Show that $O_{3}(\mathbb{R})$ spans $\mathcal{M}_3(\mathbb{R})$.

Question

It was given as an assignment problem.

Given a real vector space $Y,$ let $T: \mathcal{M}_3(\mathbb{R})\rightarrow Y$ be a linear operator such that $T$ restricted to $O_{3}(\mathbb{R}),$ the set of orthogonal matrices, is a constant mapping. Show that $T \equiv 0 .$ Can the result be extended to dimension $n \neq 3 ?$

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My attempt:

Suppose the constant is c then, T(I)=c and T(-I) is c. $\implies$ T(0) = c $\implies$ c is 0.

So T is identically zero on $O_{3}(\mathbb{R})$.

If I show $O_{3}(\mathbb{R})$ spans $\mathcal{M}_3(\mathbb{R})$ then we are done. I don't know where to start, I was trying to use SVD but couldn't succed.

Answer 1

Another proof: we note that the function $\langle A,B \rangle = \operatorname{tr}(AB^T)$ defines an inner-product over $M_n(\Bbb R)$. It suffices to show that relative to this inner product, $O_n(\Bbb R)^\perp = \{0\}$.

Indeed, suppose that $A \in O_n(\Bbb R)^\perp$. The existence of the real polar decomposition ensures that there exists a $U \in O_n(\Bbb R)$ for which $AU^T$ is positive semidefinite. However, since $A \in O_n(\Bbb R)^\perp$, we must have $$ \operatorname{tr}(AU^T) = \langle A,U \rangle = 0. $$ On the other hand, $AU^T$ is positive semidefinite, which means that $\operatorname{tr}(AU^T) \implies AU^T = 0$. Finally, $U$ is invertible, so $AU^T = 0 \implies A = 0$.