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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.

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Andreo
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1 Answers1

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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.

user8675309
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  • Could you please explain what you mean by "to maintain (conjugate) symmetry" and why this is better? – Andreo Jan 21 '20 at 19:58
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    $(\sqrt{B} + \sqrt{A} + \epsilon I)(\sqrt{B} - \sqrt{A} + \epsilon I) = \big(B-A\big)-\sqrt{B}\sqrt{A}+\sqrt{A}\sqrt{B} + 2\epsilon\sqrt{B} +\epsilon^2 I$ which isn't obviously Hermitian and is unpleasant. If you add it's conjugate transpose to it, i.e. $(\sqrt{B} - \sqrt{A} + \epsilon I)(\sqrt{B} + \sqrt{A} + \epsilon I)$ you do get something Hermitian and nice -- the ugly cross terms $\sqrt{B}\sqrt{A}$ and $\sqrt{A}\sqrt{B}$ are gone. Your update to OP suggests you took a different route on invertibility I'll think about. In general (conj) symmetrization is a very useful technique. – user8675309 Jan 21 '20 at 20:09
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    I see, the expression $D = (\sqrt{B} + \sqrt{A} + \epsilon I)(\sqrt{B} - \sqrt{A} + \epsilon I) + (\sqrt{B} - \sqrt{A} + \epsilon I)(\sqrt{B} + \sqrt{A} + \epsilon I)$ is strictly positive and therefore invertible. If we consider quadratic form $(Dx, x)$ we will get zero, contradiction. – Andreo Jan 21 '20 at 20:16
  • Could you describe when to apply conjugate symmetrization technique and how to identify those patterns. – Andreo Jan 21 '20 at 21:26
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    I think it's best to focus on the Real case (with obvious generalization to Hermitian matrices). The idea is that symmetry is nice and simplifies things -- especially many inequalities in many areas of math. E.g. in probability -- "Many messy arguments can be avoided by symmetrization" (Feller vol 2 p.149 2nd ed). In this case: your book's hint seemed ugly so I tried to 'fix' that by symmetrization of some kind. With repeated exposure you get better at recognizing opportunities for this. A nice example -- see (end of) miniature 8: https://kam.mff.cuni.cz/~matousek/stml-53-matousek-1.pdf – user8675309 Jan 21 '20 at 22:42