Linear subspaces of rotation matrices

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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)

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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.