Eigenvalues of sums of almost commuting Hermitian matrices I am considering $n \times n$ Hermitian matrices $A, B,$ and $C$ such that $A + B = C$ and with eigenvalues $a_{i}$, $b_{i}$, $c_{i}$ ordered so that $a_{1} \geq a_{2} \geq \cdots \geq a_{n}$ etc... I think (from the spectral theorem) that if $A$ and $B$ commute then they are simultaneously diagonalisable and that each eigenvalue of $C$ should equal to a sum of eigenvalues of the summands, in particular $c_{1} = a_{1} + b_{1}$ and $c_{n} = a_{n} + b_{n}$. From Weyl's inequality I also have that for $c_{1} \leq a_{1} + b_{1}$ and $c_{n} \geq a_{n} + b_{n}$ (with equality in the case that $A$ and $B$ commute as mentioned above). I've been reading a bit about "almost commuting" Hermitian matrices (in the sense that the (Frobenius) norm of the commutator of $A$ and $B$ is small) and have seen that if $A$ and $B$ almost commute than they are "almost simultaneously diagonalisable". Does this mean that if $A$ and $B$ almost commute then $c_{1} \approx a_{1} + b_{1}$ and $c_{n} \approx a_{n} + b_{n}$. Could anyone point me towards more information on this, and potentially if there are any bounds on the error of approximating the maximum/minimum eigenvalues of a sum of almost-commuting Hermitian matrices ($C$) as the sum of the maximum/minimum eigenvalues of the two matrices ($A$ and $B$), as described above.
 A: I think that other assumptions are needed before progress can be made. My concern is the following example. Take symmetric $n$-by-$n$ matrices $A$ and $B$ that are far from commuting in the relative sense, i.e.,
$$ \frac{\|AB-BA\|_F}{\|A\|_F \|B\|_F} $$
cannot be regarded as small. We now build slightly larger matrices $A'$ and $B'$ through the simple formulas
$$ A' = \begin{bmatrix} t &  \\  & A \end{bmatrix}, \quad B' = \begin{bmatrix} t &  \\  & B \end{bmatrix}$$
Here $t$ is a just a real number, so we have merely added a row and a column to $A$ and $B$. However, as $t$ tends to infinity, $A'$ and $B'$ will eventually be accepted as commuting in the relative sense, because $$\|A'B' - B'A'\|_F = \|AB-BA\|_F$$ and $$\|A'\|_F^2 = t^2 + \|A\|_F^2, \quad \|B'\|_F^2 = t^2 + \|B\|_F^2.$$ However, while the largest eigenvalue of $C' = A'+B'$ will equal 2t for $t$ sufficiently large, the remaining eigenvalues of $C'$ will eventually equal the eigenvalues of $A + B$. Since $A$ and $B$ are far from commuting, I think that additional assumptions are needed before progress can be made.
