If I have a linear image of the room where v1 and v2 is an image of theirselves and v3 is an image of the null vector.
If that gives me the matrix A=(a, b, c; d, e, f; g, h, i;) then A^n = A because you've already applied a vector to the plane.
However, if v3 is an reflection (image of -v3). If this gives me the matrice B=(a, b, c; d, e, f; g, h, i;), what would B^n result?
My first thought was that it'd be the same as the first (A) matrix, but it seemed to be wrong.
Why is it not the same as the first matrix and what would the result be?
Sorry if my translation of mathematics terms is incorrect.
If $A$ is a reflector along the subspace spanned by $v_3$ (or with respect to the subspace spanned by $v_1$ and $v_2$), i.e.,
$$Av_1 = v_1, \quad Av_2 = v_2, \quad Av_3 = -v_3,$$
for some linearly independent vectors $v_1$, $v_2$, and $v_3$, then
$$A^n = \begin{cases} A, & \text{$n$ is odd}, \\ I, & \text{$n$ is even}. \end{cases}$$
If $A$ is a projector on a subspace spanned by $v_1$ and $v_2$, i.e.,
$$Av_1 = v_1, \quad Av_2 = v_2, \quad Av_3 = 0,$$
for some linearly independent vectors $v_1$, $v_2$, and $v_3$, then $A^n = A$ for all $n \ge 1$.
If I misunderstood your question, please edit it with further explanations about what you want to know.