Meaning of matrix with respect to a basis I have the matrix $\left( \begin{array} {cc} 1 && 1 \\ -1 && -1 \end{array} \right)$. When applied to the bases $\left( \begin{array} {c} 1 \\ -1 \end{array} \right)$ and $\left( \begin{array} {c} 1 \\ 0 \end{array} \right)$ I get a resulting matrix with respect to this basis which is $\left( \begin{array} {cc} 0 && 1 \\ 0 && 0 \end{array} \right)$.
My question is: what does it mean to have a matrix with respect to another basis and why is it significant?
I know that this does not mean the matrices are equivalent, otherwise a simple Gaussian elimination would have done that. 
Furthermore, applying each of the matrices to, say, the vector $\left( \begin{array} {c} 1 \\ 2 \end{array} \right)$ you get different answers. 
So if my resulting matrix is not just a simplified version of the other, what is it exactly?
 A: Suppose that $A$ has the matrix $B$ with respect to the basis $\{v_1,v_2\}$.  What this means is that whenever
$$
A(a v_1 + b v_2) = cv_1 + dv_2
$$
we also have
$$
B \pmatrix{a\\b} = \pmatrix{c\\d}
$$
Put another way, this means that "$v_1$ and $v_2$ act under $A$ in the exact same way that the standard basis vectors $(1,0)$ and $(0,1)$ under $B$".  Yet another way of saying this: "$B$ is the same linear transformation as $A$, with $v_1$ 'relabeled' as $(1,0)$ and $v_2$ 'relabeled' as $(0,1)$".
Note that all of this only really makes sense when $A$ is a square matrix.  In particular, as a linear transformation, it takes the span of $v_1$ and $v_2$ (which is $\Bbb R^2$) to the span of $v_1$ and $v_2$, which means that both the inputs and outputs can be (and are) relabeled in the same way.
So, for example, if $A$ is equal to $B$, then $A$ and $B$ will have the same rank, determinant, trace, and as you will soon discover, eigenvalues.  If $A$ is row-reduced to get $B$, then all $A$ and $B$ have in common is their null space.
