# Math Help - [SOLVED] Linear Transformations

1. ## [SOLVED] Linear Transformations

Hi. I need help understanding this question:

The question tells us to state why the following is not a linear transformation:

$T: R^2 \rightarrow R^2$
$T(x,y) = (x^2, xy)$

Their answer is simple enough:
There's a property: T(-v) = -T(v)
however, T(-(1,1) ǂ -T(1,1) $\Longrightarrow$ (by the way, is there a way to post the not equal sign in Latex. I just copy pasted from character map)
therefore, it is not a linear transformation.

But I am having trouble solving the question using the main two properties:
T(a+b) = T(a) + T(b)
T( $\lambda a$) = $\lambda T(a)$

I started like this:

T(x,y) = (x^2, xy)
T(a+b) = T( $\alpha_1 + \beta_1, \alpha_2 + \beta_2$)=
=( $(\alpha_1 + \beta_1)^2$, $(\alpha_1 + \beta_1)(\alpha_2 + \beta_2)$)=
= ( $\alpha_1^2 + 2\alpha_1\beta_1 + \beta_1^2$), ( $\alpha_1\alpha_2 + \alpha_1\beta_2 + \beta_1\alpha_2 + \beta_1\beta_2$)

and from here I am having trouble proving or disproving if it equals T(a) + T(b).
Can someone help?

Thanks.

2. Originally Posted by jayshizwiz
Hi. I need help understanding this question:

The question tells us to state why the following is not a linear transformation:

$T: R^2 \rightarrow R^2$
$T(x,y) = (x^2, xy)$

Their answer is simple enough:
There's a property: T(-v) = -T(v)
however, T(-(1,1) ǂ -T(1,1) $\Longrightarrow$ (by the way, is there a way to post the not equal sign in Latex. I couldn't find it in the tutorial. I just copy pasted from character map)
therefore, it is not a linear transformation.

But I am having trouble solving the question using the main two properties:
T(a+b) = T(a) + T(b)
T( $\lambda a$) = $\lambda T(a)$

I started like this:

T(x,y) = (x^2, xy)
T(a+b) = T( $\alpha_1 + \beta_1, \alpha_2 + \beta_2$)=
=( $(\alpha_1 + \beta_1)^2$, $(\alpha_1 + \beta_1)(\alpha_2 + \beta_2)$)

and from here I am having trouble proving or disproving if it equals T(a) + T(b).
Can someone help?

Thanks.
The given solution used the second property, setting $\lambda = -1$.

Anyway, assume the first condition holds. Then this means that $T(a, b) + T(\alpha, \beta) = T(a+\alpha, b + \beta)$, which means that

$(a^2, ab) + (\alpha^2, \alpha\beta) = ((a+\alpha)^2, (b+\beta)^2)$.

Do you understand how I got to this point? Can you see how to show that this gives you a contradiction? (Essentially, $(p+q)^2 \neq p^2 + q^2$.)

Also, I would like to point out that a map can fulfill one condition but not the other.

3. Thanks for such a quick response. I must've edited my post 5 times while you answered, so my original post is a bit different than what you read.

Anyway, assume the first condition holds. Then this means that , which means that

.

Do you understand how I got to this point? Can you see how to show that this gives you a contradiction? (Essentially, .)

I can see. Your technique is much better than my book's technique.

you did:
and got
as in, you did both sides at once to prove they are unequal.

The way my book answers the questions is by starting with the first step: T(a+b), doing all the work from that, and finally coming to the conclusion that it equals T(a) + T(b). For example, the way I tried to answer was like this:
T(a+b) = T( $\alpha_1 + \beta_1, \alpha_2 + \beta_2$)
=( $(\alpha_1 + \beta_1)^2$, $(\alpha_1 + \beta_1)(\alpha_2 + \beta_2)$)=
= ( $\alpha_1^2 + 2\alpha_1\beta_1 + \beta_1^2$), ( $\alpha_1\alpha_2 + \alpha_1\beta_2 + \beta_1\alpha_2 + \beta_1\beta_2$)

and now I am having trouble proving or disproving if this equals T(a) + T(b).

Also, I would like to point out that a map can fulfill one condition but not the other.
It can, but it would not be a linear transformation
Correct?

4. Originally Posted by jayshizwiz
Thanks for such a quick response. I must've edited my post 5 times while you answered, so my original post is a bit different than what you read.

I can see. Your technique is much better than my book's technique.

you did:
and got
as in, you did both sides at once to prove they are unequal.

The way my book answers the questions is by starting with the first step: T(a+b), doing all the work from that, and finally coming to the conclusion that it equals T(a) + T(b). For example, the way I tried to answer was like this:
T(a+b) = T(\alpha_1 + \beta_1, \alpha_2 + \beta_2)=
=((\alpha_1 + \beta_1)^2, (\alpha_1 + \beta_1)(\alpha_2 + \beta_2))=
= (\alpha_1^2 + 2\alpha_1\beta_1 + \beta_1^2), (\alpha_1\alpha_2 + \alpha_1\beta_2 + \beta_1\alpha_2 + \beta_1\beta_2)

and now I am having trouble proving or disproving if this equals T(a) + T(b).

It can, but it would not be a linear transformation
Correct?
Yes. My point is that what you are trying to do might have been fruitless...!

I am a tad confused with what your problem is. Can you not just do what I did above?

5. Sorry, again I edited it while you were responding. I should probably click post preview before I click post.

The point is I didn't know how to go from

(), () to T(a) + T(b).

If I just would've opened up T(a) + T(b), I would've seen that there are no 2alphabeta things in it, which means it is not a linear combination.

My book never opens up T(a) +T(b). They just do everything in T(a+b) and reorganize to show that it equals T(a) +T(b).

It's kinda hard to explain, but the point is, that this thread is solved!

Thanx!!

6. Actually, it's almost a solved thread. Just one thing remaining...

is there latex code for ǂ ???

7. Originally Posted by jayshizwiz
My book never opens up T(a) +T(b). They just do everything in T(a+b) and reorganize to show that it equals T(a) +T(b).
...but, it doesn't equal T(a)+T(b)...

8. ...but, it doesn't equal T(a)+T(b)...
Correct, so I guess that's why the book didn't use that method, it just inserted numbers and proved that it is not true...

So thanks for showing me a different way to answer this type of question instead of just inserting numbers.

9. You answered the thread while I was in the middle of posting something else, so you may have skipped over it:

is there latex code for the symbol ǂ ???

10. Originally Posted by jayshizwiz
You answered the thread while I was in the middle of posting something else, so you may have skipped over it:

is there latex code for the symbol ǂ ???
Yeah, \neq. I find this site quite useful.