Apologies - I know there are some similar questions on this site, but I couldn't quite apply what they are saying here.
Let's say I have the following matrices: $$A=\begin{bmatrix}-1&6\\-2&6\end{bmatrix}; B=\begin{bmatrix}1&2\\-1&4\end{bmatrix}$$ I conclude that they are similar (without finding $P$ such that $PAP^{-1} = B$) because both have eigenvalues 2,3 and thus will be similar to $D=\begin{bmatrix}2&0\\0&3\end{bmatrix}$.
If we let $x=\begin{bmatrix}2\\3 \end{bmatrix}$, we have: $$Ax=\begin{bmatrix}16\\14\end{bmatrix}; Bx=\begin{bmatrix}8\\10\end{bmatrix}$$
I was told, however, that if two matrices are similar then they induce the same linear transformation. How can they do this if they map the same vector to a different vector? What is meant by 'same' here?
Thanks for help and sorry if an elementary question.
When you write
$$x = \begin{bmatrix} 2\\3\end{bmatrix}$$
what you are really referencing is the linear combination $2v_1 + 3v_2$ where $v_1, v_2$ is a basis for $\mathbb{R}^2$. Usually, we assume that
$$v_1 = \begin{bmatrix} 1\\0\end{bmatrix}\qquad v_2 = \begin{bmatrix} 0\\1\end{bmatrix}, $$
but we don’t need to. Given a vector $x$ and a basis $\alpha = \{\alpha_1, \alpha_2\}$, if $x = a\alpha_1 + b\alpha_2$, I’ll be explicit and write that
$$[x]_{\alpha} = \begin{bmatrix} a\\b\end{bmatrix}. $$
Your statement, that similar matrices induce the same linear transformation, is true assuming you choose the correct basis. That is, there are bases $\alpha$ and $\beta$ such that given any $x\in\mathbb{R}^2$
$$A[x]_{\alpha} = B[x]_{\beta}.$$