Mfrhtyj

Let $A$ and $B$ be $ntimes n$ matrices with complex entries such that
$AB - BA$ is a linear combination of $A$ and $B$.

I'd like to prove that there exists a non-zero vector $v$ that is an eigenvector of both $A$ and $B$.

It is assumed throughout that

The given matrices $A$ and $B$ are linearly independent in the $mathbb C$-vector space $M_n(mathbb C)$,

thus ignoring the unremarkable setup where one matrix is a multiple of the other.

My next guess is that the question doesn't focus upon a vanishing commutator of $A$ and $B$, hence consider
$$0:neq:[A,B]:=:AB-BA:=:alpha A+beta B$$
and at least one of the scalars $alpha,beta $ is not zero. Divide by a scalar, say $alphaneq0$, then set
$$B'=frac1alpha B;text{ and }; A'= A+beta B'quadtext{to obtain };
big[,A',B'big]:=:A',.$$
Thus in the sequel we may start w.l.o.g. from
$,mathbf{AB-BA=Aqquadqquad (star)}$

I decided to spill some more MathJax ink here & now, to give examples and a bit Lie algebra background as well. Who's not willing to go through all this should ...

simply jump: Goto The shared eigenvector !

Examples

how to satisfy the relation above:

• If $,n=2$, then $,A=begin{pmatrix}0&1\ 0&0end{pmatrix}:$ and
  $;B=frac 12begin{pmatrix} -1&0\ 0&1end{pmatrix};$ do the job.
  $:begin{pmatrix}1\ 0end{pmatrix}$ is a joint eigenvector.




• For $n=3$ one may consider
  $$A:=:begin{pmatrix}0&a_2&a_3\ 0&0&0\ 0&0&0end{pmatrix};,quad
  B:=:begin{pmatrix}b&0&0\ 0&b+1&0\ 0&0&b+1end{pmatrix}$$
  where $a_2,a_3,binmathbb C$,and $a_2,a_3$ not both zero.

It is not by accident that $A$ is represented by a nilpotent matrix since this cannot be avoided:

For each $ninmathbb N$ one has
$$big[,A^n,Bbig]:=:n,A^n,$$
the proof by induction using
$,[AA^{n-1},B] = A[A^{n-1},B] + [A,B],A^{n-1}$ (derivation property) is straightforward. Let $|cdot|$ be a submultiplicative matrix norm, then
$$n|A^n|:=:|A^nB-BA^n|;leq;2|B| |A^n|quadforall,n$$
which implies $A^n=0,$ at the latest when $,n>2|B|$. Thus $A$ is nilpotent.

The shared eigenvector

for $B$ and $A$ does exist: Pick an eigenvector $wneq 0$ to an eigenvalue $mu$ of $B$ such that $,mu-1,$ is no eigenvalue of $B$. Insert $w$ into $(star)$
$$BAw:=:ABw - Aw:=:(mu-1)AwquadLongrightarrow; Aw:=:0,,$$
thus $w$ is also an eigenvector of $A$, to the eigenvalue zero.

Lie algebra background
$A$ and $B$ span a complex Lie algebra with Lie bracket given by $(star)$. It is up to (Lie algebra) isomorphism the unique complex Lie algebra, which is 2-dimensional and nonabelian. It is solvable, but not nilpotent.

Lie's theorem guarantees (in any matrix representation) the existence of a shared eigenvector for all elements of a solvable Lie algebra. One implication is that all matrices of a representation may be chosen to be triangular.

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