Linear transformation question

• Dec 21st 2009, 11:27 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
• Dec 22nd 2009, 03:19 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
• Dec 22nd 2009, 10:04 PM
dhammikai
Hi thanks for the help, Now I got an idea for the solution, thanks again..