# 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)

-
 Let me phrase this a little differently. Are the set of generating matrices a generating set for $\text{Mat}_3(\mathbb{R})$? If this isn't true then the answer to your question is yes. – Alex Youcis Jan 12 '12 at 19:39 what is the largest linear subspace of $\mathbb{R}^{3\times3}$ containing $SO(3)$? – yoyo Jan 12 '12 at 20:57 @yoyo Presumably you mean "smallest" since largest would be no help. In that case, it's basically the same issue as what I had said. If the OP can prove that the rotation matrices are contained in some proper subspace of $\text{Mat}_3(\mathbb{R})$ he's done. – Alex Youcis Jan 12 '12 at 21:13 My feeling tells me that it is not possible so the proof might be not as easy as finding a proper subspace. I guess one should try to proof it by contradiction. – Fabian Jan 12 '12 at 22:12

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.

-
 Thanks for the answer. I am not sure whether this really proves that all the rotation matrices can't be spanned by fewer than 9 matrices. I keep on thinking... – Jonas Östlund Jan 13 '12 at 7:27