Show $S^2 = T$ is positive definite

Question

Let $(V, ⟨, ⟩)$ be a real inner product space. Let a self-adjoint transformation $T : V → V$ be positive definite, that is $⟨Tv,v⟩ > 0$ for all $v \ne 0$. I have shown that all eigenvalues of T are positive. How can I show that there is a positive definite self-adjoint $S:V →V$ with $S^2 =T$?

Answer 1

Hint: try to take the (positive) square root of the eigenvalues.

• What do they represent?
• How can you use them to construct $S$?

Second hint: Try diagonalizing $T$. What do you get?

Answer 2

Consider the Laplace transform \begin{align} \mathscr{L}\{t^{-1/2}\}(s) = \int_0^\infty e^{-st}t^{-1/2}dt&=\sqrt{\pi}s^{-1/2} \\ \frac{s}{\sqrt{\pi}}\int_0^\infty e^{-st}t^{-1/2}dt&=s^{1/2} \\ -\frac{1}{\sqrt{\pi}}\int_0^\infty \frac{d}{dt}(e^{-st}-1)t^{-1/2}dt&=s^{1/2} \\ \frac{1}{\sqrt{\pi}}\int_0^\infty(e^{-st}-1)(-1/2)t^{-3/2}dt&=\sqrt{s} \\ \frac{1}{2\sqrt{\pi}}\int_0^{\infty}(1-e^{-st})t^{-3/2}dt&=\sqrt{s} \end{align} So, one way to define $S=\sqrt{T}$ is by using the $C_0$ semigroup: $$ \frac{1}{2\sqrt{\pi}}\int_0^{\infty}(1-e^{-tT)})t^{-3/2}dt = \sqrt{T} $$ This definition makes it easier to show that $\sqrt{T} \ge 0$.