Linear Algebra (Linear Transformation)

[DanielFaraday]

Homework Statement

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

Homework Equations

None

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]

Homework Statement

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

P.S. Nice use of latex

 

[DanielFaraday]

Much better

P.S. Nice use of latex

Thanks!