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Take a look at the following problem:

Let $V = \{v_1, v_2, v_3\} $ with $v_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$. $v_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$, $v_3 = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$ be a basis of $\mathbb R^3$.
Let $f: \mathbb R^3 \to \mathbb R^3$ be a homomorphism with $f(v_1) = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$, $f(v_2) = \begin{pmatrix} a \\ b \\ c - 1 \end{pmatrix}$, $f(v_3) = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$. Identify the transformation matrix $A$ of $f$ with respect to $V$.

Now the solution happens to be $$A = \begin{pmatrix} 1 & -1 & 1 \\ 1 & 0 &-1 \\ 0 & 1 & 0 \end{pmatrix}.$$ The problem I have is that $$Av_1 = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \neq \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = f(v_1).$$ I know that this could arise from the fact that we're dealing with a different basis than the unit basis, but I would appreciate an explanation for this. What vector would I have to multiply $A$ with to get the result I want?

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You're right this arise because you are using the coordinate matrix for $v_1$ wrt the standard basis, but $A$ is the transformation matrix for $V$. In fact wrt to $V$ we have: $v_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}_{V}$. Hence:

$$[A]_V[v_1]_V = \begin{pmatrix} 1 & -1 & 1 \\ 1 & 0 &-1 \\ 0 & 1 & 0 \end{pmatrix}_V\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}_{V} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}_{V} = [A(v_1)]_V$$

If you want to switch basis to the standard basis then you have to multiply the transition matrix from $V$ to the standard basis by $A$. And the transition matrix is $T = \begin{pmatrix} 0 & 1 & a \\ 0 & 0 & b \\ 1 & 0 & c \end{pmatrix}$.

And indeed $T[A]_VT^{-1}v_1 = f(v_1)$, where the basis is the standard one, unless specified.

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    $\begingroup$ Awesome, that's the information I wanted. Thanks! $\endgroup$ – Staki42 Feb 9 '17 at 23:59

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