Inner product spaces and isometry

Question

In an inner product space $V$ and $T \in Hom(V,V)$ the following conditions are equivalent:-

b) $T^*T=I$ where $T^*$ is adjoint of $T$ and $I$ is identity operator

c) $T$ is an isometry How to prove this

Answer 1

From b) to c) is easier. If $T^* T = I$, then $$\|Tx\|^2 = \langle Tx, Tx \rangle = \langle T^* Tx, x \rangle = \langle x, x\rangle = \|x\|^2,$$ proving $T$ is an isometry.

On the other hand, if $T$ is an isometry, then using the polarisation identity, we see that $T$ also preserves inner products, i.e. $\langle Tx, Ty\rangle = \langle x, y \rangle$ for all $x, y$. If we fix $x$, then we get $$\langle x, y \rangle = \langle Tx, Ty \rangle = \langle T^* Tx, y\rangle \implies \langle T^*T x - x, y\rangle = 0$$ for all $y$. That is, $T^* T x - x$ is a vector orthogonal to every vector. The only such vector is $0$ (consider $y = T^* T x - x$), hence $$T^* T x - x = 0 \implies T^* T x = x,$$ i.e. $T^* T = I$.