Suppose $T$ is a contraction on a Hilbert space $H$ (separable, if you wish). $D_T=(I-T^*T)^{1/2}$ and $D_{T^*}=(I-TT^*)^{1/2}$.
I want to show that $TD_T=D_{T^*}T$. I had done this before using a messy argument and the fact that, for any polynomial $p$ on $\mathbb{C}$, $p(T)D_T^2=D_{T^*}^2p(T)$. It seems like there should be a simpler way, perhaps using the spectral theorem?
Since $T$ is a contraction, $I-T^*T\geq 0$, so the spectral theorem applies to $I-T^*T$ (i.e. the correspondence between $C(\sigma(I-T^*T))$ and $C^*(I-T^*T)$, the smallest norm closed $*$-invariant sub-algebra of $B(H)$ containing $I-T^*T$ and $I$) and $I-TT^*$. Could I then make claims about $Tf(I-T^*T)=f(I-TT^*)T$ for some continuous $f$ (specifically, $f(z)=z^{1/2}$) on, say the intersection of their spectrums? Does it matter that the two don't necessarily commute?
Because $T$ is a contraction, then $\sigma(TT^{\star})\subseteq[0,1]$ and $\sigma(T^{\star}T)\subseteq[0,1]$. Choose a sequence of real polynomials $\{ p_{n}(t)\}_{n=1}^{\infty}$ that converges uniformly to $\sqrt{t}$ on $[0,1]$. Then $p_{n}(I-T^{\star}T)$ converges to $(I-T^{\star}T)^{1/2}$ and $p_{n}(I-TT^{\star})$ converges to $(I-TT^{\star})^{1/2}$ in the topology of $\mathcal{L}(H)$.
Next notice that $T(T^{\star}T)^{n}=(TT^{\star})^{n}T$ holds for $n=0,1,2,3,\cdots$, which yields $$ Tp_{n}(I-T^{\star}T)=p_{n}(I-TT^{\star})T. $$ Therefore, letting $n\rightarrow\infty$ gives $$ T(I-T^{\star}T)^{1/2}=(I-TT^{\star})^{1/2}T. $$ The result extends to more general functions of $TT^{\star}$ and $T^{\star}T$.