# $\mathcal{O}(n,\mathbb R)$ spans $\mathcal{M}(n,\mathbb R)$

Let $n\geq 3$. One can show that the orthogonal group of degree $n$ over the real field, $\mathcal{O}(n,\mathbb R)$, spans the entire vector space of real $n\times n$ matrices, $\mathcal{M}(n,\mathbb R)$. More precisely if $k(n)$ denotes the smallest integer such that each $M\in \mathcal{M}(n,\mathbb R)$ can be written as $$M=\sum_{i=1}^{k(n)} \lambda_i O_i, \quad \text{with}\quad (\lambda_i, O_i)\in (\mathbb R, \mathcal{O}(n,\mathbb R)).$$ After showing that $\forall n>2, k(n)\leq 4$, can one find the integers such that $k(n)=4$?

• Is this a problem from a book or something like that? Oct 25 '11 at 22:49
• Every $M \in {\cal M}(n,{\mathbb R})$ has a Singular Value Decomposition $M = U \Sigma V$ where $U$ and $V$ are orthogonal matrices and $\Sigma$ is diagonal. Since $\Sigma = \sum_i \lambda_i O_i$ implies $M = \sum_i \lambda_i U O_i V$, we can wlog assume $M$ is diagonal. Now for $n=2$ or 3 it is easy to write the diagonal matrices as linear combinations of 2 or 3 diagonal orthogonal matrices, so $k(2) \le 2$ and $k(3) \le 3$. I suspect $k(n) = 4$ for all $n \ge 4$, but I don't see how to prove it. Oct 25 '11 at 23:28
• @MarianoSuárez-Alvarez, it may be in a book, but I don't know which/where, it came through a discussion. Oct 26 '11 at 0:01
• Imean: are you asking if «one can find such an $n$ after showing that for all $n>2$ one has $k\leq 4$», or you know that $k(n)\leq 4$ for such $n$ somehow and you want to know if you do get $k=4$ for some $n$? Oct 26 '11 at 0:15
• The following paper may be helpful: Chi-Kwong Li and Edward Poon, Additive Decomposition of Real Matrices, Linear and Multilinear Algebra, 50(4):321-326, 2002. It essentially says that $k(n)\le 4$, but whether the bound is tight is an open problem. Oct 27 '11 at 8:06

The following paper may be helpful: Chi-Kwong Li and Edward Poon, Additive Decomposition of Real Matrices, Linear and Multilinear Algebra, 50(4):321-326, 2002. It essentially says that $k(n)\le4$, but whether the bound is tight is an open problem.