Dimension of subspace of a vector space.

Question

Consider $W_{B}$={$B^{t}AB:A\in M_{n}(\mathbb{R})$}, Where $B$ is orthogonal matrix and $B^{t}$ denote the transpose of $B.$ Then $W_{B}=M_{n}(\mathbb{R})$ and dim$W_{B}$=rank($B$)rank($B^{t}$).Please help me to solve. Thanks in adbance.

Answer 1

Hint: Find a vector space isomorphism $f:M_n(\mathbb{R})\rightarrow W_B$; what about $A\mapsto B^tAB$ ? Since $B$ and $B^t$ are invertible, we have $rank(B)=rank(B^t)=n$. Because of $W_B\subseteq M_n(\mathbb{R})$ and $\dim W_B=\dim M_n(\mathbb{R})=n^2$, we obtain equality, $W_B=M_n(\mathbb{R})$.