Yes, that looks right! Good job.

In summary, the problem involves finding the matrix [L] for a linear transformation L:R3->R3. L is a dilation 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({1,0,1}) by a factor of 3. The equations provided include z = 0, L(1,0,0) = (3,0,0), L(0,1,0) = (0,3,0), and L(span(1,0,1)) = (1/3,0,1/3). The solution involves setting [L]e1 = e1, [
  • #1
Snoogx
22
0

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?
 
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  • #2
Snoogx said:

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)
[tex][L]= \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}[/tex]
So
[tex][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}[/tex]

Similarly
[tex][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}[/tex]

Finally, we have
[tex][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}[/tex]
 
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  • #3
HallsofIvy said:
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]en = xnvn; 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)
[tex][L]= \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}[/tex]
So
[tex][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}[/tex]

Similarly
[tex][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}[/tex]

Finally, we have
[tex][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}[/tex]

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] = [tex]\begin{bmatrix}3 & 0 & 1/3 \\ 0 & 3 & 0 \\ 0 & 0 & 1/3\end{bmatrix}[/tex] ?
 

1. What is a linear transformation check?

A linear transformation check is a method used to determine if a given function or mapping preserves the properties of linearity, such as preserving parallelism and proportionality.

2. Why is linear transformation check important?

Linear transformations are used in various fields of science, such as physics and engineering, to model real-life situations. Checking for linearity ensures that the model accurately represents the system being studied.

3. How do you perform a linear transformation check?

To perform a linear transformation check, you would need to apply the function to two vectors and see if the resulting image vectors satisfy the properties of linearity. These include the property of addition and scalar multiplication.

4. What are the properties of linearity?

The properties of linearity include the preservation of addition, scalar multiplication, and the zero vector. Additionally, a linear transformation must preserve parallelism and proportionality between vectors.

5. Can a function be both linear and non-linear?

No, a function can either be linear or non-linear. A linear function preserves the properties of linearity, while a non-linear function does not. It is not possible for a function to exhibit both of these properties at the same time.

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