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The polar decomposition of a bounded operator states that every bounded $T:H\to H$ is uniquelywhere $V$ is a partial isometry with initial space $\overline{\mathrm{Ran}|T|}=(\ker T{)}^{\perp }$ and final space $\overline{\mathrm{Ran}T}$. Its initial and final orthogonal projections are $V^{\ast }V$ and $V{V}^{\ast }$.

Existence of $|T|$ was proved in (a). For every $\xi$,Thus the map $|T|\xi \mapsto T\xi$ is well defined and isometric on $\mathrm{Ran}|T|$. It extends uniquely to an isometry between its closed range and $\overline{\mathrm{Ran}T}$. Extend it by zero on $\ker |T|=\ker T$ to obtain $V$. The equation and the projection assertions follow immediately. Any other normalized partial isometry with the same product must agree on the dense range of $|T|$ and vanish on its orthogonal complement, giving uniqueness. The kernel normalization is essential; without it an action on the kernel could be added.

A useful consequence for a Von Neumann algebra $M$ is that the polar factors of $T\in M$ also belong to $M$. The polynomial construction in (a) puts $|T|$ in the norm-closed algebra $M$. For $\epsilon >0$, the inverse of $|T|+\epsilon I$ belongs to $M$ by a convergent Neumann series, andThese operators are contractions, since $\Vert |T|\xi \Vert \le \Vert (|T|+\epsilon I)\xi \Vert$. On vectors $|T|\xi$ the difference from $V$ has norm at most $\epsilon \Vert \xi \Vert$, and on the kernel it is zero. Density and the uniform bound prove the strong limit. Strong closedness of $M$ then gives $V\in M$. This justifies the use of algebra-valued polar factors in Question 7(c).

Apply part (a) to the compact positive operator $T^{\ast }T$. 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 $x$, $\Vert Tx{\Vert }^2=⟨T^{\ast }Tx,x⟩=\Vert Bx{\Vert }^2$.

Consequently $U(Bx)=Tx$ is a well-defined isometry on $\mathrm{ran}B$. 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 $\overline{\mathrm{ran}T}$, andFor uniqueness, the positive factor in any such polar decomposition of a bounded operator satisfies $B^2=T^{\ast }T$. The positive-square-root uniqueness proved in 4(b) fixes $B$. The relation $U(Bx)=Tx$ then fixes $U$ on the dense subset $\mathrm{ran}B$ 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 $U\cdot 0$ with many different unitary $U$.

An element $x$ of a unital C-star algebra is positive when $x=y^{\ast }y$ for some $y$, equivalently when $x=x^{\ast }$ and $\sigma (x)\subseteq [0,\infty )$. The continuous functional calculus for the nonnegative function $t\mapsto \sqrt{t}$ defines a positive element $x^{1/2}$ satisfying $(x^{1/2}{)}^2=x$. If a positive $z$ also satisfies $z^2=x$, functional calculus for $z$ gives $z=\sqrt{z^2}=\sqrt{x}$, proving uniqueness. For a positive operator $T=S^{\ast }S$ on ${\ell }^2$,

For arbitrary $T$, put $|T|=(T^{\ast }T{)}^{1/2}$. Thenso $\ker |T|=\ker T$. Defineon $\mathrm{im}|T|$. The kernel identity makes this well-defined, and the norm identity makes it an isometry. Extend it continuously toand set it equal to zero on $\ker T$. The resulting $U$ is a partial isometry, has $\ker U=\ker T$, and satisfies the polar decomposition of a bounded operator $T=U|T|$.

Finally $U^{\ast }U$ is the orthogonal projection onto $(\ker T{)}^{\perp }$, which contains the range of $|T|$. Therefore