If $0 \le A \le B$ then $\sqrt{A} \le \sqrt{B}$

Question

If $0 \le A \le B$ then $\sqrt{A} \le \sqrt{B}$, where $A, B$ are linear transformations.

Hint: compute $(\sqrt{B} + \sqrt{A} + \epsilon)(\sqrt{B} - \sqrt{A} + \epsilon)$.

I used a hint from the book and proved that expression $C = (\sqrt{B} - \sqrt{A} + \epsilon)$ is invertible for $\epsilon > 0$.

Now I don't know how to show that $C$ is positive?

Once $0 \le C$, we can write $\sqrt{A} \le \sqrt{B} + \epsilon$, for all $\epsilon > 0$, and conclude that $\sqrt{A} \le \sqrt{B}$. Also for me it is not clear if we can apply the "techniques with limits" to the linear transformations.

Update: Proving that $C = (\sqrt{B} - \sqrt{A} + \epsilon)$ is invertible. Let

$$(\sqrt{B} + \sqrt{A} + \epsilon)(\sqrt{B} - \sqrt{A} + \epsilon) = X + i\cdot Y$$

where: $$X = B - A + 2\epsilon\sqrt{B} + \epsilon^2$$

$$Y = -i(\sqrt{A}\sqrt{B} - \sqrt{B}\sqrt{A})$$

Now $X$ is strictly positive for $\epsilon > 0$ and $Y$ is self-adjoint/Hermitian and $X + iY$ is invertible, which follow from previous exercise in the book.

Answer 1

I assume that $A$ and $B$ are n x n Hermitian (equivalently real symmetric). Your post says you've already proven $C := (\sqrt{B} - \sqrt{A} + \epsilon I)$ is invertible so skip to 3

1. A better hint is to try to maintain (conjugate) symmetry, so consider proving $(\sqrt{B} + \sqrt{A} + \epsilon I)(\sqrt{B} - \sqrt{A} + \epsilon I) + (\sqrt{B} - \sqrt{A} + \epsilon I)(\sqrt{B} + \sqrt{A} + \epsilon I) \succeq 2 \epsilon^2 I \succ 0$

2. We know $(\sqrt{B} - \sqrt{A} + \epsilon I) = C$ has a trivial nullspace -- i.e. if there was some $\mathbf x \neq \mathbf 0$ such that $C\mathbf x = \mathbf 0$ consider the quadratic form implications which imply a contradiction with (1).

3. Since you've proven C invertible for any $\epsilon \gt 0$ that is equivalent to proving
   $\det\Big(\big(\sqrt{B} - \sqrt{A}\big)-\lambda I \Big) = (-1)^n \cdot \det\Big(\lambda I - \big(\sqrt{B} - \sqrt{A}\big) \Big) \neq 0$
   since rescalaing by plus or minus 1 doesn't change whether a number is zero, we can just say $p\Big(\lambda \Big) = p\Big(-\epsilon \Big) \neq 0$ for any $\lambda \lt 0$ (where we assign $\lambda:= -\epsilon$)
   and $p$ is the characteristic polynomial of $\big(\sqrt{B} - \sqrt{A}\big)$, which is evidently Hermitian and hence all eigenvalues are real but none can be negative, which proves $\big(\sqrt{B} - \sqrt{A}\big)\succeq \mathbf 0$

4. You mention concerns about 'techniques with limits' -- as shown in 3 that isn't really what's going on here. The positive $\epsilon$ in effect acts as a negative number to evaluate under the image of the characteristic polynomial of $\big(\sqrt{B} - \sqrt{A}\big)$ and we show this negative number can never be a root (eigenvalue). Techniques involving continuity can be used, however, though some care is needed. On the other hand continuity related techniques can often dramatically simplify proofs.