An apparently new 4-by-4 counterexample.

An apparently new 4-by-4 counterexample:
https://mathoverflow.net/questions/508592/product-of-a-matrix-with-the-centralizer-of-another-one

Consider the following MathOverflow conjecture:

If A and B are invertible complex n-by-n matrices that are not simultaneously triangularizable, then C(A)B contains a diagonalizable matrix with n distinct eigenvalues.

Here C(A) denotes the set of invertible matrices that commute with A.

Take

A =

[1  1  0  0]
[0  1  0  0]
[0  0  2  1]
[0  0  0  2]

and

B =

[ 0  -1   0   0]
[-1   1   1   1]
[ 0   0  -1   0]
[ 0   0   0  -1].

Both matrices are invertible, since det(A) = 4 and det(B) = -1.

They are not simultaneously triangularizable.

The eigenspaces of A are

ker(A - I) = span(e1)

and

ker(A - 2I) = span(e3).

Therefore, every eigenvector of A is a multiple of either e1 or e3. However,

B e1 = -e2

and

B e3 = e2 - e3.

Thus neither e1 nor e3 is an eigenvector of B. Consequently, A and B have no common eigenvector.

If two complex matrices are simultaneously upper triangularizable, then they must have a common eigenvector. Therefore, A and B are not simultaneously triangularizable.

Now consider the centralizer of A.

Because A consists of two 2-by-2 Jordan blocks having distinct eigenvalues, every invertible matrix X commuting with A has the form

X =

[x  y  0  0]
[0  x  0  0]
[0  0  z  w]
[0  0  0  z],

where x and z are nonzero.

Multiplying X by B gives

XB =

[-y  -x+y   y    y]
[-x    x     x    x]
[ 0    0    -z   -w]
[ 0    0     0   -z].

The characteristic polynomial of XB is

chi_XB(lambda)
= (lambda + z)^2
(lambda^2 + (y - x)lambda - x^2).

Therefore, -z is an eigenvalue of XB with algebraic multiplicity at least 2 for every X in C(A).

It follows that no matrix in C(A)B can have four distinct eigenvalues.

Hence A and B are not simultaneously triangularizable, but C(A)B contains no matrix with four distinct eigenvalues.

This gives a counterexample to the conjecture in dimension 4.