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

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