Linear subspaces of rotation matrices A rotation matrix for two dimension is contained within the space of matrices on the form
$\left( \begin{array}{cc}a & -b\\ b & a\end{array} \right)$
where $a, b \in \mathbf{R}$. This matrix represents a rotation and a scaling. We can write it as a linear combination of two basis matrix like this:
$\left( \begin{array}{cc}a & -b\\ b & a\end{array} \right) = a\left( \begin{array}{cc}1 & 0\\ 0 & 1\end{array} \right) + b\left( \begin{array}{cc}0 & -1\\ 1 & 0\end{array} \right)$
The question: Can we represent any $3 \times 3$ rotation matrix using a linear combination of a number (less than 9) of constant basis matrices, that is,
$R = \sum_{i = 1}^{n}\lambda_i B_i$
where $n < 9$, $R$ can be any $3 \times 3$ rotation matrix, $B_i$ are constant basis matrices for all $R$ and $\lambda_i$ are scalar weights that are different for different $R$?
(Of course we can represent other matrices too that are not pure rotations, what is important is that the space of all rotation matrices is contained within the space of all matrices that we can generate as a linear combination of the basis matrices)
 A: Let's see if we can get 9 degrees of freedom together. First include the identity and the permutation matrices. This is a total of 6 matrices but they only include 5 degrees of freedom (as the matrix with all elements identical is included twice):
$$\left(\begin{array}{ccc}a&b&c\\c&a&b\\b&c&a\end{array}\right),\;\;\;\;\;\;
\left(\begin{array}{ccc}d&e&f\\e&f&d\\f&d&e\end{array}\right).$$
Infinitesimal rotation angles give the identity (included above) plus elements from the Lie algebra (which are the anti-symmetric real matrices):
$$\left(\begin{array}{ccc}0&g&h\\-g&0&i\\-h&-i&0\end{array}\right).$$
Another set of matrices in the 3x3 rotation group are the diagonal matrices with two -1s and one 1:
$$\left(\begin{array}{ccc}+1&0&0\\0&-1&0\\0&0&-1\end{array}\right),\;\;\;\;\;
\left(\begin{array}{ccc}-1&0&0\\0&+1&0\\0&0&-1\end{array}\right),\;\;\;\;\;
\left(\begin{array}{ccc}-1&0&0\\0&-1&0\\0&0&+1\end{array}\right)$$
These last three matrices, plus the identity, give you all the degrees of freedom on the diagonal. And the other matrices give you the symmetric and anti-symmetric off diagonal elements. Thus all nine degrees of freedom are covered.
So there is no linear combination of less than nine matrices that include all the 3x3 rotation matrices.that is, a vector that is annihilated by the above 9x9 matrix.
A: You can have a basis made up only of “signed permutation” matrices as shown below.
For $x,y,z \in {\mathbb R}$, let
$$
M_1(x,y,z)=
\left(
\begin{array}{ccc}
x & y & -z \\
y & z & -x \\
z & x & -y \\
\end{array}
\right),
M_2(x,y,z)=
\left(
\begin{array}{ccc}
-y & z & x \\
-z & x & y \\
-x & y & z \\
\end{array}
\right),
M_3(x,y,z)=
\left(
\begin{array}{ccc}
z & -x & y \\
x & -y & z \\
y & -z & x \\
\end{array}
\right)
$$
For each $k$, the $M_k(x,y,z)$ span a three-dimensional subspace of
${\cal M}_{3,3}({\mathbb R})$, that we will denote by $V_k$. For any matrix 
$A=(a_{ij})_{1\leq i,j \leq 3}$, we have $A=\frac{A_1+A_2+A_3}{2}$ where
$$
\begin{array}{lcl}
A_1 &=& M_1(a_{11}+a_{32},a_{12}+a_{21},a_{22}+a_{31}) \\
A_2 &=& M_2(a_{13}+a_{22},a_{23}+a_{32},a_{12}+a_{33}) \\
A_3 &=& M_3(a_{21}+a_{33},a_{13}+a_{31},a_{11}+a_{23})
\end{array}
$$
So we deduce that ${\cal M}_{3,3}({\mathbb R})$ can be decomposed 
 as $V_1 \oplus V_2 \oplus V_3$. Now, consider the nine matrices 
$$
 M_k(1,0,0), M_k(0,1,0), M_k(0,0,1) \ (1\leq k \leq 3)
 $$
These are all in $SO(3)$ (with the additional luxury that all coefficients
 are equal to $0,1$ or $-1$), and by what has just been shown, they form
 a basis of ${\cal M}_{3,3}({\mathbb R})$ as wished.
A: This is a complement to my other answer to this question.
 The other answer gives an explicit basis when $n=3$.
 This answer shows the generation property (but does not give an explicit basis) when $n\geq 3$.
  I thought it best to post it as a separate answer. 
For any $n\geq 3$, ${\cal M}_{n,n}(\mathbb R)$ is indeed generated by
$SO(n)$, it is in fact already generated by the signed permutation matrices 
whose determinant is $+1$ (let us denote by $Z(n)$ the set of all such 
matrices).
(Recall that a signed permutation matrix is a matrix which has exactly
 one nonzero coefficient in each row and column, and this coefficient
 is always $\pm 1$).
To check this, denote by $W$ the subspace generated by those matrices.
 Denote by $E_{ij}$ the matrix all of those coefficents are zero except
the one in place $(i,j)$, equal to 1 (so $(E_{ij})_{1\leq i,j \leq n}$
is the canonical basis of ${\cal M}_{n,n}(\mathbb R)$).
Any two $E_{i,i}$ are conjugate, and any two $E_{i,j} (i\neq j)$ are 
  conjugate. Since $SO(n)$ (and hence $W$ also) are invariant by 
  conjugation, it will suffice to show that $E_{1,1}$ and $E_{1,2}$ are
  in $W$.
In fact, it will suffice to show that $E_{1,2}\in W$, because once
we know that $E_{1,2}\in W$, we have an element
$T=(t_{ij})\in Z(n)$ with $t_{11}=1,t_{j(j+1)}=\pm 1$ 
(for $2\leq j \leq n-1$),$t_{n2}=\pm 1$. Then $E_{1,1}-T$ is a sum
of precisely $n-1$ matrices of the form $E_{xy}$ with $x\neq y$ ; and we deduce
that $E_{1,1} \in T$ also.
So let us show that $E_{1,2}\in W$.
For any sequence of signs $\varepsilon=
(\varepsilon_1,\varepsilon_2, \ldots, 
\varepsilon_n) \in \lbrace\pm 1\rbrace ^n$, consider the following two 
signed permutation matrices :
$$
A(\varepsilon)=\left(
\begin{array}{ccccccc}
0 & \varepsilon_1 & 0 & 0 & \ldots & 0 & 0 \\
0 & 0 & \varepsilon_2  & 0 & \ldots & 0 & 0 \\
0 & 0 & 0 & \varepsilon_3  & \ldots & 0 & 0\\
\vdots & \vdots & \vdots & \vdots  & \ddots & \vdots & \vdots \\
0 & 0 & 0 & 0  & \ldots &  \varepsilon_{n-2} & 0\\
0 & 0 & 0 & 0  & \ldots & 0 & \varepsilon_{n-1}\\
\varepsilon_n & 0 & 0 & 0  & \ldots & 0 & 0 \\
\end{array}\right) ,
B(\varepsilon)=\left(
\begin{array}{ccccccc}
0 & \varepsilon_1 & 0 & 0 & \ldots & 0 & 0 \\
0 & 0 & \varepsilon_2  & 0 & \ldots & 0 & 0 \\
0 & 0 & 0 & \varepsilon_3  & \ldots & 0 & 0\\
\vdots & \vdots & \vdots & \vdots  & \ddots & \vdots & \vdots \\
0 & 0 & 0 & 0  & \ldots &  \varepsilon_{n-2} & 0\\
\varepsilon_n & 0 & 0 & 0  & \ldots & 0 & 0 \\
0 & 0 & 0 & 0  & \ldots & 0 & \varepsilon_{n-1}\\
\end{array}\right)
$$
so that $B(\varepsilon)$ has been obtained from $A(\varepsilon)$ by swapping
the last two rows. If we put $p=\prod_{k}\varepsilon_k$ , the determinant
of $A(\varepsilon)$ is $(-1)^{n-1}p$, and the determinant of $B$ is
$(-1)^np$. 
When $n$ is odd, we simply have 
$E_{12}=\frac{B(1,-1,-1,\ldots,-1)+B(1,1,1,\ldots,1)}{2}$. 
When $n$ is even, things are slightly more complicated : if we put 
$$C=\bigg\lbrace (\varepsilon_2,\varepsilon_3, \ldots, 
\varepsilon_{n}) \in \lbrace\pm 1\rbrace ^{n-1}
\bigg| \ \prod_{k=2}^{n}\varepsilon_k=(-1)\bigg\rbrace,
$$ 
$$
C'=\bigg\lbrace( 1,\varepsilon_2,\varepsilon_3, \ldots, 
\varepsilon_{n}) \bigg| (\varepsilon_2,\varepsilon_3, \ldots, 
\varepsilon_{n}) \in C \bigg\rbrace
$$
then 
$$
E_{12}=\frac{\displaystyle\sum_{\varepsilon \in C'}A(\varepsilon)}{2^{n-1}}
$$
In both cases, we have written $E_{12}$ as a linear combination
of matrices in $Z(n)$, so we are done.
