Define the inner product: $\langle \cdot $ , $\cdot \rangle$ over real matrices: $\langle A,B \rangle$ = $tr(B^tA)$. Given a matrix Q over the reals, define the linear operator $T_Q$: $T_Q(X)=QX$. Show that if $T_Q$ is an isometry over the real matrices, then Q is an orthogonal matrix.
My attempt:
Consider the matrices X and Y over the reals. Since $T_Q$ is isometry, it follows that: $\langle QX, QY \rangle = \langle X,Y \rangle.$ Hence, by the inner product we have that:
$tr((QY)^{t}QX) = tr(Y^{t}Q^tQX)=tr(Y^tX)$
I cannot go further than here. Having $Q^tQ$ inside the trace is pretty suggestive, but none of trace properties let me conclude that $Q^tQ= I$. Any suggestions?
If $Q$ is orthogonal, $Q^tQ=I$, $tr((QY)^tQX)=tr(Y^t(Q^tQ)X)=tr(Y^tX)$. Thus $T_Q$ is an isometry.
Suppose that $T_Q$ is an isometry, $tr((QY)^tQX)=tr(Y^t(Q^tQ)X)=tr(Y^tX)$ for every $X,Y$ in particular if $Y=I_n$, $tr(Q^tQX)=tr(X)$, this implies that $tr((Q^tQ-I)X)=0$ this is equivalent to saying that $Q^tQ-I_n$ is orthogonal to $M(n,R)$ the space of $n\times n$-real matrices i.e $Q^tQ-I_n=0$.