prove $\dim(\operatorname{range}(T)) = \dim(\operatorname{range}(\sqrt{T^*T}))$

Question

I'm a student and I'm studying linear algebra. in Polar Decomposition we have:

for a linear operator $T$, there exist a linear isometry $S$ that: $$ T =S\sqrt{T^*T}$$

so if $S$ is a linear transformation then it must be $\dim(\operatorname{range}(T)) \leq \dim(\operatorname{range}(\sqrt{T^*T}))$.

But why?

edit: I think it must be equal, I mean:

$$\dim(\operatorname{range}(T)) = \dim(\operatorname{range}(\sqrt{T^*T}))$$

I know that $\dim(\operatorname{range}(T)) = \dim(\operatorname{range}(T^*))$ but because of the square root I cannot prove that $\dim(\operatorname{range}(T)) = \dim(\operatorname{range}(\sqrt{T^*T}))$.

Answer 1

As @egreg pointed out, for any linear maps $S,T$, the rank of $ST$ is always less than or equal to that of $T$. To see this, note that by dimension theorem $$\dim \text{ran} L =\dim \operatorname{dom} L-\dim\ker L\le\dim \operatorname{dom} L$$ for any linear map $L$. Now, the image of $ST$ can be seen as the image of $$ S\big|_{\text{ran} T} :\text{ran} T\to V, $$ we can see that $\dim \text{ran} (ST)\le \dim \text{ran} T$ as wanted. Moreover, equality holds if $\dim \ker S\big|_{\text{ran}T}=\dim [\ker S\cap\text{ran}T]=0$. (Here, $V$ denotes the vector space where $S,T$ are defined.)

To see that $\dim \text{ran}T =\dim \text{ran}(T^*T)$, note that $Tx = 0$ if and only if $T^*Tx =0$, which is saying that $\ker T = \ker (T^*T)$. Now, by dimension theorem, we have $$ \dim \text{ran} T = \dim V - \dim \ker T = \dim V - \dim \ker (T^*T) = \dim\text{ran}(T^*T). $$ (Or we can use the fact that $S$ is an isometry.)

Answer 2

If $T\colon U\to V$ and $S\colon V\to W$ are linear maps between finite dimensional vector spaces, then $$\DeclareMathOperator{\range}{range} \dim\range{ST}\le\dim\range(T) $$ This follows from the rank-nullity theorem: \begin{align} \dim U&=\dim\range(T)+\dim\ker(T) \\ \dim U&=\dim\range(ST)+\dim\ker(ST) \end{align} Therefore $$ \dim\range(ST)=\dim\range(T)+\dim\ker(T)-\dim\ker(ST)\le\dim\range(T) $$ because from $\ker(T)\subseteq\ker(ST)$ we have $\dim\ker(T)\le\dim\ker(ST)$.

In your case you can conclude that $$ \dim\range(T)\le\dim\range(\sqrt{T^*T}) $$ On the other hand, $S$ is an isometry, so it is invertible and $$ \sqrt{T^*T}=S^{-1}T $$ The same argument as before implies $$ \dim\range(\sqrt{T^*T})\le\dim\range(T) $$