of nxn matrices? Here there is a nice story,having to do with Siegel modular forms and powers of the Siegel theta function; see e.g. A.N.Andrianov's book "Quadratic forms and Hecke operators." – JSE Apr 10 '10 at 2:29
up vote 13 down vote accepted

The answer is YES if $n$ is even. But if $n$ is odd, then the answer is NO since $-I$ is not a sum of two squares. See

Griffin and Krusemeyer, Matrices as sums of squares, Linear and Multilinear Algebra 5 (1977/78), no. 1, 33-44

for the proofs of these facts and generalizations.

share|cite|improve this answer
Thanks Bjorn, this is useful. – Portland Apr 10 '10 at 1:17

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.