[SOLVED] Linear Transformations

**jayshizwiz** · May 31st 2010, 02:16 AM

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.

**Swlabr** · May 31st 2010, 02:37 AM

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.

**jayshizwiz** · May 31st 2010, 03:03 AM

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?

**Swlabr** · May 31st 2010, 03:10 AM

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?

**jayshizwiz** · May 31st 2010, 03:20 AM

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!!

**jayshizwiz** · May 31st 2010, 03:23 AM

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

is there latex code for ǂ ???

**Swlabr** · May 31st 2010, 03:23 AM

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)...

**jayshizwiz** · May 31st 2010, 03:26 AM

...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.

**jayshizwiz** · May 31st 2010, 03:40 AM

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 ???

**Swlabr** · May 31st 2010, 04:18 AM

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.

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