If two invertible matrices agree on a vector, does this imply their determinant agrees as well?

Question

As stated, if we let $A, B \in M_n(\mathbb{R})$ be invertible and there is some $v\in R^n$ such that $$Av = Bv$$ does it follow that $\det(A) = \det(B)$?

Additionally, does this hold if we let $A, B \in M_n(\mathbb{C})$?

Answer 1

No. As a counterexample consider $$A= \left(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) $$

$$ B=\left( \begin{matrix} 2 & 0 \\ 0 & 1 \end{matrix} \right) $$ and $$v=\left( \begin{matrix} 0 \\ 1 \end{matrix} \right) $$

Answer 2

Hint: Consider eigenvectors and eigenvalues and diagonalizable invertible matrices. If $A$ and $B$ have one eigenvector $v$ in common with the same eigenvalue, so that $Av = Bv$, does that mean that all the other eigenvalues are equal, or for that matter, the products of eigenvalues are equal? (which is the determinant)

Answer 3

No. Counter eg:$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \left[ \begin{array}{c} 2 \\ 1 \end{array} \right] = \begin{bmatrix} 0 & 2 \\ 1 & -1 \end{bmatrix} \left[ \begin{array}{c} 2 \\ 1 \end{array} \right]$

Answer 4

Consider the matrices $$A= \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix},B= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

Can you find a vector $v$ such that $Av=Bv$? Do $A$ and $B$ have the same determinant?