Linear Transformation

Shapeshift · #1 $Old$ November 3rd, 2009, 05:44 PM

If

is a linear transformation such that

1

2

11

14 6 and

4

-1

17

2 6 ,
what is the standard matrix of T?

Any help is appreciated

hjortur · #2 $Old$ November 3rd, 2009, 06:57 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:

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

math2009 · #3 $Old$ November 3rd, 2009, 08:00 PM

As above, I repeat equations.

$T(\vec{x})=A\vec{x},A=[T(\vec{e}_1)\ \ T(\vec{e}_2)]$

$T(\vec{e}_1)=\frac{1}{9}\cdot\begin{bmatrix} *45* \\18\\18\end{bmatrix}=\begin{bmatrix} 5 \\2\\2\end{bmatrix}\ ,\ T(\vec{e}_2)=\begin{bmatrix} 3 \\6\\2\end{bmatrix}\ ,\ A=\begin{bmatrix} 5&3\\2&6\\2&2\end{bmatrix}$