Let $P,Q$ be Hermitian operators in the algebra of linear maps $H\rightarrow H$, with $H$ a Hilbert space. Further $P$ and $Q$ have orthogonal support (they are orthogonal).
How do I show that $||P+Q||_{\text{tr}}=||P||_{\text{tr}}+||Q||_{\text{tr}}$?
What I know:
The trace norm of an operator $P$ is $||P||_{\text{tr}}=\text{tr}(|P|)$, where we can write $|P|=\sqrt{P^\dagger P}=\sum_i |p_i||i\rangle\langle i|$ the spectral decomposition.
Thus $$||P+Q||_{\text{tr}}=\text{tr}(\sum_i |p_i||i\rangle\langle i|\sum_j |q_j||j\rangle\langle j|)=\sum_{i,j}|p_iq_j|\text{tr}(|i\rangle\langle i||j\rangle\langle j|)=\sum_i|p_iq_i|$$ and $$||P||_{\text{tr}}+||Q||_{\text{tr}}=\text{tr}(\sum_i |p_i||i\rangle\langle i|)+\text{tr}(\sum_j |q_j||j\rangle\langle j|)=\sum_i|p_i|+\sum_j|q_j|.$$
These don't seem to be equal. What am I doing wrong?
You have to use the fact that they have orthogonal support. That would mean that you can find an orthonormal basis $\{ p_k \}$ so that $Pp_k = 0$ or $Qp_k=0$ for all $k$. Then also $|P|$ and $|Q|$ have orthogonal support in the same basis, which leads to $$ (|P|+|Q|)^2=|P|^2+|Q|^2=(P+Q)(P+Q)=|P+Q|^2. $$