# Transformations R 3 to R 2

1. Mar 11, 2010

### Dustinsfl

1. The problem statement, all variables and given/known data
Determine if this is a linear transformation from R3 to R2

2. Relevant equations
L(x) = (1 + x1, x2)

3. The attempt at a solution
Whenever I perform addition and scalar multiplication, I obtain this is closed under both. The book says this isn't a transformation though.

2. Mar 11, 2010

### tiny-tim

Hi Dustinsfl!

Is L(2x) = 2L(x) ?

3. Mar 11, 2010

### Dustinsfl

I don't see how that doesn't work.

(2 + 2*x1, 2*x2) = 2*(1 + x1, x2)

4. Mar 11, 2010

### tiny-tim

No, L(2x) = L((2x1,2x2,2x3)) = (1 + 2x1,2x2)

5. Mar 11, 2010

### Dustinsfl

Based on that then, the other problems the book says are transformations would fail then too.

6. Mar 11, 2010

### Dustinsfl

Here is a similar problem which is a transformation:
L(x)= (x2, x3)

Since there is no x1, this one would also fail on that premise but the book states this one is a transformation.

7. Mar 11, 2010

### Staff: Mentor

I think you are missing something. To show that a transformation L is linear, show that L(x + y) = L(x) + L(y), and that L(cx) = cL(x).

For L(x) = (x2, x3), where x = (x1, x2, x3),

L(x + y) = L(x1 + y1, x2 + y2, x3 + y3) = ( x2 + y2, x3 + y3) = (x2, x3) + (y2, y3) = L(x) + L(y)

L(cx) = L(cx1, cx2, cx3) = (cx2, cx3) = c(x2, x3) = cL(x).

Therefore, L is a linear transformation.

8. Mar 11, 2010

### Dustinsfl

So looking at a 2 dimension example, this still doesn't make any sense. You are saying x=(x1, x2).

If L(x) = (-x1, x2), then verifying scalar multiplication we would obtain (cx1, cx2)=c(x1, x2) which doesn't equal c*L(x).

The problem with this conclusion is that this example I just did is a linear transformation and is closed under addition and multiplication.

9. Mar 11, 2010

### Staff: Mentor

No, I'm not. I specifically said for that problem, that x = (x1, x2, x3).
You're leaving out some pretty important stuff, and are arriving at incorrect conclusions. The whole story for this new problem is
L(cx) = L(cx1, cx2) = (-cx1, cx2) = c(-x1, x2) = cL(x).
The problem with your conclusiong was that it was wrong. Do you see now why it was wrong?

10. Mar 11, 2010

### Dustinsfl

So if that is the case,looking at my original post transformation, then how is it not a linear transformation?

11. Mar 11, 2010

### Staff: Mentor

Work it through, following my work as a template. For that transformation, and using tiny-tim's hint, what is L(2x)? Is L(2x) = 2L(x)?

12. Mar 11, 2010

### Dustinsfl

Here is where both of your explanations are esoteric. We are saying x=(x1,....,xn).

Next, we are using x times the scalar and then comparing that with L(x). Unless we plug in L(x) or x is equal to L(x), it will never be closed. Hence, I used the this as my example to show that:

"So looking at a 2 dimension example, this still doesn't make any sense. You are saying x=(x1, x2).

If L(x) = (-x1, x2), then verifying scalar multiplication we would obtain (cx1, cx2)=c(x1, x2) which doesn't equal c*L(x).

The problem with this conclusion is that this example I just did is a linear transformation and is closed under addition and multiplication."

Now looking at what you said next you are using L(x) not x.
"You're leaving out some pretty important stuff, and are arriving at incorrect conclusions. The whole story for this new problem is
L(cx) = L(cx1, cx2) = (-cx1, cx2) = c(-x1, x2) = cL(x)."

If we do they same with the original problem, it will be closed.

13. Mar 11, 2010

### Staff: Mentor

Maybe you are confusing a vector x (note the use of bold) with its components, x1, x2, x3, ...

I can't tell what it is that you're not getting, but something is not clicking.

You didn't use L here. The equation (cx1, cx2)=c(x1, x2) is true for any vector in R2, due to the vector space properties of scalar multiplication.

Most of what you said in your last reply I don't understand. Instead of me trying to understand what you meant, see if you can work through what I asked you to do, using the transformation in your first post.

Here, x = (x1, x2, x3), and L(x) = (1 + x1, x2)

L(2x) = ?

14. Mar 11, 2010

### Dustinsfl

I understand what you are saying but the way we addressing this problem makes this one not closed then too.

x=(x1, x2) and L(x)=(-x1, x2)

L(2*x)=(2*x1, 2*x2)

Which doesn't equal L(x) when 2 is factored since x1 is positive.

15. Mar 11, 2010

### Staff: Mentor

Again, you are not using the transformation. This is what you should have.

L(2*x) = L(2*x1, 2*x2) = (-2x1, 2x2) = 2(-x1, x2) = 2L(x).

It's a simple matter to show that this works for any scalar c.

L(c*x) = L(c*x1, c*x2) = (-cx1, cx2) = c(-x1, x2) = cL(x).

So this transformation is closed under scalar addition.

Now work the one that tiny-tim and I have been asking you to work.

16. Mar 11, 2010

### Dustinsfl

L(cx)=(cx1, cx2, cx3)=(c(1+x1), cx2)=c(1+x1, x2)=cL(x)

If it isn't suppose to be closed, I don't know how working this one was suppose to verify that.

17. Mar 11, 2010

### vela

Staff Emeritus
This problem has nothing to do with closure. It's just asking you to check if the transformations satisfy the defining properties of a linear transformation.

You're making some really basic math errors. This is just scalar multiplication of a vector:

cX=(cx1,cx2,cx3)

It has nothing to do with L. This is just substitution:

L(cX)=L(cx1,cx2,cx3)

Now use the definition of L to write down what L(cx1,cx2,cx3) is equal to:

L(cx1,cx2,cx3)=(1+cx1,cx2)

Do you see how this is different from what you've written above?

18. Mar 11, 2010

### Staff: Mentor

L(cx)=(cx1, cx2, cx3) $\neq$ (c(1+x1), cx2).

What does L in this transformation do to an input vector? It adds 1 to the first component, keeps the second component, and discards the third component.

Here's what you should have:
L(cx)=(cx1, cx2, cx3)= (1 + cx1, cx2)

Can you still assert that L(cx) = cL(x)?

19. Mar 12, 2010

### tiny-tim

(Mark44 seems to go to bed slightly later than i do! :zzz:)
No!

L(cx) is not equal to (cx1, cx2, cx3) …

it is equal to L(cx1, cx2, cx3).

And then L(cx1, cx2, cx3) is not equal to (c(1+x1), cx2) …

it is equal to (1 + cx1, cx2).

You see?