Linear Transformations Change of Basis I am not sure how this one gets computed.
Let $\alpha = \{(1, 1, 1), (1, 1, 0), (1, 0, 0)\}$ be a basis for $\mathbb{R}^3$.
Let $T$ satisfy $T((1, 1, 1)) = (2, 2, 2)$, $T((1, 1, 0)) = (3, 3, 0)$, and $T((1, 0, 0)) = (-1, 0, 0)$.
Find $[T]_\beta^\beta$ for $\beta$ the standard basis for $\mathbb{R}^3$.
I know the final answer is $\begin{bmatrix} -1 & 4 & -1 \\ 0 & 3 & -1 \\ 0 & 0 & 2 \end{bmatrix}$ but am unsure how to compute that.
 A: Thanks for some help in the comments I determined I must find $(T(β1)|T(β2)|T(β3))$. 
Where $T(β_i)$ are the standard basis vectors mapped to $\beta$ in $R^3$.
I can already see that T(β1)= (-1, 0, 0) from my basis $\alpha$.
T(β2)= T((1, 1, 0)) - T((1, 0, 0)) = (3, 3, 0) - (-1, 0, 0) = (4, 3, 0)
T(β3)= T((1, 1, 1)) - T((1, 1, 0)) = (2, 2, 2) - (3, 3, 0) = (-1, -1, 2)
So [$T]^\beta_\beta$ = $\begin{bmatrix}-1 & 4 & -1 \\0 & 3 & -1 \\ 0 & 0 & 2\end{bmatrix}$
A: In general, for a linear map $T:V\rightarrow W$ and bases $\beta = \{v_1, ...,v_n\}$ and $\gamma = \{w_1,...,w_m\}$ for $V$ and $W$ respectively, we have
$$T(v_j)=\sum^m_{i=1}{([T]^\gamma_\beta)}_{ij}w_i$$
Hence
$$[T]^\gamma_\beta=([T(v_1)]_\gamma\,|\,...\,|\,[T(v_n)]_\gamma)$$
where $[v]_B$ denotes the components of a vector in basis $B$.

In this question, $V=W=\Bbb R^3$ and $\beta=\gamma=\{e_1, e_2, e_3\}$, the standard basis.
So we have
$$T(e_j)=\sum^3_{i=1}{([T]^\beta_\beta)}_{ij}e_i$$
and hence
$$[T]^\beta_\beta\,\,\,(=[T]_\beta)=\begin{pmatrix}T(e_1)\,|\,T(e_2)\,|\,T(e_3)\end{pmatrix}$$
A: Let $e_1,e_2,e_3$ denote the vectors of the standard basis, and let $v_1,v_2,v_3$ denote the vectors of $\alpha$.  Now, let's compute the third column of the matrix $[T]_\beta^\beta$, remembering that $\beta = \{e_1,e_2,e_3\}$. What we're looking for are coefficients $a_1,a_2,a_3$ such that
$$
T(e_3) = a_1 e_1 + a_2 e_2 + a_3 e_3 = \pmatrix{a_1\\a_2\\a_3}_\beta = \pmatrix{a_1\\a_2\\a_3}
$$
However, we can't access what $T$ does to $e_3$ directly.  It would be helpful if we could find coefficients $b_1,b_2,b_3$ such that
$$
e_3 = b_1 v_1 + b_2 v_2 + b_3 v_3
$$
In this case, we would find that
$$
T(e_3) = b_1 T(v_1) + b_2 T(v_2) + b_3T(v_3)
$$
Now, let's describe this process in terms of matrix multiplication: we have 
$$
\pmatrix{a_1\\a_2\\a_3}_{\beta} = \pmatrix{T(v_1)_\beta \quad T(v_2)_\beta \quad T(v_3)_\beta} \pmatrix{b_1\\b_2 \\b_3}
$$
where the $b_i$ solve the equation
$$
\pmatrix{v_1 \quad v_2 \quad v_3} \pmatrix{b_1\\b_2\\b_3} = e_3 = \pmatrix{0\\0\\1} \implies\\
\pmatrix{b_1\\b_2\\b_3} = \pmatrix{v_1 \quad v_2 \quad v_3}^{-1}e_3
$$
So, the third column of the desired matrix will be given by
$$
\pmatrix{a_1\\a_2\\a_3}_{\beta} = \pmatrix{T(v_1)_\beta \quad T(v_2)_\beta \quad T(v_3)_\beta} \pmatrix{v_1 \quad v_2 \quad v_3}^{-1}e_3
$$
And from there, perhaps you can see that the matrix $\pmatrix{T(v_1)_\beta \quad T(v_2)_\beta \quad T(v_3)_\beta} \pmatrix{v_1 \quad v_2 \quad v_3}^{-1}$ is in fact the matrix of the transformation.
You may find it helpful to break future problems down in a similar way.
