A non-negative operator has a unique non-negative square root

By [L1], there is an orthonormal basis $(e_1,\dots ,e_n)$ of $V$ and numbers ${\lambda }_1,\dots ,{\lambda }_n\ge 0$ such that $T{e}_j={\lambda }_j{e}_j$. By [L2], each ${\lambda }_j$ has a nonnegative square root. Define $R$ by $R{e}_j=\sqrt{{\lambda }_j}{e}_j$. Then $R^2{e}_j={\lambda }_j{e}_j=T{e}_j$ for every $j$, so $R^2=T$; the same basis shows that $R$ is self-adjoint with nonnegative eigenvalues, hence non-negative by [L1].

Let $Q$ be another non-negative square root of $T$. Because $Q^2=T$, each eigenspace $E_{\lambda }(T)$ is $Q$-invariant. On $E_{\lambda }(T)$ one has $Q^2=\lambda I$, and [L1] applied to the non-negative operator $Q{\mathrm{\mid }}_{E_{\lambda }(T)}$ shows that all its eigenvalues are nonnegative; by [L2], the only nonnegative number whose square is $\lambda$ is $\sqrt{\lambda }$, so $Q=\sqrt{\lambda }I$ on $E_{\lambda }(T)$. Thus $Q$ acts on every basis vector $e_j$ exactly as $R$ does, and therefore $Q=R$.