# Linear Algebra help

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• July 6th 2008, 08:23 PM
JCIR
Linear Algebra help
Let T: R^2 --> R^3 be defined by T(a1,a2)=(a1-a2,a1,2a1+a2) Let beta be the standard basis for R^2 and gamma={(1,1,0),(0,1,1),(2,2,3)}.
compute [T]_beta to gamma
• July 7th 2008, 09:27 AM
Isomorphism
Quote:

Originally Posted by JCIR
Let T: R^2 --> R^3 be defined by T(a1,a2)=(a1-a2,a1,2a1+a2) Let beta be the standard basis for R^2 and gamma={(1,1,0),(0,1,1),(2,2,3)}.
compute [T]_beta to gamma

"beta be the standard basis for R^2" => T(1,0) = (1,1,2) and T(0,1) = (-1,0,1)

The question essentially asks:

We want to write (a1-a2,a1,2a1+a2) as a linear combination of gamma. Let the coefficients of this linear combination be x,y,z in that order.

Then given (a1,a2) in R^2, we want to find a matrix that maps it to T(a1,a2) which is written as a linear combination of gamma.

Thus:

T(a1,a2) = (a1-a2,a1,2a1+a2) = x(1,1,0)+y(0,1,1)+z(2,2,3)

But observe that:

$(a1-a2,a1,2a1+a2) = \begin{pmatrix}1 & -1 \\ 1 & 0 \\ 2 & 1\end{pmatrix} \begin{pmatrix}a1\\ a2\end{pmatrix}$

$x(1,1,0)+y(0,1,1)+z(2,2,3) = \begin{pmatrix}1 & 0 & 2 \\ 1 & 1 & 2 \\ 0 & 1 & 3\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}$

So:
$\begin{pmatrix}1 & -1 \\ 1 & 0 \\ 2 & 1\end{pmatrix} \begin{pmatrix}a1\\ a2\end{pmatrix} = \begin{pmatrix}1 & 0 & 2 \\ 1 & 1 & 2 \\ 0 & 1 & 3\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}$

This means,

$\begin{pmatrix}1 & 0 & 2 \\ 1 & 1 & 2 \\ 0 & 1 & 3\end{pmatrix}^{-1}\begin{pmatrix}1 & -1 \\ 1 & 0 \\ 2 & 1\end{pmatrix} \begin{pmatrix}a1\\ a2\end{pmatrix} = \begin{pmatrix}x\\ y\\ z\end{pmatrix}$

Thus the matrix $\begin{pmatrix}1 & 0 & 2 \\ 1 & 1 & 2 \\ 0 & 1 & 3\end{pmatrix}^{-1}\begin{pmatrix}1 & -1 \\ 1 & 0 \\ 2 & 1\end{pmatrix}$ maps (a1,a2) to (x,y,z)

I will let you compute that matrix....
• July 7th 2008, 03:22 PM
Jhevon
Quote:

Originally Posted by Isomorphism
The question essentially asks:

We want to write (a1-a2,a1,2a1+a2) as a linear combination of gamma.

thanks for that. for some reason, i do not recall the notation the OP used, so i was wondering what they were asking for (Wondering)