Suppose $A,B$ are positive operators with $AB=0$,what is the norm of $A+B$?

Question

Suppose $A,B$ are positive operators with $AB=0$,what is the norm of $A+B$?

Answer 1

Sorry that I misread the OP's question as confining to the finite dimensional case. Here's finite dimensional approach and general approach.

(Finite dimension approach) Observe that $O=(AB)^*=B^*A^*=BA$, hence $AB=BA =O$. Since $A$ and $B$ commute, they are simultaneously diagonalizable, i.e. there exists a unitary matrix $P$ such that $$ A = PDP^*,\quad B=PEP^* $$ where $D,E$ are positive diagonal matrix. If we write $D=\text{diag}(d_1,d_2,\ldots,d_n)$ and $E=\text{diag}(e_1,e_2,\ldots,e_n)$, then we have $d_ie_i =0$ for every $i$ since $DE=P^*ABP=O$. Also, it holds that $\|A\| =\max_i d_i$ and $\|B\|=\max_i e_i$. This gives $$\max_i (d_i+e_i) = \max\{\max_i d_i,\max_i e_i\} = \max \{\|A\|,\|B\|\},$$ hence $$\|A+B\|=\max_i (d_i+e_i) =\max\{\|A\|,\|B\|\}.$$

General approach: Observe that for every self-adjoint operator $A$, $\|A^2\|=\|A^*A\|=\|A\|^2$ holds. Inductively, we have $\|A^{2^k}\|=\|A\|^{2^k}$ for every $k\ge 1$. Now, assume without loss of generality that $\|A\|\ge \|B\|$, then we have $$ \|(A+B)\|^{2^k}=\|(A+B)^{2^k}\|=\|A^{2^k}+B^{2^k}\|\le \|A^{2^k}\|+\|B^{2^k}\|\le2\|A\|^{2^k}. $$ Thus $$ \|A+B\|\le 2^{1/2^k}\|A\|\xrightarrow[]{k\to\infty} \|A\|. $$ Since $\|A\|\le \|A+B\|$ is obvious from $\displaystyle \|Ax\|\le \|(A+B)x\| $, we get $$\|A+B\|=\|A\|=\max\{\|A\|,\|B\|\}$$ as desired.

Answer 2

Here is another argument. Since we also have $BA=0$ (from $(AB)^*=0$), we have $\ker A\subset\ker B$ and viceversa, and the respective range projections are pairwise orthogonal. So we may choose an orthonormal basis so that $$ A=\begin{bmatrix} A_0&0&0\\ 0&0&0\\ 0&0&0\end{bmatrix} , \ \ B=\begin{bmatrix} 0&0&0\\ 0&B_0&0\\ 0&0&0\end{bmatrix}. $$ Then $$ A+B=\begin{bmatrix} A_0&0&0\\ 0&B_0&0\\ 0&0&0\end{bmatrix}, $$ and we clearly have $$ \|A+B\|=\max\{\|A_0\|,\|B_0\|\}=\max\{\|A\|,\|B\|\}. $$