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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 $(X+Y)(X-Y)$.

Julien
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wemblem
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    Positive operators can be defined on infinite dimensional Hilbert spaces also. You cannot assume that the space is finite dimensional so the approved answer is invalid. – geetha290krm Feb 01 '23 at 23:12

3 Answers3

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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.)

user1551
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    Also, is it always true that if $A\leq B$, then $I\leq A^{-1}B$ for $A\geq 0$,$B\geq 0$ – dineshdileep Apr 04 '13 at 13:16
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    "$I\le A^{-1}B$" does not make any sense if $A^{-1}B$ is not Hermitian. Anyway, $A\le B$ iff $v^TAv \le v^TBv$ for all $v$, which is equiv. to $u^Tu \le u^TA^{-1/2}BA^{-1/2}u$ for all $u=A^{1/2}v$, which is equiv. to $I\le A^{-1/2}BA^{-1/2}$. Since $A^{-1/2}BA^{-1/2}$ is similar to $A^{-1}B$, if $A^{-1}B$ happens to be Hermitian, it is $\ge I$. – user1551 Apr 04 '13 at 17:05
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    And yes, I assumed that $X$ and $Y$ are Hermitian. At least for matrices, this is the standard convention when talking about positive semidefiniteness. – user1551 Apr 04 '13 at 18:08
  • @geetha290krm Of course positive operators can be defined on infinite-dimensional Hilbert spaces, but I have stated very clearly in the opening sentence that I only deal with finite-dimensional Hilbert spaces (i.e., when the operators can be represented by PSD matrices) in my answer. So, I don't understand what you are getting at. – user1551 Feb 02 '23 at 00:49
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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.

Julien
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  • Good answer. I am surprised at the voting on this page. Results in FA are generally very easy in the finite dimensional case, so it is this answer that should have received all the upvotes and the approval. – geetha290krm Feb 02 '23 at 09:34
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Define $Y = X + k$, where $k$ is $\ge 0$. Then $Y^2 = X^2 + 2kX + k^2$

Jonathan Rich
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