# 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 $f\colon U\to V$ and $g\colon V\to W$ are linear maps, then the rank of $g\circ f$ cannot be greater than the ranks of $g$ and $f$. – egreg Jan 6 at 17:24

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.)

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