Linear algebra, det, isometry

Question

Prove or disprove:

$T\in \mathcal L(V)$ , $T$ is an isometry, and $V$ is an inner product space over $\mathbb{R}$, then $|$det$T$|$=1$.

My answer to this is the statement is right, since if $T$ is an isometry then $T^{-1}$=$T^*$, but the problem is I have some intuition in my mind, but I don't know how to write down the proof. Can someone help me?

Answer 1

If $T$ is an isometry then $T^*T=I$, and also $T^*=T^t$ since $V$ is real. Therefore $$ 1=\det(T^tT)=\det(T^t)\det(T)=\det(T)^2 $$ so $\det(T)=\pm 1$.

Answer 2

Note that $T B(0,1) = B(0,1)$ hence the change of variables theorem gives: $\int_{B(0,1)} 1 dx = \int_{B(0,1)} 1 |\det T| dx$, from which we get $|\det T| = 1$.