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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 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
    
Carl Brannen has not been seen since February so my comment is not likely to be noticed. So, I have started a bounty as I am interested in a general answer. –  Rahul Oct 16 '13 at 20:40

3 Answers 3

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.

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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... –  Rulle Jan 13 '12 at 7:27
    
I'm not sure this works: some of the permutation matrices have determinant $-1$ and so are not rotation matrices. In other words, you have shown that the span of $O(3)$ is $\mathbb R^{3\times3}$, but the span of $SO(3)$ may be smaller. (To help the intuition, observe that $SO(2)$ and $O(2)$ do have different spans.) –  Rahul Oct 16 '13 at 17:15

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.

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+1: It looks like this works. How did you pick this particular construction? It looks quite non-obvious. Also, I'm curious if you have any thoughts on the $n$-dimensional case. –  Rahul Oct 17 '13 at 17:33
    
@RahulNarain : actually, it’s rather natural in some sense. The $M_k$ are all conjugate matrices, where the conjugation comes from a circular permutation of the basis elements. The central fact is that (in general, for any $n \geq 3$ )the signed permutation matrices with det $+1$ span the whole of $M_n$ ; the main difficulty is selecting a sub-family so that it is a basis. One tries to make this sub-family have as many symmetries as possible. –  Ewan Delanoy Oct 17 '13 at 18:08
    
"in general, for any $n\ge3$... the signed permutation matrices with det $+1$ span the whole of" $\mathcal M_{n\times n}$ (I assume you mean)? That's what I wanted to know! Is there a reference for this I can cite in case I need to use it in a paper? –  Rahul Oct 17 '13 at 18:29
    
@RahulNarain the second answer I posted should answer your second query. –  Ewan Delanoy Oct 18 '13 at 13:34

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.

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