One more Linear Transformation

[says]

Homework Statement

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?

Homework 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.

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.

 

[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

 

[Samy_A]

Homework Statement

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?

Homework 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.

The Attempt at a Solution

let u=(u1,u2)
let v=(v1,v2)

f(u+v) = ((u2+v2) + (u1²x+v1²x))

The way you compute $f(u+v)$ is not correct: in general, $(u_1+v_1)² \neq {u_1}²+{v_1}²$.

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

This is not correct, you should have a $c²{u_1}²x$ in the right hand side expression.

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.

The polynomial has a degree of 1 in $x$.

 

[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.

 

[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.

 

[Samy_A]

f(u+v) = [(u2+v2)+(u1+v1)²x]

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

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.

The never ending joy of learning.

 

[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.

 

[Mark44]

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.

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

 

[says]

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

*Convincing myself*

f(u+v) = [(u2+v2)+(u1+v1)²x]

f(u) + f(v) = u2 + u1²x + v2 + v1²x

(u1+v1)² ≠ u1² + v1²

 

[Mark44]

*Convincing myself*

f(u+v) = [(u2+v2)+(u1+v1)²x]

f(u) + f(v) = u2 + u1²x + v2 + v1²x

(u1+v1)² ≠ u1² + v1²

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)