One more Linear Transformation

1. Dec 3, 2015

says

1. The problem statement, all variables and given/known data
I've posted a few of these recently. I have one question about this problem -- hopefully my calculations are correct.

f: R2 to P1, f(a,b)=b+a2x

Is this a linear transformation?

2. Relevant equations
f(u+v) = f(u) + f(v)
f(cu) = cf(u)

where u and v are vectors in R2 and c is a scalar.

3. The attempt at a solution
let u=(u1,u2)
let v=(v1,v2)

f(u+v) = ((u2+v2) + (u12x+v12x))

f(u) + f(v) = (u2 + u12x) + (v2 + v12x)

So f(u+v) = f(u) + f(v) (First condition met)

c(u1,u2) = (cu1,cu2)

f(cu) = cu2 + cu12x

cf(u) = c(u2 + u12x)
= cu2 + cu12x

Second condition met.

However, my question is -- The question asks is this a linear transformation from R2 to p1. If my calculations are correct, both conditions are met, but the polynomial has a degree of 2, not 1, so it's not a linear transformation from R2 to p1.

2. Dec 3, 2015

DuckAmuck

The squared u1 and v1 will have a c^2 coefficient, not c. So second condition is not met.
if v1 -> c*v1
then v1^2 -> c^2*v1^2

3. Dec 3, 2015

Samy_A

The way you compute $f(u+v)$ is not correct: in general, $(u_1+v_1)² \neq {u_1}²+{v_1}²$.
This is not correct, you should have a $c²{u_1}²x$ in the right hand side expression.
The polynomial has a degree of 1 in $x$.

4. Dec 3, 2015

says

I see what you mean with the c2 on the right hand side of the equation, I forgot to put a parenthesis around it (cu12)x

Just looking at the first part now. Thanks for your replies.

5. Dec 3, 2015

says

f(u+v) = [(u2+v2)+(u1+v1)2x]

I can see now (with @DuckAmuck's reply) that the second condition is not met.

Oh yes, you are correct with the polynomial having a degree of 1 as well! I got my u,v and x variables mixed up there!

I'm still making a few mistakes, but I'm a lot more confident with these now than I was a couple of days ago.

6. Dec 3, 2015

Samy_A

Working that out will show that the first condition isn't met either.

The never ending joy of learning.

7. Dec 3, 2015

says

f(u+v) = [(u2+v2)+(u1+v1)2x]

f(u) + f(v) = u2 + u12x + v2 + v12x

That condition isn't met either! Totally skipped over that one.

8. Dec 3, 2015

Staff: Mentor

Pretend that you're trying to convince an idiot, and show why the first line above isn't equal to the second line.

9. Dec 3, 2015

says

*Convincing myself*

f(u+v) = [(u2+v2)+(u1+v1)2x]

f(u) + f(v) = u2 + u12x + v2 + v12x

(u1+v1)2 ≠ u12 + v12

10. Dec 3, 2015

Staff: Mentor

Better:
f(u+v) = (u2+v2)+(u1+v1)2x = $u_2 + v_2 + (u_1^2 + 2u_1v_1 + v_1^2)x$
f(u) + f(v) = u2 + u12x + v2 + v12x = $u_2 + v_2 + (u_1^2 + v_1^2)x$
Since $u_1^2 + v_1^2 \ne u_1^2 + 2u_1v_1 + v_1^2$, if $u_1 \ne v_1$, then f(u + v) $\ne$ f(u) + f(v)

Last edited: Dec 3, 2015