According to which basis is a vector represented when mapped? Let us suppose we are given a linear map $\tau : V \to W$ with associated matrix $A$. According to which basis of $W$ is $\tau (u)=Au$ represented?
 A: If $A$ represents a linear map $\tau\colon V\to W$, then this is always with respect to a given basis of $V$ and a given basis of $W$. Note that for $u\in V$ you can write $\tau(u)$ but usually not $Au$. This is because $u$ is an element of $V$, not a column vector of real numbers (or whatever your base field is). So what you do is write $u$ according to the aforementioned basis of $V$ as a column vector, multiply the result with $A$, and the resultiung column vector represents $\tau(u)$ according to the aforementioned basis of $W$.

Example: Let $V$ be the space of functions $f\colon \Bbb R\to\Bbb R$ with the condition that $f''(x)=-f(x)$ for all $f$. Let $W$ be the space of real polynomials of degree $\le 2$. and let $\tau$ map a smooth function to the unique even polynomial in $W$ that is tangent to $f$ at $x=\pi$. Note that I specifically tried to avoid any reference to bases of $V$ or $W$ or to describe $\tau$ according to such bases.
It turns out that $V\cong \Bbb R^2$ and $W\cong \Bbb R^3$. 
A suitable basis for $V$ might be $x\mapsto\sin x$, $x\mapsto \cos x$ and a suitable basis for $W$ might be $1,X,X^2$.
You might verify that with respect to these bases we can represent $\tau$ by the matrix $$A=\begin{pmatrix}\frac\pi2&-1\\0&0 \\-\frac1{2\pi}&0\end{pmatrix}$$
Thus if we are given an element $u$ of $V$, such as $u(x)=17\sin(x+42)$, we first notice that $\sin(x+42)=\sin x\cos42+\cos x\sin42$ and hence $u$ is prepresented with respect to the basis mentioned above by the column vector $\begin{pmatrix}17\cos42\\17\sin42\end{pmatrix}$. After multiplication with $A$ this becomes $\begin{pmatrix}\frac{17\pi}{2}\cos42-17\sin 42\\0\\-\frac{17}{2\pi}\cos42\end{pmatrix}$ as column vector. If we interprete that column vector according to the basis mentioned above we ultimately find $\tau(u)$ as element of $W$:
$$-\frac{17\cos42}{2\pi}\cdot X^2+\left(\frac{17}2\cos 42-17\sin 42\right) $$
We might have found a different basis if $V$ and/or $W$ more suitable. That would have produced a different $A$ and hence a different way to interprete the final resulting column vector ...
