On proving Ker $(T^*T)$=Ker $((T^*T)^\frac{1}{2})$

Question

Can we prove Ker $(T^*T)$=Ker $((T^*T)^\frac{1}{2}))$ without using functional calculus? One inclusion Ker $((T^*T)^\frac{1}{2}))\subset$ Ker $(T^*T)$ is clear. For the other inclusion I want to prove without using functional calculus.

Answer 1

Let $S=T^{*}T$. Then $\|S^{1/2}x\|^{2}= \langle S^{1/2}x, s^{1/2} x \rangle = \langle Sx, x \rangle$ so $Sx=0$ implies $S^{1/2}x=0$