TITLE: If $A, B$ are positive definite and $\|A+B\| = \|A\|+\|B\|$, they have a common eigenvector
QUESTION [6 upvotes]: Suppose $A$ and $B$ are two positive definite matrices such that $\|A+B\| = \|A\|+\|B\|$. Show that $A$ and $B$ have a common eigenvector, where $\|A\|$ is the operator norm of $A$.

I'm wondering if there is a general strategy to solving problems of this form, I've been struggling with the problems for a while and can't get it. Any help is appreciated, Thank you!

REPLY [6 votes]: Suppose $A$ and $B$ are positive operators on a finite-dimensional inner product space $V$ and $\|A + B\| = \|A\| + \|B\|$.
Because $A$ and $B$ are positive, $A+B$ is also a positive operator. Thus there exists $x \in V$ such that$$\|x\| = 1 \quad \text{and} \quad\|A+B\| = \langle (A+B)x, x \rangle.
$$
Now
\begin{align*}
\|A+B\| &= \langle (A+B)x, x \rangle\\[3pt]
&= \langle Ax, x \rangle + \langle Bx, x \rangle\\[3pt]
&\le \|A\| + \|B\|\\[3pt]
& = \|A+B\|.
\end{align*}
Because the first term above equals the last term above, the inequality must be an equality. In other words,
$$
\langle Ax, x \rangle = \|A\| \quad \text{and} \quad  \langle Bx, x \rangle = \|B\|.
$$
The equations above imply that $x$ in an eigenvector of $A$ and $x$ is an eigenvector of $B$. Hence we have shown that $A$ and $B$ have a common eigenvector, as desired.