# Why is square root of hermitian operator on hilbert space unique

How do I prove that the square root of a Hermitian positive definite operator on a Hilbert space is unique?

I'm specifically interested in infinite dimensional Hilbert spaces.

Also, I'm working through a proof of the spectral theorem, so any answers citing the spectral theorem are not helpful.

To answer this in an algebraic way we have to consider one possible construction of the square root operator as for example described here. The following argument also is taken from Section 104 of the book "Functional Analysis" by Riesz & Nagy (1990, Dover publications).

Let $T,T_1,T_2\in\mathcal B^+(\mathcal H)$ (positive semi-definite bounded operators on any Hilbert space $\mathcal H$) be given such that $T_1^2=T_2^2=T$ and $T_1$ comes from the construction I linked above. We now have to show $T_1=T_2$.

First observe that $T_2$ commutes with $T$ since $T_2T=T_2^3=TT_2$ so $T_2$ commutes with polynomials in $T$ as well as their limits, so in particular $T_1T_2=T_2T_1$ by the construction of $T_1$. This gives us the very important property

$$(T_1+T_2)(T_1-T_2)=T_1^2+T_1T_2-T_2T_1-T_2^2=T_1^2-T_2^2=T-T=0.$$

Now let any $x\in\mathcal H$ be given. Define $y:=(T_1-T_2)x$ and consider

$$\langle y,T_1y\rangle+\langle y,T_2y\rangle=\langle y,(T_1+T_2)(T_1-T_2)x\rangle=0$$

as stated above. This implies

$$T_1,T_2\geq 0\Rightarrow\langle y,T_1y\rangle,\langle y,T_2y\rangle\geq 0\Rightarrow \langle y,T_1y\rangle=\langle y,T_2y\rangle=0.$$

Since $T_1,T_2$ are positive semi-definite that means $T_1y=T_2y=0$. Note that in the answer just linked one only needs the existence of a square root which is secured by $T_1,T_2\geq 0$ - but not its uniqueness (which is important as we're trying to prove exactly that). Now this yields

$$\Vert (T_1-T_2)x\Vert^2=\langle x,(T_1-T_2)^2x\rangle=\langle x,(T_1-T_2)y\rangle=0$$

so $T_1x=T_2x$. As $x\in\mathcal H$ was chosen arbitrarily this concludes the proof of the square-root operator being unique.