Yes, that looks right! Good job.

[Snoogx]

Homework Statement

L:R³->R³ is a dialiation by a factor of 3 of points in the plane W given by the equation z = 0 and a contraction along the line L = span({(1,0,0)}) by a factor of 3.
Find [L], but I'm mainly concerned with finding L(e³₁)

Homework Equations

z = 0
L(1,0,0) = (3,0,0)
L(0,1,0) = (0,3,0)
L(span(1,0,1)) = (1/3,0,1/3)

The Attempt at a Solution

L(e₁) = a(1,0,0) + b(0,1,0) + c(1,0,1)
e₁ = aL(1,0,0) + bL(0,1,0) + cL(1,0,1)
e₁ = a(3,0,0) + b(0,3,0) + c(1/3,0,1/3)
e₁ = [(3,0,0),(0,3,0),(1/3,0,1/3)][a,b,c]
[(1/3,0,0),(0,1/3,0),(-1/3,0,3)]e₁ = [a,b,c]
[(1/3,0,0),(0,1/3,0),(-1/3,0,3)][1,0,0] = [a,b,c]
[a,b,c] = [1/3,0,0]

Can anyone check my work?

 

[HallsofIvy]

Homework Statement

L:R³->R³ is a dialiation by a factor of 3 of points in the plane W given by the equation z = 0 and a contraction along the line L = span({(1,0,0)}) by a factor of 3.

Would you please recheck the problem? The line, L, is in the plane, W. You can't "dilate by a factor of 3" and "contract by a factor of 3" the same points!

Find [L], but I'm mainly concerned with finding L(e³₁)

Homework Equations

z = 0
L(1,0,0) = (3,0,0)
L(0,1,0) = (0,3,0)
L(span(1,0,1)) = (1/3,0,1/3)

Aha! L is spanned by (1, 0, 1), not (1, 0, 0). But you don't mean to say "L(span(1,0,1))= (1/3,0,1/3)" What is true is that L(1, 0, 1)= (1/3, 0, 1/3) and that L(a, 0, a)= (a/3, 0, a/3) for any a.

The Attempt at a Solution

L(e₁) = a(1,0,0) + b(0,1,0) + c(1,0,1)
e₁ = aL(1,0,0) + bL(0,1,0) + cL(1,0,1)
e₁ = a(3,0,0) + b(0,3,0) + c(1/3,0,1/3)
e₁ = [(3,0,0),(0,3,0),(1/3,0,1/3)][a,b,c]
[(1/3,0,0),(0,1/3,0),(-1/3,0,3)]e₁ = [a,b,c]
[(1/3,0,0),(0,1/3,0),(-1/3,0,3)][1,0,0] = [a,b,c]
[a,b,c] = [1/3,0,0]

Can anyone check my work?

Looks like a good start- but you don't want just "a b c", you are looking for a 3 by 3 matrix. In a slightly better notation (using LaTeX)
$$[L]= \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}$$
So
$$[L]e_1= \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}= \begin{bmatrix}a \\ d\\ g\end{bmatrix}= \begin{bmatrix}3 \\ 0 \\ 0\end{bmatrix}$$

Similarly
$$[L]e_2= \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}= \begin{bmatrix} b \\ e \\ i\end{bmatrix}= \begin{bmatrix}0 \\ 3\\ 0\end{bmatrix}$$

Finally, we have
$$[L]e_2= \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}= \begin{bmatrix} a+ c \\ d+ f \\ g+ h\end{bmatrix}= \begin{bmatrix}1/3 \\ 0\\ 1/3\end{bmatrix}$$

 

[Snoogx]

Would you please recheck the problem? The line, L, is in the plane, W. You can't "dilate by a factor of 3" and "contract by a factor of 3" the same points!

That's exactly what the problem said. I believe it means that the line is contracted by a factor of 3 only and the points are dilated by 3.

Aha! L is spanned by (1, 0, 1), not (1, 0, 0). But you don't mean to say "L(span(1,0,1))= (1/3,0,1/3)" What is true is that L(1, 0, 1)= (1/3, 0, 1/3) and that L(a, 0, a)= (a/3, 0, a/3) for any a.

What I had thought was that I was supposed to pick 2 points from W and then use the span as the third point. And then like you did set [L]e_n = x_nv_n; where v is the vectors and x is the dilate/contract.

Looks like a good start- but you don't want just "a b c", you are looking for a 3 by 3 matrix. In a slightly better notation (using LaTeX)
$$[L]= \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}$$
So
$$[L]e_1= \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}= \begin{bmatrix}a \\ d\\ g\end{bmatrix}= \begin{bmatrix}3 \\ 0 \\ 0\end{bmatrix}$$

Similarly
$$[L]e_2= \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}= \begin{bmatrix} b \\ e \\ i\end{bmatrix}= \begin{bmatrix}0 \\ 3\\ 0\end{bmatrix}$$

Finally, we have
$$[L]e_2= \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}= \begin{bmatrix} a+ c \\ d+ f \\ g+ h\end{bmatrix}= \begin{bmatrix}1/3 \\ 0\\ 1/3\end{bmatrix}$$

I know I'm looking for the whole matrix, I was just concerned with L(e₁) for the moment. What I had thought was that I would just apply the same steps for L(e₂) and L(e₃).

I'm still a little confused when you calculated [L]e₂. For the last one, I'm not sure if you meant [L]e₃ or [L]e₂?

So would this make the matrix [L] = $$\begin{bmatrix}3 & 0 & 1/3 \\ 0 & 3 & 0 \\ 0 & 0 & 1/3\end{bmatrix}$$ ?