# Thread: Find matrix for T for bases B and B'

1. ## Find matrix for T for bases B and B'

Let T: $\Re^2 --> \Re^3$ be defined by

T $\begin{pmatrix}
x_1\\
x_2

\end{pmatrix}

$
= $\begin{pmatrix}
x_1 - x_2\\
x_1 + 2x_2\\
2x_1 - x_2
\end{pmatrix}$

Find the matrix for T with respect to the bases B = { $u_1, u_2$} and B' = { $v_1, v_2, v_3$}

where $u_1 = \begin{pmatrix}
1\\
2
\end{pmatrix}$

$u_2 = \begin{pmatrix}
-1\\
1
\end{pmatrix}$

and
$v_1 = \begin{pmatrix}
1\\
0\\
1
\end{pmatrix}$
, $v_2 = \begin{pmatrix}
0\\
1\\
1
\end{pmatrix}$
, $v_3 = \begin{pmatrix}
1\\
1\\
0
\end{pmatrix}$

I know that

T $(u_1) = \begin{pmatrix}
-1\\
5\\
0
\end{pmatrix}$
and

T $(u_2) = \begin{pmatrix}
-2\\
1\\
-3
\end{pmatrix}$

but i dont know how to do the rest.

2. Originally Posted by flaming
Let T: $\Re^2 --> \Re^3$ be defined by

T $\begin{pmatrix}
x_1\\
x_2

\end{pmatrix}

$
= $\begin{pmatrix}
x_1 - x_2\\
x_1 + 2x_2\\
2x_1 - x_2
\end{pmatrix}$

Find the matrix for T with respect to the bases B = { $u_1, u_2$} and B' = { $v_1, v_2, v_3$}

where $u_1 = \begin{pmatrix}
1\\
2
\end{pmatrix}$

$u_2 = \begin{pmatrix}
-1\\
1
\end{pmatrix}$

and
$v_1 = \begin{pmatrix}
1\\
0\\
1
\end{pmatrix}$
, $v_2 = \begin{pmatrix}
0\\
1\\
1
\end{pmatrix}$
, $v_3 = \begin{pmatrix}
1\\
1\\
0
\end{pmatrix}$

I know that

T $(u_1) = \begin{pmatrix}
-1\\
5\\
0
\end{pmatrix}$
and

T $(u_2) = \begin{pmatrix}
-2\\
1\\
-3
\end{pmatrix}$

but i dont know how to do the rest.
next you need to write $T(u_1), T(u_2)$ as a linear combination of $v_1,v_2,v_3.$ that is, $T(u_1)=-3v_1+3v_2+2v_3$ and $T(u_2)=-3v_1+v_3.$ so $[T(u_1)]_{B'}=\begin{pmatrix}-3 \\ 3 \\ 2 \end{pmatrix}$ and $[T(u_2)]_{B'}=\begin{pmatrix}-3 \\ 0 \\ 1 \end{pmatrix}.$

therefore the matrix you're looking for is: $[T]_B^{B'}=\begin{pmatrix}[T(u_1)]_{B'} & [T(u_2)]_{B'} \end{pmatrix}=\begin{pmatrix}-3 & -3 \\ 3 & 0 \\ 2 & 1 \end{pmatrix}.$