I am reading Axler's Linear Algebra Done Right and I came across a point in his proof of the Polar Decomposition Theorem that I didn't fully understand. His wording of the theorem is as follows (adapted slightly to explain his notation, which I'm not sure is ubiquitous):
7.45 Polar Decomposition Theorem: Suppose $T$ is a linear operator on a finite-dimensional inner product space $V$. Then there exists an isometry $S$ (also a linear operator on $V$) such that $T=S \sqrt{T^*T}$.
I've attached the first page of the two-page proof here.
It is long, hence why it's a picture. I do not follow his assertion that $dim(range \sqrt{T^*T})=dim (range T)$. To be clear, the Fundamental Theorem of Linear Maps (3.22) alluded to is:
Fundamental Theorem of Linear Maps (3.22): Suppose $V$ is finite-dimensional and $T$ is a linear operator from $V$ to another vector space $W$. Then $rangeT$ is finite-dimensional and $dimV = dim(nullT) + dim(rangeT).$
In particular and in this case, it would seem that we have, using 3.22 and the injectivity of $S_1$, $$dim(range \sqrt{T^*T})=dim(nullS_1)+dim(rangeS_1)=0+dim(rangeS_1)=dim(rangeS_1).$$
Now $rangeS_1$ is necessarily a subset of the codomain into which it is mapping, namely $range T$. But I don't see why we should have that their dimensions are equal as stated by Axler at the bottom of the page provided. That is, how does Axler add on the last equality in the string below?
$$dim(range \sqrt{T^*T})=dim(nullS_1)+dim(rangeS_1)=0+dim(rangeS_1)=dim(rangeS_1)=dim(rangeT).$$
The equality you're asking about isn't just an equality of dimensions, we actually have $\operatorname{range}(S_1) = \operatorname{range}(T)$. This follows from 7.47, because any element of $\operatorname{range}(T)$ will have the form $Tv$ for some $v$, and then $S_1(\sqrt{T^*T}v) = Tv$.