# Find matrix for T for bases B and B'

• Mar 30th 2009, 11:26 PM
flaming
Find matrix for T for bases B and B'
Let T:$\displaystyle \Re^2 --> \Re^3$ be defined by

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

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

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

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

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

I know that

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

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

but i dont know how to do the rest.
• Mar 31st 2009, 01:15 AM
NonCommAlg
Quote:

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

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

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

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

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

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

I know that

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

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

but i dont know how to do the rest.

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

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