Linear transformation question

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December 21st 2009, 10:27 PM
dhammikai

Linear transformation question

Plz help me for answer for the following question;

T is a linear transformation of $\mathbb{R} ^3$ into $\mathbb{R} ^3$ , such that

$T \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ 4 \end{pmatrix} , $

$T \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 6 \\ 7 \\ 8 \end{pmatrix} , $

$T \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 2 \end{pmatrix} $

Find T and Kernal of T

When answering the question, I found taking linear transformation
$T(x_1 , x_2 ,x_3)= x_1 T(1,1,1)+ x_2 T(0,1,0)+ x_3 T(0,0,1) $
$= x_1(3,4,4)+ x_2 (6,7,8)+ x_3 (2,3,2) =(3 x_1 +6 x_2+2 x_3 , 4x_1+8x_2+3x_3, 4x_1+7x_2+2x_3)$

After this how I going to calculate T and kernal of T
December 22nd 2009, 02:19 AM
tonio

Quote:

Originally Posted by dhammikai

Plz help me for answer for the following question;

T is a linear transformation of $\mathbb{R} ^3$ into $\mathbb{R} ^3$ , such that

$T \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ 4 \end{pmatrix} , $

$T \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 6 \\ 7 \\ 8 \end{pmatrix} , $

$T \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 2 \end{pmatrix} $

Find T and Kernal of T

When answering the question, I found taking linear transformation
$T(x_1 , x_2 ,x_3)= x_1 T(1,1,1)+ x_2 T(0,1,0)+ x_3 T(0,0,1) $
$= x_1(3,4,4)+ x_2 (6,7,8)+ x_3 (2,3,2) =(3 x_1 +6 x_2+2 x_3 , 4x_1+8x_2+3x_3, 4x_1+7x_2+2x_3)$

After this how I going to calculate T and kernal of T

Well, you did most of the hardest part: now you can take the matrix of T wrt the standard basis: $\begin{pmatrix}3&6&2\\4&8&3\\4&7&2\end{pmatrix}$ and bring it to echelon form: that'll will give you a solution to the associated homogeneous linear system, i.e. the kernel of the matrix (of T), and then you can take a basis for the row space for the image of T (all the time, wrt the standard basis) .

Tonio
December 22nd 2009, 09:04 PM
dhammikai

Hi thanks for the help, Now I got an idea for the solution, thanks again..