For bounded operators $A,B$ on a Hilbert space $\mathbb{H}$ prove that for any $c \geq 0$, $A^*A \leq cB^*B \Leftrightarrow \lVert A \rVert \leq \sqrt{c}\lVert B \rVert$.

This is an exercise from Wolfgang Sherer's book, Mathematics of Quantum Computing. Here $A^*$ denotes the adjoint operator of $A$, $\lVert A \rVert$ denotes the operator norm and for operators $A, B$ the statement $A \leq B$ is defined as $B-A \geq 0$ (i.e for all $|\psi\rangle \in \mathbb{H}$, $\langle \psi | (B-A)\psi\rangle \geq 0$).

My work so far: Using the definitions $A^*A \leq cB^*B$ is equivalent to $\langle \psi | (cB^*B-A^*A)\psi\rangle \geq 0$ for all $|\psi\rangle \in \mathbb{H}$. Expanding the inner product this is equivalent to $\lVert A\psi\rVert \leq \sqrt{c}\lVert B\psi\rVert$. Now taking supremums over all $|\psi\rangle$ with $\lVert \psi \rVert = 1$ we get the $\Rightarrow$ direction.

My question: How can I proceed with the $\Leftarrow$ direction? In the solutions on the back of the book it states that $\lVert A\psi\rVert \leq \sqrt{c}\lVert B\psi\rVert \Leftrightarrow \lVert A \rVert \leq \sqrt{c}\lVert B \rVert$ citing equation $(2.57)$ which is found later in the book and is just the definition of trace of an operator. I don't think that this is correct though.


1 Answer 1


I believe the converse direction is false.

Here is an explicit example. Consider $M_2(\mathbb{C}) = B(\mathbb{C}^2)$ and $$X= \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}, \quad Y = \begin{pmatrix}1 &0 \\ 0 & 0\end{pmatrix}$$

Then $X^*X = X$ and $Y^*Y = Y.$ We have $$\|X\| = \|Y\| = 1$$ so $\|X\| \le \|Y\|.$ However, $$Y-X = \begin{pmatrix}0 & 0 \\ 0 & -1\end{pmatrix}$$ is not a positive matrix, so $X \not\le Y$.

More generally, let $H$ be any Hilbert space and $X,Y$ be distinct orthogonal projections in $B(H)$. Then $\|X\| = \|Y\| = 1$ and $X \not\leq Y$ or $Y \not\le X$ is true, so this yields large families of counterexamples.

  • $\begingroup$ I see now. Thanks! $\endgroup$ Oct 16, 2021 at 12:02
  • $\begingroup$ @GiorgosGiapitzakis You are welcome! $\endgroup$
    – J. De Ro
    Oct 16, 2021 at 12:02

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