Square root is operator monotone

Question

This is a fact I've used a lot, but how would one actually prove this statement?

Paraphrased: given two positive operators $X, Y \geq 0$, how can you show that $X^2 \leq Y^2 \Rightarrow X \leq Y$ (or that $X \leq Y \Rightarrow \sqrt X \leq \sqrt Y$, but I feel like the first version would be easier to work with)?

Note: $X$ and $Y$ don't have to commute, so $X^2 - Y^2$ is not necessarily $(X+Y)(X-Y)$.

Answer 1

If $X$ and $Y$ are $n\times n$ positive semidefinite matrices over $\mathbb{C}$ or $\mathbb{R}$, we can prove your statement as follows.

We first assume that $X$ is positive definite. Let $A=YX^{-1}$. Since $A$ is similar to $X^{-1/2}YX^{-1/2}$, all eigenvalues of $A$ are positive. Let $v$ be a unit eigenvector corresponding to $\lambda_\min(A)$. As $X^2\le Y^2$, we have $I \le X^{-1}Y YX^{-1} = A^TA$ and hence $1\le v^TA^T Av=\lambda_\min(A)^2$. Thus all eigenvalues of $X^{-1/2}YX^{-1/2}\sim A$ are bounded below by $1$, i.e. $I\le X^{-1/2}YX^{-1/2}$. Hence $X\le Y$.

Now the positive semidefinite case can be obtained as a limiting case of positive definite cases.

Afternote: In the above proof, the only irreversible step is $I\le A^TA\Rightarrow1\le\lambda_\min(A)$. This is because we have $\sigma_\min(A)\le|\lambda|_\min(A)$ in general, but strict inequality can hold. The irreversibility of this step indicates that, while the square root function is operator monotone in our case, the square function is not. (Thanks for julien for pointing out that.)

Answer 2

Define $Y = X + k$, where $k$ is $\ge 0$. Then $Y^2 = X^2 + 2kX + k^2$

Answer 3

Here is a proof which works more generally for $x,y\ge 0$ in a $C^*$-algebra, where the spectral radius satisfies $r(z)\le \|z\|=\sqrt{\|z^*z\|}$ for every element $z$, and $r(t)=\|t\|$ for every normal element $t$. The main difference with @user1551's argument is that we will use the invertibility of $y$. Other than that, the idea is essentially the same.

If $y$ is not invertible, replace $y$ by $y+\epsilon1$ in the following argument, and then let $\epsilon$ tend to $0$ at the very end.

Now assume that $y$ is invertible and note that $$x^2\le y^2\quad\Rightarrow\quad y^{-1}x^2y^{-1}\le y^{-1}y^2y^{-1}=1\quad\Rightarrow\quad\left\|y^{-1}x^2y^{-1}\right\|\le 1.$$

Since $t=y^{-1/2}xy^{-1/2}$ and $z=xy^{-1}$ are similar, they have the same spectral radius and therefore $$\left\|t\right\|=r(t)=r\left(z\right)\le\|z\|=\sqrt{\|z^*z\|}=\sqrt{\left\|y^{-1}x^2y^{-1}\right\|}\le 1.$$ It follows that $y^{-1/2}xy^{-1/2}\le 1$, whence $x\le y$ as desired.