Let $H$ and $K$ be Hilbert spaces over the same scalar field, and let $T \in \mathcal{L}(H,K)$ be bounded. Let $T^* \in \mathcal{L}(K,H)$ denote the Hilbert-space adjoint of $T$, and let
be the unique non-negative self-adjoint square root of the non-negative self-adjoint operator $T^*T \in \mathcal{L}(H)$. Then there exists a partial isometry $U \in \mathcal{L}(H,K)$ such that
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\begin{align*}
T = U|T|.
\end{align*}
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Here $U$ being a partial isometry means that there is a closed subspace $M \subset H$ such that $U$ restricts to an isometry $M \to U(M)$ and $U=0$ on $M^\perp$; this subspace $M$ is the initial space, and $U(M)$ is the final space. For the operator constructed below, the initial space is
Moreover, $U$ is uniquely determined on $(\ker T)^\perp$, and under the standard choice it satisfies $U=0$ on $\ker T$. In particular, this applies when $T$ is compact.