# Linear Transformations p2 to R2

1. Nov 30, 2015

### says

1. The problem statement, all variables and given/known data
f: p2 to R2, f(ax2+bx+c) = (a+b, b+c) = V1

2. Relevant equations
Create a p2 to R2 polynom and R2 equation that is equivalent to the above statement, so:

f(dx2+ex+g) = (d+e, e+g) = V2

Therefore a=d b=e c=g

f(ax2+bx+c) = (a+b, b+c) = f(dx2+ex+g) = (d+e, e+g)

3. The attempt at a solution

T(V1+V2) = T(V1) + T(V2)

T(V1+V2) = T(V1) +T(V2) for a linear transformation

T(V1+V2) =T(a+d, b+e) + (b+e, c+g) = T(a+d+b+e, b+e+c+g)

T(V1) + T(V2) = (a+b, b+c) + (d+e, e+g) = T(a+b+d+e, b+c+e+g)

T(a+d+b+e, b+e+c+g) = T(a+b+d+e, b+c+e+g) FIRST CONDITION MET!

T(wV1) = Tw(V1)

T(w(a+b,b+c) = Tw(a+b,b+c)
T(wa+wb, wb+wc) = (Twa+Twb, Twb +Twc)
(Twa+Twb, Twb +Twc) = (Twa+Twb, Twb +Twc) SECOND CONDITION MET!

This is a linear transformation from p2 to R2.

I was hoping someone could help me out just to make sure I'm on the right track. I get a bit confused with vectors and column vector notation in linear algebra.

2. Nov 30, 2015

### HallsofIvy

The attempt at a solution to what problem? Are you trying to show that this is a linear transformation? You didn't say that!

This makes no sense. V1 and V2 are the vectors above so T(V1+ V2) makes sense but then T(a+d, b+e) does not. T is not applied to pairs of numbers.
You are not distinguishing between the linear transformation, T, you are applying and the result of applying it.

Same criticism here. Certainlt "(Twa+ Twb, Twb+ Twc)" makes no sense and saying it is equal to itself is meaningless.

3. Nov 30, 2015

### says

Yes, sorry I'm trying to show that it is a linear transformation.

So T (V1 + V2) = T(a+b + d+e) in R2 = T(a+b, d+e) ?

4. Nov 30, 2015

### Staff: Mentor

The work you show is very confusing. In post 1, you have a function (transformation) named f, where f: P2 → R2. After that, you don't use f any more, but change to T.

I'm assuming that f and T represent the same transformation.

With regard to your equation just above, T(V1 + V2) makes no sense. From your first post, V1 is apparently a vector in R2, so it makes no sense to use it as an input to the transformation.

T(a + b + d + e) also makes no sense. a + b + d + e represents a single number. The domain of T is polynomials of degree less than or equal to 2. .
T(a + b, d + e) also makes no sense for the same reason.

Try to be more careful in what you write. According to your definition in post 1, T(ax2 + bx + c) = <a + b, b + c>. T (or f as you originally used) is a map from polynomials of degree 2 to vectors in R2.

One of the things you need to prove is that T(p1 + p2) = T(p1) + T(p2), where p1 and p2 are arbitrary polynomials of degree 2.

When you write stuff like T(V1 + V2), it would appear that you don't understand how this transformation works.

Last edited: Nov 30, 2015
5. Dec 1, 2015

### says

Yeah, I'm so confused with linear transformations, image, range, kernel, etc. and it's so hard to find something online that goes through the process step by step without being inundated with a whole bunch of new terms...

6. Dec 1, 2015

### Staff: Mentor

Your textbook should have definitions of all of these terms.

For this problem, though, image, range, and kernel are not relevant. The only term of those you listed that is relevant is "linear transformation."

7. Dec 2, 2015

### says

f: p2 to R2, f(ax2+bx+c) = (a+b, b+c)

p1: ax2+bx+c
p2: dx2+ex+g

(p1+p2) = ((a+d)+(b+e)+(c+g))
f(p1+p2) = (a+b+d+e, b+c+e+g)

f(p1) = (a+b, b+c)
f(p2) = (d+e, e+g)

f(p1+p2) = f(p1) + f(p2)
f(p1+p2) = (a+b+d+e, b+c+e+g) = (a+b, b+c) + (d+e, e+g)

First condition met!

replacing the term c with the letter j.

f(jp1) = j(a+b, b+c)
fj(p1) = j(ax2+bx+c)

j(a+b, b+c) ≠ j(ax2+bx+c)

Second condition not met, unless there is some other way to write the j(ax2+bx+c) polynomial which I'm unaware of.

8. Dec 2, 2015

### HallsofIvy

No, (p1+ p2)= (a+d)x2+ (b+ e)x+ (c+ g)

"fj(p1)" makes no sense. It looks like you have a new function "fj" that you have not defined. If you mean "j times f(p1)" that would be jf(p1) or f(p1)j.

It is difficult to understand what you are doing because your notation is so bad. What you have here is NOT that f(jp1) is not equal to jf(p1) but that jf(p1) is not equal to jp1- and that has nothing to do with f being a linear transformation.

Given p1= ax2+ bx+ c then jp1= ajx2+ bjx+ cj and then f(jp1)= (aj+ bj, bj+ cj). Of course, jf(p1)= j(a+ b, b+ c)= (j(a+b),j(b+ c))= (aj+ bj, bj+ cj).[/sup][/sup]

Last edited by a moderator: Dec 2, 2015
9. Dec 2, 2015

### Staff: Mentor

You're skipping steps again. p1 + p2 is the sum of the two polynomials, so that ( p1 + p2)(x) = (a + d)x2 + (b + e)x + (c + g)
Don't write the above, since that's what you're trying to show. Calculate f(p1 + p2) and f(p1) + f(p2) separately. If these two expressions turn out to be equal, you will have established that f(p1+p2) = f(p1) + f(p2).
Again, you're skipping steps, plus your notation is hard to understand (HallsOfIvy couldn't follow it).

Instead of j, use a different letter for the scalar, say, k.
You want to show that f(k*p1) = k * f(p1).
Do this by calculating f(k * p1), and then k * f(p1).

For the first one, don't skip steps. Calculate k * p1, and then f(k * p1). Compare that to k * f(p1) to see whether they are equal.

10. Dec 2, 2015

### says

Sorry, I knew they were the sum of the two polynomials, i just didn't write them in.

(p1 + p2) = (a + d)x2+ (b + e)x + (c + g)

f(p1) = (ax2+bx, bx+c)
f(p2) = (dx2+ex, ex+g)

f(p1+p2) = (ax2+dx2+bx+ex, bx+ex+c+g)

(ax2+dx2+bx+ex, bx+ex+c+g) = (ax2+bx, bx+c) + (dx2+ex, ex+g)

First condition met.

Last edited by a moderator: Dec 2, 2015
11. Dec 2, 2015

### says

p1=ax2+bx+c

k*p1 = kax2+kbx+kc

f(k*p1) = (kax2+kbx, kbx+kc)

f(p1) = (ax2+bx, bx+c)

k*f(p1) = (kax2+kbx, kbx+kc)

Second condition met.

12. Dec 2, 2015

### Staff: Mentor

OK, except the left side is really (p1 + p2)(x).
No. f(p1) is a vector in R2, as is f(p2). x should not appear on the right sides of these two equations.
You need to recalculate f(p1) and f(p2).
No.
You have (p1 + p2)(x) = $(a + d)x^2 + (b + e)x + (c + g)$, and $f(rx^2 + sx + t) = <r + s, s + t>$, based on how f is defined in post #1.
Try again at calculating f( (p1 + p2)(x)).

13. Dec 2, 2015

### Staff: Mentor

No. You're still skipping steps.
f(k * p1) = $f(kax^2 + kbx + kc)$ = ?
What does f do to an input polynomial? Again, x should NOT appear in the output of this function.
Nope.

Last edited: Dec 2, 2015
14. Dec 2, 2015

### says

f(k*p1) = (ka+kb, kb+kc)

f(p1) = (a+b, b+c)

k*f(p1) = k(a+b, b+c) = (((k(a+b),(k(b+c)))

15. Dec 2, 2015

### Staff: Mentor

So is f(k*p1) = k * f(p1)?

What about the other part, showing that f(p1 + p2) = f(p1) + f(p2)?

16. Dec 3, 2015

### says

Yes, the second condition is met.

f(ax2+bx+c+dx2+ex+g) = (a+d+b+e, b+e+c+g)

f(p1) = (a+b, b+c)
f(p2) = (d+e, e+g)

f(p1) + f(p2) = (a+b, b+c) + (d+e, e+g) = (a+b+d+e, b+c+e+g)

The first condition is met.

17. Dec 3, 2015

### Staff: Mentor

Better would be to actually show your work that supports saying that f(k*p1) = k * f(p1).
Yes. You could summarize what you found by saying that you have shown that f(p1 + p2) = f(p1) + f(p2)