If $Tv=\mu v$ for some $\mu>0$, then $v\in\ker(T^{1/2})^\perp$

Question

Let

• $V$ be a separable $\mathbb R$-Hilbert space
• $T$ be a bounded, linear, nonnegative and symmetric operator on $V$
• $(v_n)_{n\in\mathbb N}$ be an orthonormal basis of $V$ with $$Tv_n=\mu_nv_n\;\;\;\text{for all }n\in\mathbb N$$ for some $(\mu_n)_{n\in\mathbb N}\subseteq[0,\infty)$

How can we prove that $$v_n\in\ker(T^{1/2})^\perp$$ for all $n\in\mathbb N$ with $\mu_n>0$?

I'm not sure how I need to prove the statement. Obviously, if $n\in\mathbb N$ with $\mu_n>0$, then $$v_n\not\in\ker T\supseteq\ker(T^{1/2})\;,$$ but I don't see how that helps.

Answer 1

Start with $$ \overline{\mathcal{R}(T)}=\mathcal{N}(T)^\perp. $$ As you noted, $\mathcal{N}(T^{1/2})\subseteq\mathcal{N}(T)$. Because $T^{1/2}$ is selfadjoint, $$ \|T^{1/2}x\|^2=(T^{1/2}x,T^{1/2}x)=(Tx,x). $$ Therefore $\mathcal{N}(T)\subseteq\mathcal{N}(T^{1/2})$. So, $$ \overline{\mathcal{R}(T)}=\mathcal{N}(T)^{\perp}=\mathcal{N}(T^{1/2})^{\perp} $$ Every eigenvector $v$ with non-zero eigenvalue $\mu$ is in the range of $\mathcal{R}(T)$ because $v=T(\frac{1}{\mu}v)$. So every such $v$ is orthogonal to $\mathcal{N}(T^{1/2})$.