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hjortur · #2 November 3rd, 2009, 05:57 PM

I dont know what standard matrix is, but I am going to assume it means the matrix of
T with respect to the standard basis.

Now what you need to find is what T does to the standard basis

$T\left(\begin{bmatrix}1\\2\end{bmatrix}\right)=T\left(\begin{bmatrix}1\\0\end{bmatrix}+2\cdot\begin{bmatrix}0\\1\end{bmatrix}\right)=T\left(\begin{bmatrix}1\\0\end{bmatrix}\right)+2T\left(\begin{bmatrix}0\\1\end{bmatrix}\right)=\begin{bmatrix}11\\14\\6\end{bmatrix}$

And :

$T\left(\begin{bmatrix}4\\-1\end{bmatrix}\right)=T\left(4\cdot\begin{bmatrix}1\\0\end{bmatrix}-1\cdot\begin{bmatrix}0\\1\end{bmatrix}\right)=4T\left(\begin{bmatrix}1\\0\end{bmatrix}\right)-1T\left(\begin{bmatrix}0\\1\end{bmatrix}\right)=\begin{bmatrix}17\\2\\6\end{bmatrix}$

Multiplying the lower formula by 2 and adding to the upper one gives:

$9T\left(\begin{bmatrix}1\\0\end{bmatrix}\right)+0=\begin{bmatrix}11\\14\\6\end{bmatrix}+2\cdot\begin{bmatrix}17\\2\\6\end{bmatrix}$

So that
$T\left(\begin{bmatrix}1\\0\end{bmatrix}\right)=\frac{1}{9}\cdot\begin{bmatrix}54\\18\\18\end{bmatrix}$

Now find $T\left(\begin{bmatrix}0\\1\end{bmatrix}\right)$

And the standard matrix would then be:

$\begin{bmatrix}T\left(\begin{bmatrix}1\\0\end{bmatrix}\right) & T\left(\begin{bmatrix}0\\1\end{bmatrix}\right)\end{bmatrix}$

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