If $ Av=Bv=\lambda v$, can we conclude that $A=B$?

Question

Let $A $ and $B$ be $2\times2$ matrices with integer entries. Let $v$ be an eigenvector of both $A$ and $B$ with the same eigenvalue $\lambda \neq 0$. So we have $$ Av=Bv=\lambda v. $$

My question is the following:

Under which restrictions can we conclude that $A=B$ ? Or is it always $A=B$ ? Does it depend on the singularity of $A$ and $B$ ?

In other words, if a vector $v$ is an eigenvector with a non-zero eigenvalue for some matrix $A$, is $A$ unique?

Answer 1

Take $A$ to be the identity matrix (i.e. 1 on the diagonal zero otherwise). Then clearly every non-zero vector is an eigenvector with eigenvalue $1$.

In this case, would you say that any matrix $B$ with an eigenvalue $1$ is the identity matrix? (The answer is no, see @ Floris' answer).

---

You can't deduce that $A=B$ if they only have one common eigenvalue (unless you have $1\times 1$ matrices). However if you show that $n\times n$ matrices $A,B$ have $n$ common eigenvalues then you will be able to show that $A=B$ (up to a change of basis).

Answer 2

Counterexample: $$A=\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix},B=\begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}\text{ and }v=\begin{pmatrix}1\\ 0\end{pmatrix}.$$

Answer 3

No. Take $A= \begin{bmatrix} 1 & 0\\ 0 & 1\\ \end{bmatrix}$ and $B= \begin{bmatrix} 1 & 0\\ 0 & 2\\ \end{bmatrix}$. For both $\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}$ is an eigenvector with eigenvalue $\lambda=1$.