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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)?

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  • $\begingroup$ According to Wikipedia, every matrix is the average of two unitaries (en.wikipedia.org/wiki/Unitary_matrix#Properties). $\endgroup$
    – Steve Huntsman
    Commented Apr 9, 2010 at 23:56
  • $\begingroup$ 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. $\endgroup$
    – Bjorn Poonen
    Commented Apr 10, 2010 at 0:28
  • $\begingroup$ Wikipedia missed the requirement that the norm of the matrix should be ≤ 1. $\endgroup$
    – François G. Dorais
    Commented Apr 10, 2010 at 0:46
  • $\begingroup$ It's fixed now. $\endgroup$
    – Steve Huntsman
    Commented Apr 10, 2010 at 1:22
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    $\begingroup$ 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." $\endgroup$
    – JSE
    Commented Apr 10, 2010 at 2:29

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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.

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