Does there exist a base of $M(n, \mathbb{R})$ of elements of $SO(n)$?

Question

I was wondering whether or not there exist a base of $M(n, \mathbb{R})$ as a vector space over $\mathbb{R}$ made solely by matrices in $SO(n)$, or equivalently if $Span(SO(n))=M(n)$ for each natural $n$. Even partial answers, such as one answering to the same question with $O(n)$ replacing $SO(n)$, would be greatly appreciated.

Answer 1

Any matrix $A$ in the span of $SO(2)$ has $A_{21} = -A_{12}$.

Answer 2

Note that if $p$ is an orthogonal projection then $I - 2 p$ is an isometry. Since $I$is also an isometry, you can write any orthogonal projection as a linear combination of isometries, and so any $T \cdot \pi$, where $T$ is an isometry and $T$ is a projection. It should be clear now how to get all the $E_{ij}$'s.