Linear Algebra (Linear Transformation)

[DanielFaraday]

1. The problem statement, all variables and given/known data
True or False:

T(x,y)=(2x+5y,-x+2)\text{ is a linear transformation from }\mathbb{R}^2\text{ to }\mathbb{R}^2.



2. Relevant equations
None


3. The attempt at a solution
I thought the answer was true, but the correct answer is false. Here is my reasoning for true:

T depends only on x and y and the transformation depends only on x and y, so it must be in the same space.

[VeeEight]

Your reasoning is not detailed enough. Do you know how to check if a transformation is linear?

[DanielFaraday]

I think I do. If a transformation is linear, then:

T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v })
and
T(a\mathbf{u})=aT(\mathbf{u})

[n!kofeyn]

1. The problem statement, all variables and given/known data
True or False:

T(x,y)=(2x+5y,-x+2)\text{ is a linear transformation from }\mathbb{R}^2\text{ to }\mathbb{R}^2.



You need to show your answer though. A function T:V\to W is a linear transformation if for all \vec u, \vec v\in V and for all scalars r\in\mathds{R},
T(\vec u+\vec v) = T(\vec u)+T(\vec v)
T(r\vec u) = rT(\vec u)
(You can actually combine these two requirements into one, but I think it is usually best to leave them separate early on.)

In your case, let \vec u=(x,y) and \vec v = (u,v). (Don't get confused between my re-use of u and v.) You need to try and show that
T(\vec u+\vec v) = T(x+u,y+v) = T(x,y)+T(u,v) = T(\vec u)+T(\vec v)
[tex] T(r\vec u) = T(rx,ry) = rT(x,y) = rT(\vec u)[/itex]

If you can show this, then T is a linear transformation. If either of these two statements are not true, then T is not a linear transformation.

[DanielFaraday]

Do I pick arbitrary values for u and v and then test them, or is there a more systematic way to go about it?

[n!kofeyn]

You pick arbitrary values just as I mentioned at the end of my last post.

[DanielFaraday]

Okay, so this is obviously not a linear transformation because the first component involves both an x and a y, correct?.

[mlarson9000]

Okay, so this is obviously not a linear transformation because the first component involves both an x and a y, correct?.

I'm not sure what you mean by an x and a y. It's not a linear transformation because T(u+v) is not equal to T(u)+T(v). You probably need to illustrate that by picking two arbitrary vectors, and performing the transformations. You will then see that they are not equal.

[mlarson9000]

I forgot, you need to show a counterexample.

[DanielFaraday]

I get that they are equal. What am I doing wrong?


\text{Let }\overset{\rightharpoonup }{u}=\left(
\begin{array}{c}
1 \\
0
\end{array}
\right)\text{ and }\overset{\rightharpoonup }{u}=\left(
\begin{array}{c}
0 \\
1
\end{array}
\right)



T(u+v)=\left(
\begin{array}{c}
2+5 \\
2-1
\end{array}
\right)=\left(
\begin{array}{c}
7 \\
1
\end{array}
\right)



T(u)+T(v)=\left(
\begin{array}{c}
2 \\
2-1
\end{array}
\right)+\left(
\begin{array}{c}
5 \\
0
\end{array}
\right)=\left(
\begin{array}{c}
7 \\
1
\end{array}
\right)

[DanielFaraday]

Okay, I just happened to pick a case that works. Here is one that doesn't:


\text{Let }\overset{\rightharpoonup }{u}=\left(
\begin{array}{c}
2 \\
0
\end{array}
\right)\text{and }\overset{\rightharpoonup }{u}=\left(
\begin{array}{c}
0 \\
1
\end{array}
\right)



T(u+v)=\left(
\begin{array}{c}
4+5 \\
-2+2
\end{array}
\right)=\left(
\begin{array}{c}
9 \\
0
\end{array}
\right)



T(u)+T(v)=\left(
\begin{array}{c}
4 \\
-2+2
\end{array}
\right)+\left(
\begin{array}{c}
5 \\
2
\end{array}
\right)=\left(
\begin{array}{c}
9 \\
2
\end{array}
\right)


Thanks for your help everyone!

[Hootenanny]

I get that they are equal. What am I doing wrong?


\text{Let }\overset{\rightharpoonup }{u}=\left(
\begin{array}{c}
1 \\
0
\end{array}
\right)\text{ and }\overset{\rightharpoonup }{u}=\left(
\begin{array}{c}
0 \\
1
\end{array}
\right)



T(u+v)=\left(
\begin{array}{c}
2+5 \\
2-1
\end{array}
\right)=\left(
\begin{array}{c}
7 \\
1
\end{array}
\right)



T(u)+T(v)=\left(
\begin{array}{c}
2 \\
2-1
\end{array}
\right)+\left(
\begin{array}{c}
5 \\
{\color{red}0}
\end{array}
\right)=\left(
\begin{array}{c}
7 \\
1
\end{array}
\right)

You might want to check the highlighted value ...

[DanielFaraday]

Oops, you are right. Is this correct?


\text{Let }\overset{\rightharpoonup }{u}=\left(
\begin{array}{c}
1 \\
0
\end{array}
\right)\text{ and }\overset{\rightharpoonup }{u}=\left(
\begin{array}{c}
0 \\
1
\end{array}
\right)



T(u+v)=\left(
\begin{array}{c}
2+5 \\
2-1
\end{array}
\right)=\left(
\begin{array}{c}
7 \\
1
\end{array}
\right)



T(u)+T(v)=\left(
\begin{array}{c}
2 \\
2-1
\end{array}
\right)+\left(
\begin{array}{c}
5 \\
2
\end{array}
\right)=\left(
\begin{array}{c}
7 \\
3
\end{array}
\right)

[Hootenanny]

Oops, you are right. Is this correct?


\text{Let }\overset{\rightharpoonup }{u}=\left(
\begin{array}{c}
1 \\
0
\end{array}
\right)\text{ and }\overset{\rightharpoonup }{u}=\left(
\begin{array}{c}
0 \\
1
\end{array}
\right)



T(u+v)=\left(
\begin{array}{c}
2+5 \\
2-1
\end{array}
\right)=\left(
\begin{array}{c}
7 \\
1
\end{array}
\right)



T(u)+T(v)=\left(
\begin{array}{c}
2 \\
2-1
\end{array}
\right)+\left(
\begin{array}{c}
5 \\
2
\end{array}
\right)=\left(
\begin{array}{c}
7 \\
3
\end{array}
\right)

Much better :approve:

P.S. Nice use of latex

[DanielFaraday]

Much better :approve:

P.S. Nice use of latex

Thanks!