# Is the linear span of special orthogonal matrices equal to the whole space of $N\times N$ matrices?

(Disclaimer : I know very well that $SO(N)$ has a Lie algebra of dimension $N(N-1)/2$ etc. This absolutely not the point of my question.)

To make my problem more understandable, I start with the example of $SO(2)$. All $SO(2)$ matrices $M$ can be written as ($\theta\in [0,2\pi[$) $$M=\begin{pmatrix}\cos\theta & \sin\theta\\ -\sin\theta&\cos\theta\end{pmatrix}.$$ Using the basis of $2\times2$ real matrices $\sigma_0=\begin{pmatrix}1 & 0\\0&1\end{pmatrix}$, $\sigma_1=\begin{pmatrix}0 & 1\\1&0\end{pmatrix}$, $\sigma_2=\begin{pmatrix}0 & 1\\-1&0\end{pmatrix}$, $\sigma_3=\begin{pmatrix}1 & 0\\0&-1\end{pmatrix}$, one find that

$$M=\cos\theta\;\sigma_0+\sin\theta\;\sigma_2.$$ Clearly, $M$ does not have components along $\sigma_1$ and $\sigma_3$, so the dimension of the smallest linear subspace of $\mathrm{M}_2(\mathbb{R})$ that contains $SO(2)$ is $2$.

How to articulate the reasoning (for the cases $N>2$ in particular) is not completely clear. I guess that we can say that the component along $\sigma_0$ and $\sigma_2$ are independent because $\cos\theta$ and $\sin\theta$ are independent functions (in a functional analysis sense).

Assuming that made sense, we can try to increase $N$. For example, an $SO(3)$ matrix can be written as $$M=\left(\begin{matrix} \cos\varphi\cos\psi - \cos\theta\sin\varphi\sin\psi & -\cos\varphi\sin\psi - \cos\theta\sin\varphi\cos\psi & \sin\varphi\sin\theta\\ \sin\varphi\cos\psi + \cos\theta\cos\varphi\sin\psi & -\sin\varphi\sin\psi + \cos\theta\cos\varphi\cos\psi & -\cos\varphi\sin\theta\\ \sin\psi\sin\theta & \cos\psi\sin\theta & \cos\theta \end{matrix}\right)\,$$ with $(\phi,\psi)\in [0,2\pi[^2$ and $\theta\in [0,\pi[$. Now, if I look at each matrix element one by one, they are all independent in a functional sense.$^*$ Does that mean that the ''dimension of the matrix space'' that $SO(3)$ matrices live in is $9$ ? Is there a way to generalize that to arbitrary $N$ ?

In the end, does any of what I wrote above make any sense ?

$^*$ It is slightly more subtle than that for the $\theta$ dependence, because in the end I am interested in doing integral over the Haar measure, which means that one should look at $x=\cos(\theta)\in[-1,1]$. But $x$ and $\sqrt{1-x^2}$ are orthogonal, so all should be fine.

• [I've tried to make a couple of edits to the question (including the title) to make it clearer. If such edits are not in the direction you want, you are welcome to undo them! :) ] Sep 11, 2018 at 10:00
• @Qfwfq : I think you are understanding my question better than me, so that's indeed better :)
Sep 11, 2018 at 10:27
• By the way: Your title says "span of the orthogonal matrices", but your question is really about the span of the special orthogonal matrices. I think that this is what confused Denis. Sep 11, 2018 at 10:43
• @RobertBryant I've modified the title accordingly, thanks.
Sep 11, 2018 at 11:30

Elementary proof. The linear space $E$ spanned by $SO_n$ is the orthogonal of those matrices $M$ such that $\langle M,Q\rangle:={\rm Tr}(MQ)=0$ for every $Q\in SO_n$. Let $M=SR$ be a polar decomposition, where $S\in Sym_n^+$ and $R\in O_n$. This decomposition is unique with $S\in SPD_n$ if $M$ is non-singular, but in general it exists and might be non-unique. If $R\in SO_n$, then ${\rm Tr}(SQ)=0$ for every $Q\in SO_n$ ; chosing $Q=I_n$, we have ${\rm tr}\,S=0$, which implies $S=0_n$. If on the contrary $R\in O_n^-$, we have ${\rm Tr}(SQ)=0$ for every $Q\in O_n^-$. Diagonalize $S$ in an orthogonal basis, the matrix of $D$ eigenvalues satisfies ${\rm Tr}(DQ)=0$ for every $Q\in O_n^-$. Chosing $Q$ the symmetry with respect to hyperplane $x_j=0$, we obtain ${\rm Tr}\,D=2d_j$ for every $j$. There follows $n\,{\rm Tr}\,D=2\,{\rm Tr}\,D$. Whence ${\rm Tr}\,D=0$, $d_j=0$ if $n\ge3$. This yields $M=0_n$, hence $E$ is the full space $M_n$.

When $n=2$, this gives only $d_1=d_2$ and we recover $M\in O_2^-$.

• This doesn't explain why it doesn't work for $n=2$. Sep 11, 2018 at 10:12
• @RobertBryant. You are unfair: when I posted my answer, the title was about the orthogonal group, which does span $M_n$, even when $n=2$. Sep 11, 2018 at 12:12
• To make this constructive: is there a deep reason why $SO(2)$ is different from the other $N$, and why it is not the case of $O(N)$ ?
Sep 11, 2018 at 12:50
• @Dennis Serre I did not know this description of the orthogonal group as the extremal set of the unit sphere in $M_n(\mathbb{R})$. Can you indicate a reference? Thanks you. Sep 11, 2018 at 13:09
• @LiviuNicolaescu this is a classical result. See Exercise 108 of my list: perso.ens-lyon.fr/serre/DPF/exobis.pdf . Mind that my first name is Denis. Sep 11, 2018 at 14:16

For $n>2$, the span of the matrices in $\mathrm{SO}(n)$ is the full space $M_n(\mathbb{R})$ of $n$-by-$n$ matrices with real entries.

One proof is using representation theory: If we let $S\subset M_n(\mathbb{R})$ denote the span of $\mathrm{SO}(n)$, then notice that this span is invariant under the action of $\mathrm{SO}(n)\times\mathrm{SO}(n)$ defined by $$(A,B)\cdot C = ACB^{-1}.$$ Thus, it suffices to show that this representation of $\mathrm{SO}(n)\times\mathrm{SO}(n)$ on $M_n(\mathbb{R})$ is irreducible, since $S$ is clearly not the zero subspace.

When $n>2$, the center of $\mathrm{SO}(n)$ is discrete (it is either the identity (when $n$ is odd) or $\pm$ the identity (when $n$ is even), and the (irreducible) representation of $\mathrm{SO}(n)$ on $\mathbb{R^n}$ has commuting ring isomorphic to $\mathbb{R}$. Since the above representation of $\mathrm{SO}(n)\times\mathrm{SO}(n)$ on $M_n(\mathbb{R})$ is clearly the tensor product of the irreducible $\mathbb{R}^n$-representations of the two $\mathrm{SO}(m)$-factors of $\mathrm{SO}(n)\times\mathrm{SO}(n)$, it is irreducible.

• Could you please expand the answer and/or give a reference ? For example, it cannot be the full space, since each matrix element is between -1 and 1.
Sep 11, 2018 at 9:54
• I just did that. You asked for the span of the matrices in $\mathrm{SO}(n)$, in particular, that includes all multiples of any element of $\mathrm{SO}(n)$, so there will be elements of the span with arbitrarily large absolute values. Sep 11, 2018 at 9:58
• Fair enough. Unfortunately, I don't really understand the proof (e.g. why do you use this action O(n)xO(n) ? Why is it sufficient that it is irreducible ?) Is there a more pedestrian way (maybe in the spirit of what I wrote in my question) to do this ? It might help my physicist's mind to understand... :)
Sep 11, 2018 at 10:32
• This is surely a valid proof, but it’s clear from the phrasing of the question that OP does not have the background to follow it. You could consider (at the very least) stating the theorems you use in the various steps. Then possibly recommend to OP a source? Sep 13, 2018 at 7:02
• On the other hand, I think that the site math.stackexchange was more appropriate for this question. See mathoverflow.net/tour and the closely related question math.stackexchange.com/a/2301105/25917
– BS.
Sep 16, 2018 at 13:13

Let $S$ be the span of $SO(N)$ .

Then it's obvious that if $A \in S$ and $D_1, D_2 \in SO(N)$ then $D_1^{-1} A D_2 \in S$ .

Therefore it's enough to show that $B := diag(1,0,...0) \in S$ .

If $N$ is odd and $> 1$ then we can write $$B = \frac{1}{2} (\left[\begin{matrix}1&0\cr0&I_{N-1}\end{matrix}\right] + \left[\begin{matrix}1&0\cr0&-I_{N-1}\end{matrix}\right])$$ .

If $N$ is even and $> 2$ then we can write $$B = \frac{1}{4} (\left[\begin{matrix}1&0&0\cr0&1&0\cr0&0&I_{N-2}\end{matrix}\right] + \left[\begin{matrix}1&0&0\cr0&1&0\cr0&0&-I_{N-2}\end{matrix}\right] + \left[\begin{matrix}1&0&0\cr0&-1&0\cr0&0&E\end{matrix}\right] + \left[\begin{matrix}1&0&0\cr0&-1&0\cr0&0&-E\end{matrix}\right])$$ ,

where we choose $E \in O(N-2)$ with $det E = -1$ .

Clearly the span of $SO(n)$ contains the tangent space of $SO(n)$, the antisymmetric matrices. But it also contains the rotations by $\pi$ in any 2-plane, so contains the diagonal matrices with even numbers of $-1$ entries and all other entries $1$. Linear combinations of such easily contain the diagonal matrices. All symmetric matrices are diagonalizable by rotation matrices. So we get all symmetric matrices as well. Hence all $n \times n$ matrices.

Here is an elementary proof but less elegant than the one indicated by @Robert Bryant. $\newcommand{\bR}{\mathbb{R}}$ $\DeclareMathOperator{\Mat}{Mat}$ $\DeclareMathOperator{\SO}{SO}$

Denote by $\Mat_n(\bR)$ the vector space of $n\times n$ matrices with real entries. Then $\SO(n)\subset \Mat_n(\bR)$ and I assume that you are asking what is the dimension of the smallest vector subspace of $\Mat_n(\bR)$ that contains $\SO(n)$. Denote by $V_n$ this subspace. We want to show that $V_n=\Mat_n(\bR)$ if $n\geq 3$.

For $i<j<k$ let $\Mat_n^{i,j,k}(\bR)$ be the subspace of $\Mat_n(\bR)$ consisting of matrices $A$ such that $$a_{pq}=0,\;\;\forall p,q\not\in\{i,j, k\}.$$

Note that the span of the union of the subspaces $\Mat_n^{i,j, k}(\bR)$ is $\Mat_n(\bR)$.

We claim that $\Mat_n^{i,j, k}(\bR)\subset V_n$. For simplicity we assume $i=1, j=2, k=3$. $\newcommand{\be}{\boldsymbol{e}}$

For $R\in \SO(3)$ let $A_R\in\SO(n)$ be the orthogonal transformation defined by $\newcommand{\bone}{\boldsymbol{1}}$ $A_R=R\oplus\bone$

Then $A(R,S):=A_R-A_S\in \Mat^{1,2,3}_n(\bR)$, $\forall R,S\in\SO(3)$.

For any skew-symmetric $3\times 3$ matrix $X$ we have $$A(e^{tX},1)\in \Mat_n^{1,2,3}.$$ Thus

$$Xe^{tX}\oplus 0=\frac{d}{dt} A(e^{tx},1)\in\Mat^{1,2,3}_n.$$ We deduce that

$$X\oplus 0=\frac{d}{dt}\Big\vert_{t=0}A(e^{tX},1)\in\Mat_n^{1,2,3}.$$

Thus $V_n$ contain all the skew-symmetric matrices in $\Mat^{1,2,3}_n$. Similarly

$$X^2\oplus 0=\frac{d^2}{dt^2}\Big\vert_{t=0}A(e^{tX},1)\in\Mat_n^{1,2,3}.$$ Thus $X^2\oplus0\in\Mat_n^{1,2,3}\cap V_n$for any skew-symmetric $3\times 3$ matrix $X$. $\newcommand{\bu}{\boldsymbol{u}}$

For any orthornormal basis $\bu_1,\bu_2,\bu_3$ of $\bR^3$ there exists a skew-symmetric $3\times 3$ matrix $X=X_{\bu_1,\bu_2}$ such that

$$X^2\bu_1=-\bu_1,\;\;X^2\bu_2=-\bu_2,\;\;X^2\bu_3=0.$$

Now consider the matrix

$$Y_{\bu_1}=-\frac{1}{2}\Big( X_{\bu_1,\bu_2}-X_{\bu_2,\bu_3}+X_{\bu_1,\bu_3}\Big).$$

Note that

$$Y_{\bu_1}\bu_1=\bu_1,\;\;Y_{\bu_1}\bu_2=Y_{\bu_1}\bu_3=0.$$

Define $Y_{\bu_2}$ and $Y_{\bu_3}$ in a similar fashion. For any $t_1,t_2,t_3\in\bR$ we have

$$(t_1Y_{\bu_1}+t_2Y_{\bu_2}+t_3Y_{\bu_3})\bu_i=t_i\bu_i.$$

Since any symmetric operator on $\bR^3$ is diagonalizable in some orthonormal basis we deduce that $V_n$ contains all the symmetric matrices in $\Mat_n^{1,2,3}$. Thus $\Mat^{1,2,3}_n\subset V_n$.

• Thanks, I'm going through the proof, might have some questions. I think there are two typos : line 6 : $M^{i,j}\to M^{i,j,k}$ and line 10: $T_R\to A_R$
• Ok, I think I get it. The only implicit fact is that $A(R,S)$ (and its derivatives) are in $V_n$. PS: $X_{u_i}\to Y_{u_i}$
It is sufficient to prove the statement in the question only for matrices in the form $_{1\times 1}\oplus _{k}$, that is the rank-1 projections or $$\begin{pmatrix} 0&\lambda\\ -\lambda&0\end{pmatrix}\oplus[ 0]_k\qquad (*)$$ Assuming that the proof for such matrices is straithforward then the proof is completed.
Because every matrix is a sum of a symmetric and an anti symmetric matrix. Every symmetric matrix is diagonalizable via an orthonormal matrix and every anti symmetric matrix is orthonrmally equivalent to a direct sum of matrices in the form of $(*)$.
This would give us an elementary but less elegant proof than exisiting answers, in particular the answer based on represention theory, tangent space of $SO(n)$ and extermum points of the unit ball with operator norm.
This argument doesn't work for real orthogonal matrices as far as I can see, but it works for unitary matrices in the complex case, and is somewhat related to Robert Bryant's answer for the real case. It has appeared from time to time in the finite group representaton theory literature, though I'm not sure who first discovered it. For each integer $n > 1,$ there is a nilpotent group $H$ of order $n^{3}$ consisting of monomial matrices ( ie with one non-zero entry in each row and column) with non-zero entries $n$-th roots of unity, such that every non-scalar matrix in $H$ has trace zero (and the scalars in $H$ form the subgroup $Z(H)$ of order $n$). Furthermore, all non-scalar matrices in $H$ have trace zero. Let $S$ be a transversal to $Z(H)$ in $H$. Then ${\rm trace}(A\overline{B}^{T}) = 0$ whenever $A,B$ are distinct elements of $S$ (note that $H$ consists of unitary matrices). Hence $S$ is a set of $n^{2}$ mutually orthogonal non-zero matrices in ${\rm M}_{n}(\mathbb{C}),$ with respect to the inner-product $\langle X,Y \rangle = {\rm trace}(X\overline{Y}^{T}).$ Thus $S$ forms a basis of $M_{n}(\mathbb{C}),$ consisting of unitary matrices (of determinant $\pm 1).$