= Solution
Apply part (a) to the compact positive operator $T^*T$. Its positive square root $B=|T|$ is defined by multiplying each <eigenvector> of <eigenvalue> $\lambda$ by $\sqrt\lambda$ and vanishing on the kernel. The <eigenvalues> tend to zero, so $B$ is compact. For every $x$, $\|Tx\|^2=\langle T^*Tx,x\rangle=\|Bx\|^2$.
Consequently $U(Bx)=Tx$ is a well-defined isometry on $\operatorname{ran}B$. Extend it to the closure of that range and set it to zero on its orthogonal complement. This gives a <partial isometry> with initial space $(\ker T)^\perp$, final space $\overline{\operatorname{ran}T}$, and
$$
\boxed{T=U|T|,\qquad\ker U=\ker T.}
$$
For uniqueness, the positive factor in any such <polar decomposition of a bounded operator> satisfies $B^2=T^*T$. The positive-square-root uniqueness proved in 4(b) fixes $B$. The relation $U(Bx)=Tx$ then fixes $U$ on the dense subset $\operatorname{ran}B$ of its initial space and the kernel convention fixes it elsewhere. This proves uniqueness with the canonical initial-space condition. Without that condition the word “unique” would be false: $T=0$ can be written $U\cdot0$ with many different unitary $U$.
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