Proposition 35.2.3.label Let $H$ be a complex Hilbert space and $T \in B(H)$, then $T$ is a partial isometry if and only if $T^{*}T$ is a projection.
Proof. ($\Rightarrow$): Suppose that $T$ is a partial isometry. Let $x \in \ker(T)^{\perp}$, then $\dpn{Tx, Tx}{H}= \norm{x}_{H}^{2}$ and $\dpn{T^*Tx, x}{H}= \norm{x}_{H}^{2}$. By polarisation, for each $x, y \in \ker(T)^{\perp}$,
\begin{align*}\dpn{T^*Tx, y}{H}&= \frac{1}{4}\sum_{k = 0}^{3} i^{k} \dpn{T^*T(x + i^ky), x + i^ky}{H}\\&= \frac{1}{4}\sum_{k = 0}^{3} i^{k} \dpn{x + i^ky, x + i^ky}{H}= \dpn{x, y}{H}\end{align*}
Therefore $T^{*}T$ is idempotent. As $T^{*}T$ is self-adjoint, it is a projection.
($\Leftarrow$): Suppose that $T^{*}T$ is a projection, then for each $x \in \ker(T)^{\perp}$, $\dpn{Tx, Tx}{H}= \dpn{T^*Tx, x}{H}= \norm{x}_{H}^{2}$.$\square$
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