Linear Algebra (Linear Transformation)

DanielFaraday
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Homework Statement


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


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.
 
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Your reasoning is not detailed enough. Do you know how to check if a transformation is linear?
 
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})
 
DanielFaraday said:

Homework Statement


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

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)
T(r\vec u) = T(rx,ry) = rT(x,y) = rT(\vec u)[/itex]<br /> <br /> 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.
 
Do I pick arbitrary values for u and v and then test them, or is there a more systematic way to go about it?
 
You pick arbitrary values just as I mentioned at the end of my last post.
 
Okay, so this is obviously not a linear transformation because the first component involves both an x and a y, correct?.
 
DanielFaraday said:
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.
 
I forgot, you need to show a counterexample.
 
  • #10
I get that they are equal. What am I doing wrong?

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

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

<br /> T(u)+T(v)=\left(<br /> \begin{array}{c}<br /> 2 \\<br /> 2-1<br /> \end{array}<br /> \right)+\left(<br /> \begin{array}{c}<br /> 5 \\<br /> 0<br /> \end{array}<br /> \right)=\left(<br /> \begin{array}{c}<br /> 7 \\<br /> 1<br /> \end{array}<br /> \right)<br />
 
  • #11
Okay, I just happened to pick a case that works. Here is one that doesn't:

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

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

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

Thanks for your help everyone!
 
  • #12
DanielFaraday said:
I get that they are equal. What am I doing wrong?

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

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

<br /> T(u)+T(v)=\left(<br /> \begin{array}{c}<br /> 2 \\<br /> 2-1<br /> \end{array}<br /> \right)+\left(<br /> \begin{array}{c}<br /> 5 \\<br /> {\color{red}0}<br /> \end{array}<br /> \right)=\left(<br /> \begin{array}{c}<br /> 7 \\<br /> 1<br /> \end{array}<br /> \right)<br />
You might want to check the highlighted value ...
 
  • #13
Oops, you are right. Is this correct?

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

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

<br /> T(u)+T(v)=\left(<br /> \begin{array}{c}<br /> 2 \\<br /> 2-1<br /> \end{array}<br /> \right)+\left(<br /> \begin{array}{c}<br /> 5 \\<br /> 2<br /> \end{array}<br /> \right)=\left(<br /> \begin{array}{c}<br /> 7 \\<br /> 3<br /> \end{array}<br /> \right)<br />
 
  • #14
DanielFaraday said:
Oops, you are right. Is this correct?

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

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

<br /> T(u)+T(v)=\left(<br /> \begin{array}{c}<br /> 2 \\<br /> 2-1<br /> \end{array}<br /> \right)+\left(<br /> \begin{array}{c}<br /> 5 \\<br /> 2<br /> \end{array}<br /> \right)=\left(<br /> \begin{array}{c}<br /> 7 \\<br /> 3<br /> \end{array}<br /> \right)<br />
Much better :approve:

P.S. Nice use of latex
 
  • #15
Hootenanny said:
Much better :approve:

P.S. Nice use of latex

Thanks!
 
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