$\mathbf{A}, \mathbf{B} \in \mathbb{C}^{n \times n}$ are Hermitian and positive-definite (HPD) matrices. The following conditions are equivalent:
- $\mathbf{A}$ and $\mathbf{B}$ commute.
- $\mathbf{A}$ and $\mathbf{B}$ are simultaneously unitary diagonalizable.
- $\mathbf{A}$ and $\mathbf{B}$ have the same eigenspace.
If either of the above equivalent conditions holds, then
$$\lambda(\mathbf{A}+\mathbf{B})=\lambda(\mathbf{A})+\lambda(\mathbf{B}), \tag{1} $$
where $\lambda(\cdot) = (\lambda_1(\cdot), \dots, \lambda_n(\cdot))$ is the $n$-tuple of the eigenvalues of the corresponding matrix in decreasing order.
My question is the opposite of the above statement: If (1) holds, can we say $\mathbf{A}$ and $\mathbf{B}$ satisfy the above conditions, namely, they are simultaneously unitary diagonalizable?
If this is true, I guess a place to start to prove might be Theorem 7.6.4 of [1] (see also A property of positive definite matrices), but I am not sure how to proceed.
[1] Horn, R. A., Johnson, C. R. (1990). Matrix Analysis.
Best Answer
proof 1: via Golden-Thompson
using the exponential map
$\text{trace}\Big(e^{A+B}\Big)\leq \text{trace}\Big(e^Ae^B\Big)\leq \sum_{k=1}^n e^{\lambda_k^{(A)})}e^{\lambda_k^{(B)}}=\sum_{k=1}^n e^{\lambda_k^{(A)}+\lambda_k^{(B)}}=\text{trace}\Big(e^{A+B}\Big)$
where the inequalities are (i) Golden-Thompson Inequality (ii) von Neumann Trace Inequality. Golden-Thompson is met with equality which occurs iff $AB=BA$.
proof 2: via Triangle Inequality
Induct on $n$. The $n=1$ case is immediate.
Inductive case:
using the linear (dual) representation of the operator 2 norm, where $P, P', P''$ are rank one, trace 1 Hermitian matrices (hence idempotent), we have
$\lambda_1^{(A+B)}= \big \Vert A+ B \big \Vert_2 = \max_P \Big\{\text{trace}\big(P(A+B)\big)\Big\}={\max_P:}\Big\{ \text{trace}\big(PA\big) +\text{trace}\big(PB)\big)\Big\}$ $\leq {\max_P':}\Big\{ \text{trace}\big(P'A\big)\Big\} + {\max_{P''}:}\Big\{\text{trace}\big(P''B)\big)\Big\}= \big \Vert A\Big \Vert_2+ \Big \Vert B \big \Vert_2 =\lambda_1^{(A)} +\lambda_1^{(B)}$
Because 2 choices are better than one; with equality iff we may set $P':=P$ and $P'':=P$. This is met with equality hence $P=\mathbf {xx}^*$ for some length one $\mathbf x$ where $\mathbf x$ is an eigenvector of $(A+B)$ (by maximal characterization) and for $A$ and for $B$.
Now let $Q$ be a unitary matrix with $\mathbf x$ as its first column and consider
$\begin{bmatrix} \lambda_1^{(A)} & \mathbf 0 \\ \mathbf 0 & A' \end{bmatrix}+\begin{bmatrix} \lambda_1^{(B)} & \mathbf 0 \\ \mathbf 0 & B' \end{bmatrix}= Q^*AQ + Q^*BQ=Q^*(A+B)Q= \begin{bmatrix} \lambda_1^{(A+B)} & \mathbf 0 \\ \mathbf 0 & (A'+B') \end{bmatrix}$
and $A'B'=B'A'$ by induction hypothesis