$\vert A \vert$ is a linear operator when $A: \mathcal{H}\to \mathcal{H}$ is linear and bounded?

Question

Let $A \in \mathcal{B}(\mathcal{H})$. We define $$\vert A \vert := \sqrt{A^*A}$$

Then, I've showed that $\vert A \vert$ is a self-adjoint operator, but the book I'm using also say that, for $a,b \in \mathbb{C}$ and $\psi,\phi \in \mathcal{H}$ we have

$$\tag{1}\vert A \vert (a\psi+b\phi) = a\vert A \vert \psi + b\vert A \vert\phi$$

My question is then: How can I prove that? To me $(1)$ should not be linear, because of the square definition. Can someone give me a hint?

Answer 1

By definition, $|A|$ is the linear positive semidefinite operator with $|A||A|=A^*A$.

Regarding "should not be linear, because of the square definition": If we define $Abs:\mathcal B(\mathcal H)\to\mathcal B(\mathcal H)$ by $Abs(A)=|A|$ then $Abs$ is not linear, perhaps "because of the square definition". That has nothing to do with the fact that $Abs(A)$ is linear.

Answer 2

I think it's beneficial to consider a finite dimensional case for clarification.

Consider the matrix \begin{align} A = \begin{pmatrix} 1 & 0 & \sqrt{-1}\\ 0 & -1 & 0\\ 0 & 0 & -1 \end{pmatrix} \end{align} then we have \begin{align} A^\ast A = \begin{pmatrix} 1 & 0 & \sqrt{-1}\\ 0 & 1 & 0\\ -\sqrt{-1} & 0 & 2 \end{pmatrix} \end{align} which is Hermitian and diagonalizable. In particular, we see that \begin{align} A^\ast A = U \begin{pmatrix} 1 & 0 & 0\\ 0 & \frac{3-\sqrt{5}}{2} & 0\\ 0 & 0 & \frac{3+\sqrt{5}}{2} \end{pmatrix} U^{\ast} \end{align} where $U$ is some unitary matrix.

By definition (or functional calculus), we have that \begin{align} |A|=\sqrt{A^\ast A} := U \begin{pmatrix} 1 & 0 & 0\\ 0 & \sqrt{\frac{3-\sqrt{5}}{2}} & 0\\ 0 & 0 & \sqrt{\frac{3+\sqrt{5}}{2}} \end{pmatrix} U^{\ast} \end{align} which by definition is linear.

Remark: One should note that $A^\ast A$ has only nonnegative eigenvalues which makes defining the square root matrix possible.