Does $|Tf|=T|f|$ hold for positive operators?

Question

Let $A$ be a finite dimensional $\mathrm{C}^*$-algebra.

We have the usual absolute value:

$$|a|:=\sqrt{a^*a},$$

and the cone of positive elements:

$$A^+:=\{a\in A:\exists \,b\in A, a=b^*b\}.$$

An operator $T\in B(A)$ is said to be positive if $T(A^+)\subset A^+$.

Does it hold that $|Tf|=T|f|$ for positive $T$?.

Context:

I don't have any reason to believe this but I have a $T$-invariant finite trace $h\in A^{'}$ ($h\circ T=h$) and if I have the above I have

$$\|Tf\|_{\mathcal{L}^1}=h(|Tf|)=h(T|f|)=h(|f|)=\|f\|_{\mathcal{L}^1},$$ which would be useful to me.

Answer 1

In general, we do not have $|Tf | = T |f|$.
To see this, consider the $2 \times 2$-matrices over $\mathbb{C}$ and let $T$ be taking the transpose of a matrix.
Then $T$ is a positive operator, but for $f = \left( \begin{matrix} 1 & 0 \\ 1 & 0\end{matrix} \right)$ we have \begin{align*} |Tf| =\frac{1}{\sqrt{2}}\left( \begin{matrix} 1 & 1 \\ 1 & 1\end{matrix} \right) \neq \left( \begin{matrix} \sqrt{2} & 0 \\ 0 & 0\end{matrix} \right) = T|f|. \end{align*}

Answer 2

As Hetebrij pointed out, it's false. On a positive note, there is a nice (and useful!) characterisation of the linear maps that preserve the absolute value.

Theorem (Gardner 1979): For a linear map $\varphi\colon A \to B$ of C* algebras the following are equivalent

1. $\varphi$ preserves absolute values
2. $\varphi$ is positive and $\varphi(1)\varphi(ab)=\varphi(a)\varphi(b)$ for all $a,b\in A$.
3. $\varphi$ is $2$-positive and disjoint (i.e. if $x,y \geq 0$ and $xy=0$ then $\varphi(x)\varphi(y)=0$).

http://www.ams.org/journals/proc/1979-076-02/S0002-9939-1979-0537087-0/S0002-9939-1979-0537087-0.pdf

With Gardner's Theorem we can prove the following characterisation of the positive unital maps that preserve absolute value.

Proposition. (WW) For a positive unital map $\varphi\colon A \to B$ between von Neumann algebras the following are equivalent:

1. $\varphi$ preserves absolute values
2. $\varphi$ is a $*$-homomorphism
3. $\varphi$ preserves projections and is $2$-positive
4. $\varphi$ preserves ranges and is $2$-positive

( Follows from Proposition 47 of http://scitation.aip.org/content/aip/journal/jmp/57/9/10.1063/1.4961526 )