# Square root of compact operator

I'm trying to solve a functional analysis problem

A self-adjoint non-negative operator $A$ on a Hilbert space $H$ is compact if and only if its $\sqrt{A}$ is compact.

If $A^{1/2}$ is compact, $A = A^{1/2}A^{1/2}$ is as a composition of a compact and a bounded operator is compact.
If, on the other hand, $A$ is compact, we can write $A = \sum_{n\ge 1} \lambda_n(\cdot, x_n)x_n$ for some $\lambda_n \to 0$ and an orthonormal sequence (the $\lambda_n$ are the eigenvalues of $A$ and $x_n$ are the corresponding eigenvectors. Then $A^{1/2} = \sum_{n\ge 1} \lambda_n^{1/2}(\cdot, x_n)x_n$ is compact as it is the limit (in the norm topology) of the finite dimensional operators $B_N = \sum_{1\le n \le N} \lambda_n^{1/2}(\cdot, x_n)x_n$ since $\|B_N - A^{1/2}\| \le \lambda_N^{1/2} \to 0$.
• But why $A^{1/2}$ equals $\sum_{n\ge 1} \lambda_n^{1/2}(\cdot, x_n)x_n$ – Aram Malkhasyan May 22 '12 at 19:43
• @user31919 Let $B = \sum_{n\ge 1} \lambda_n^{1/2}(\cdot, x_n)x_n$. Then $B$ is self-adjoint and non-negative, moreover for $x\in H$ we have $$B(Bx) = \sum_{n\ge 1} \lambda_n^{1/2}(Bx, x_n)x_n = \sum_{n,m\ge 1} \lambda_n^{1/2}\lambda_m^{1/2}(x, x_m)(x_m, x_n)x_n = \sum_{n\ge 1}\lambda_n (x,x_n)x_n = Ax$$ and so $B = A^{1/2}$. – martini May 22 '12 at 20:28
• So this argument shows also that if A is compact and positive, then it has the same range as $A^{1/2}$, right? – Axesilo Jul 3 '14 at 0:16