# Prove the definitions of Linear Transformations

1. Oct 8, 2011

### jinksys

1. The problem statement, all variables and given/known data
Show that 2.1.1 is equivalent to the totality of 2.1.2 and 2.1.3.

2. Relevant equations

3. The attempt at a solution

aTx + bTy = aT(x) + bT(y) = T(ax) + T(by) = T(ax + by) ?

2. Oct 8, 2011

### Dick

It's really pretty easy. Just think clearly about what you have prove. But there is a really nasty typo in (2.1.3). It should read T(ax)=a*T(x). Is that what's confusing you?

3. Oct 8, 2011

### jinksys

I can 'get' that 2.1.1 follows from 2.1.2 and 2.1.3 and visa versa, but I'm not sure how to say it as a proof. I've never been good with proofs.

4. Oct 8, 2011

### Dick

You aren't bothered by (2.1.3) saying T(ax)=a*x?? You should be. But if you aren't good at proofs, maybe you aren't. Change (2.1.3) to T(ax)=a*T(x). The first stage of the proof is show that (2.1.1) implies (2.1.2) and (2.1.3). That's pretty easy. I'll give you a hint for the first part. Put a=1 and b=1 into (2.1.1). What do you conclude?

5. Oct 8, 2011

### jinksys

I meant to say that I understood that T(ax)=aT(x), but only because I have other linear algebra books to help me get through this class.

If a=b=1,

T(ax +by) = T(1x + 1y) = T(1x) + T(1y) = 1T(x) + 1T(y) = T(x) + T(y)

Edit:

Now do I have to do a=a, b=0 to get:

T(ax + by) = T(ax) =aT(x) ?

6. Oct 8, 2011

### Dick

Ok, that's good. So that shows (2.1.1) implies (2.1.2). Can you show it also implies (2.1.3)? The final step of the proof is showing (2.1.2) AND (2.1.3) imply (2.1.1).

7. Oct 8, 2011

### jinksys

I'm not sure if you saw my edit or not, so I'll just give it its own post.

To show 2.1.3, do I let a=a and b=0?

T(ax + by) where a=a, b=0:

T(ax + 0y) = T(ax) = aT(x).

Conversely with a=0, b=b:

T(0x + by) = T(by) = bT(y)

8. Oct 8, 2011

### Dick

Sure, that shows T(ax)=a*T(x). So (2.1.1) implies (2.1.3). Now you just have to show (2.1.2) AND (2.1.3) imply (2.1.1). Try it. You are doing fine so far.

9. Oct 8, 2011

### jinksys

I'm stumped, I can see how you would prove that 2.1.1 implies 2.1.2 and that 2.1.1 implies 2.1.3, but I'm not sure how to combine them.

10. Oct 8, 2011

### Dick

Umm. You want to show T(ax+by)=aT(x)+bT(y), right? (2.1.2) says T of the sum of two vectors is T(first vector)+T(second vector), yes? Are ax and by vectors you could apply (2.1.2) to?

11. Oct 8, 2011

### jinksys

If you let s = ax and t = by,

then T(s + t) = T(s) + T(t) = T(ax) + T(by) = aTx + bTy,

So, T(ax + by) = aTx + bTy ?

Any closer?

12. Oct 8, 2011

### Dick

Way closer. In fact, you finished it. Just say at what point you used (2.1.3).

13. Oct 8, 2011

### jinksys

So letting s=ax and t=by was the right thing to do?
Are there other ways to say 2.1.1 stems from 2.1.2 and 2.1.3?

14. Oct 8, 2011

### Dick

You just did it. T(s+t)=T(s)+T(t)=T(ax)+T(by) by (2.1.2). Now (2.1.3) says T(ax)=aT(x) and T(by)=b*T(y). Doesn't that mean you are done??