Change of basis of linear transformation $ℝ^2 \to ℝ^3$ Linear transformation $T:ℝ^2\to ℝ^3$ in bases
$\left\{ \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 3 \end{bmatrix}\right\}$ and $\left\{ \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}\right\}$ has matrix:
$\begin{bmatrix} 
2 & 5 \\
1 & 1 \\
8 & 1 \\
\end{bmatrix}$. What is the standard matrix? 
I thought this problem will be easy, but it got me horribly confused. I don't know where to start.
 A: We have the matrices
$$
B^{e\to b} = 
B^{-1} =
\begin{bmatrix} 
1 & 1\\ 
1 & 3 
\end{bmatrix}^{-1}
\quad
T^{b\to b'}
=
\begin{bmatrix} 
2 & 5 \\
1 & 1 \\
8 & 1 \\
\end{bmatrix}
\quad
B^{b'\to e'} = 
B' =
\begin{bmatrix} 
2 & 1 & 1\\
1 & 0 & -1  \\
1 & 1 & 1 
\end{bmatrix}
$$
where $e$ means the canonical basis in $\mathbb{R}^2$, $e'$ the canonical basis in $\mathbb{R}^3$, $b$ and $b'$ the other two given basis sets, so we get
\begin{align}
T^{e\to e'} 
&= B^{b'\to e'} \, T^{b\to b'} B^{e\to b} \\
&=
\begin{bmatrix} 
2 & 1 & 1\\
1 & 0 & -1  \\
1 & 1 & 1 
\end{bmatrix}
\begin{bmatrix} 
2 & 5 \\
1 & 1 \\
8 & 1 \\
\end{bmatrix}
\begin{bmatrix} 
1 & 1\\ 
1 & 3 
\end{bmatrix}^{-1}
\\
&=
\begin{bmatrix} 
2 & 1 & 1\\
1 & 0 & -1  \\
1 & 1 & 1 
\end{bmatrix}
\begin{bmatrix} 
2 & 5 \\
1 & 1 \\
8 & 1 \\
\end{bmatrix}
\begin{bmatrix} 
3/2 & -1/2\\ 
-1/2 & 1/2 
\end{bmatrix}
\\
&=
\begin{bmatrix} 
13 & 12 \\
-6 & 4 \\
11 & 7 \\
\end{bmatrix}
\begin{bmatrix} 
3/2 & -1/2\\ 
-1/2 & 1/2 
\end{bmatrix}
\\
&=
\begin{bmatrix} 
27/2 & -1/2\\ 
-11 & 5 \\
13 & -2 
\end{bmatrix}
\end{align}
A: T(1,1)=(13,-6,11) and T(1,3)=(12,4,7).
(1,0)=3/2(1,1)-1/2(1,3).
So T(1,0)=T[3/2(1,1)-1/2(1,3)]=(27/2,-11,13).You can do same for T(0,1).And combining the resulting vectors you get standard matrix.
A: let $x_1, x_2$ be the canonical basis in \mathbb R^2.
and
$y_1, y_2, y_3$ be the canonical basis in \mathbb R^3.
$T(x_1 + x_2) = T(x_1) + T(x_2) = $$(2\cdot 2 + 1\cdot 1 + 8\cdot 1) y_1 + (2\cdot 1 + 1\cdot 0 + 8\cdot -1) y_2 + (2+1+8) y_3\\
13 y_1 - 6y_2 + 11y_3$
$T(x_1) + 3T(x_2) = 12y_1 + 4y_2 + 7y_3\\
2T(x_2) = -1y_1 +10y_2-4 y_3\\
T(x_2) = \begin{bmatrix}{- \frac 12\\5\\-2}\end{bmatrix}\\
T(x_1) = \begin{bmatrix}{13\\-6\\11}\end{bmatrix}-\begin{bmatrix}{- \frac 12\\5\\-2}\end{bmatrix} = \begin{bmatrix}{13.5\\-11\\13}\end{bmatrix}$
Alternatively
$\begin{bmatrix} 2&1&1\\1&0&-1\\1&1&1\end{bmatrix}\begin{bmatrix}2&5\\1&1\\8&1\end{bmatrix}\begin{bmatrix}1&1\\1&3\end{bmatrix}^{-1} = \begin{bmatrix} 13.5 &-0.5\\-11&5\\13&-2\end{bmatrix}$
