# linear transformation

• May 6th 2008, 08:07 PM
Dr Zoidburg
linear transformation
Let u = (4, 0) and v = (0, -3)
T is a linear transformation such that:

T(u) = (4, 4, 4)
T(v) = (6, 9, 12)

a) Find the standard matrix of T

b) Find T(x, y)

Only problem I'm having here is that I can't really see the difference between the two questions.
Here's my solution to part a:
We need to solve a(4, 0) + b(0, -3) = (x, y)
This gives us the following:
4a = x
-3b = y
a = x/4
b = -y/3
Using the linear property of transforms on a(4, 0) + b(0, -3) = (x, y), we get
T(x, y) = T[a(4, 0)] + b(0, -3)] = aT(4, 0) + bT(0, -3)
T(x, y) = a(4, 4, 4) + b(6, 9 ,12) = (x, x, x) + (-2y, -3y, -4y)

T(x, y) = (x – 2y, x – 3y, x – 4y)

which is, as far as I can see, gives me the solution to part b)!

What am I missing here?
• May 6th 2008, 08:45 PM
TheEmptySet
Quote:

Originally Posted by Dr Zoidburg
Let u = (4, 0) and v = (0, -3)
T is a linear transformation such that:

T(x, y) = a(4, 4, 4) + b(6, 9 ,12) = (x, x, x) + (-2y, -3y, -4y)

T(x, y) = (x – 2y, x – 3y, x – 4y)

which is, as far as I can see, gives me the solution to part b)!

What am I missing here?

$T(x,y)=x(1,1,1)+y(-2,-3,-4)$

the two vectors above make up the columns of the matrix transfrom.

A= $\begin{bmatrix}
1 && -2 \\
1 && -3 \\
1 && -4 \\
\end{bmatrix}$

Good luck.