Let A, B be invertible square matrices which are commutative.

Question

Let A, B be invertible square matrices which are commutative. That is, AB = BA. Suppose X is an eigenvector of B. Show that AX must also be an eigenvector for B.

Solution:

Let BX = λX for some λ and X ≠ 0.

Apply A to both sides:

ABX = A(λX)

ABX = AλX

ABX = (Aλ)X

I am not really sure what else I can do or if this is the right way to start. Can anyone help?

Answer 1

You have reach

$$ABx = A \lambda x$$

$$(AB)x = \lambda Ax$$

We have $AB=BA$,

$$B(Ax)=\lambda (Ax)$$

Now you just have to argue that $Ax$ is non-zero.

Remark: We do not have $AB=BA=I$ that would have implied that $A$ and $B$ are inverse of each other.