Linear Transformation of R2 to R1: Determining Linearity of f(x,y)=xy

[says]

Homework Statement

R2 to R1 f(x,y)=xy
Determine if the transformation is linear or not

Homework Equations

T(V1+V2) = T(V1) + T(V2)
T(cV1) = Tc(V1)

The Attempt at a Solution

If the function f(x,y) = xy we can define another function f(a,b)=ab
Therefore, f(x,y) = f(a,b), so xy=ab, so all variables could = each other...

T(x*a + y*b) = T(xy) + T(ab)

Txa+Tyb = Txy + Tab

I'm not really sure if I'm on the right track here. Can anyone help me? I need to know the process for finding whether a transformation is linear or not. Usually, I write another function with a vector similar to the original one, in this case it's ab, then i prove it's one-to-one, then try and prove the other 2 equations. Linear transformations are so difficult and it's so hard to find a good resource online that teaches it simply!

 

[RUber]

For your first property, you have vectors (x,y) and (a,b). If the transformation f were linear, then you could say that f ( (x,y)+ (a,b) ) = f(x,y) + f(a,b).
This is not about whether it could be true, it is about whether or not it is always true.
Remember that the sum of two vectors (x,y) + (a,b) = (x+a, y+b).

For the scalar multiple, that is c(x,y) = (cx, cy) . So you would test if f(c(x,y) ) = c f(x,y).

You should find that neither of these hold true.

 

[Mark44]

Homework Statement

R2 to R1 f(x,y)=xy
Determine if the transformation is linear or not

Homework Equations

T(V1+V2) = T(V1) + T(V2)
T(cV1) = Tc(V1)

The Attempt at a Solution

If the function f(x,y) = xy we can define another function f(a,b)=ab
Therefore, f(x,y) = f(a,b), so xy=ab, so all variables could = each other...

Why are you doing this? You have not defined another function. All you have done is use different input arguments.

T(x*a + y*b) = T(xy) + T(ab)

Txa+Tyb = Txy + Tab

?? Use the function you started with -- f. You are doing nothing but confusing yourself by switching to T.
There is a further confusion in not being clear what x and y represent. When you write T(x*a + y*b), x and y are apparently vectors in $\mathbb{R}^2$, but in the function definition -- f(x, y) = xy -- x and y are the real number components of a vector in $\mathbb{R}^2$.

You absolutely need to be clear on the difference between a vector and its components.

I'm not really sure if I'm on the right track here. Can anyone help me? I need to know the process for finding whether a transformation is linear or not. Usually, I write another function with a vector similar to the original one, in this case it's ab, then i prove it's one-to-one, then try and prove the other 2 equations. Linear transformations are so difficult and it's so hard to find a good resource online that teaches it simply!

Let $\vec{u} = <u_1, u_2>$ and $\vec{v} = <v_1, v_2>$ be vectors in $\mathbb{R}^2$
What is $f(\vec{u})$?
What is $f(c\vec{u})$?
What is $f(a\vec{u} + b\vec{v})$?

 

[says]

I like to make things look familiar. I know f and T are the same thing, but I prefer to use T because I'm doing a linear transformation.
I need to prove f(x,y)=xy is linear

I do that by showing f(V1 + V2) = f(V1) + f(V2) but in the problem statement I only have one vector: xy, so I created a second vector, ab, so I can prove the equation.

 

[says]

The first additive rule for linear transformations states that T(x+y) = T(x) + T(y)

In just about every problem I've come across I've only been given one vector. x+y, X+1, and I read the textbook and it says to define another vector so you can prove it's one-to-one, and that the additive and cumulative qualities hold.

 

[Mark44]

I like to make things look familiar. I know f and T are the same thing, but I prefer to use T because I'm doing a linear transformation.

You're cluttering up your work by using different names than are in the problem. This clutter has the downside of possibly confusing you.

I need to prove f(x,y)=xy is linear

I do that by showing f(V1 + V2) = f(V1) + f(V2) but in the problem statement I only have one vector: xy, so I created a second vector, ab, so I can prove the equation.

xy is NOT a vector. It is a number.
As I said in a previous post, you need to be able to distinguish between the components of a vector (x and y) and the vector itself: $\begin{bmatrix} x \\ y\end{bmatrix}$

The first additive rule for linear transformations states that T(x+y) = T(x) + T(y)

In just about every problem I've come across I've only been given one vector. x+y, X+1, and I read the textbook and it says to define another vector so you can prove it's one-to-one, and that the additive and cumulative qualities hold.

cumulative?
Are you talking about multiplication of a vector by a scalar?
See what I wrote at the end of post #3.

 

[says]

Let u⃗ =<u1,u2> and v⃗ =<v1,v2> be vectors in R2
What is f(u⃗ )?
What is f(cu⃗ )?
What is f(au⃗ +bv⃗ )?

f(u) = u1*u2
f(cu) = f(c(u1*u2))
f(a(u1*u2) + b(v1*v2))

I don't understand what the a and b is in the last bit?

f(x,y)=xy is a function, and NOT a vector.

Does this mean, for this transformation to be linear:

f(V1+V2) = f(x,y) = xy
f(V1)+f(V2) = f(x,y) = xy
f(ca) = fc(a) = f(x,y) = xy

?

 

[says]

Do I then have to create another function that is equivalent to f(x,y)=xy to prove if it's linear?
I've stopped using V because it's making me think of vectors and I've started using f! W1 = f(x,y) = xy
W2 = f(a,b) = ab

f(W1+W2) = f(W1) + f(W2)

(x+a) * (y+b) ≠ x*y + a*b

I checked by using a=1 b=2 x=3 y=4

(3+1)*(4*2) ≠ 3*4 + 1*2
21 ≠ 14
Therefore, transformation is not linear.
Is this it? I think I'm finally starting to get it, maybe...

 

[Mark44]

Let u⃗ =<u1,u2> and v⃗ =<v1,v2> be vectors in R2
What is f(u⃗ )?
What is f(cu⃗ )?
What is f(au⃗ +bv⃗ )?

f(u) = u1*u2

Yes, assuming that u = <u₁, u₂>

f(cu) = f(c(u1*u2))

No. u is a vector, as is cu. Calculate cu first (IOW, find the components of cu), then apply your function to it.

f(a(u1*u2) + b(v1*v2))

I don't understand what the a and b is in the last bit?

Scalars

f(x,y)=xy is a function, and NOT a vector.

f is a function. xy is a number.

Does this mean, for this transformation to be linear:

f(V1+V2) = f(x,y) = xy
f(V1)+f(V2) = f(x,y) = xy
f(ca) = fc(a) = f(x,y) = xy
?

None of these makes any sense.

 

[says]

I think my post above this with the original problem makes more sense

 

[Mark44]

Do I then have to create another function that is equivalent to f(x,y)=xy to prove if it's linear?

No, you only have the one function, f.

I've stopped using V because it's making me think of vectors and I've started using f!

Good.

W1 = f(x,y) = xy
W2 = f(a,b) = ab

This doesn't make much sense. In part you're saying that W1 = xy and W2 = ab, which means that W1 and W2 are numbers.
In the line below you have f(W1 + W2), which implies that W1 and W2 are vectors. Which is it? They can't be both vectors and numbers.

f(W1+W2) = f(W1) + f(W2)

(x+a) * (y+b) ≠ x*y + a*b

I checked by using a=1 b=2 x=3 y=4

(3+1)*(4*2) ≠ 3*4 + 1*2
21 ≠ 14
Therefore, transformation is not linear.
Is this it? I think I'm finally starting to get it, maybe...

I don't think so.

Let u = <u₁, u₂>, and v = <v₁, v₂>
Hopefully it's clear that when you see u, we're talking about a vector, but when you see u₂, we're talking about one of the components of u.

What is u + v?
What is f(u + v)?
What is cu? (c is a scalar)
What is f(cu)?

 

[says]

u + v = u₁*v₁ + u₂*v₂
f(u + v) = u₁*u₂ + v₁*v₂
cu = c(u₁*u₂)
f(cu) = c*u₁*u₂

 

[Mark44]

u + v = u₁*v₁ + u₂*v₂

NO! This is just plain old vector addition!

f(u + v) = u₁*u₂ + v₁*v₂
cu = c(u₁*u₂)

NO! This is nothing more than multiplication of a vector by a scalar.

f(cu) = c*u₁*u₂

You're not going to get this if you don't understand the very basic concepts of vector addition and scalar multiplication.
If $\vec{u} = <u_1, u_2>$ and $\vec{v} = <v_1, v_2>$ then $\vec{u} + \vec{v} = <u_1 + v_1, u_2 + v_2>$
Also, if c is a scalar, then $c\vec{u} = <cu_1, cu_2>$
I hope that this looks familiar...

 

[says]

where does f(x,y) = xy come into this problem though? I understand the mistakes I made above, but they were because I'm trying to incorporate the original function into what you wrote...

 

[Mark44]

where does f(x,y) = xy come into this problem though? I understand the mistakes I made above, but they were because I'm trying to incorporate the original function into what you wrote...

It doesn't come into play yet, because you're still having problems working with vectors -- i.e., adding two vectors and multiplying a vector by a scalar. Until you get these concepts down, you won't have any success working with functions that take vectors as arguments.

 

[says]

u + v = (u1+v1 , u2+v2)
f(u + v) = (u1+u2 + v1+v2)
cu = c(u1,u2)= (cu1,cu2)
f(cu) = (cu1,cu2)

 

[Mark44]

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

Yes

f(u + v) = (u1+u2 + v1+v2)

No. Look at how f is defined.

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

Yes

f(cu) = (cu1,cu2)

No. Look at how f is defined.

 

[says]

f(x,y)=xy

f(u+v) = u1*u2+v1*v2

f(cu) = f(c(u1*u2)

 

[says]

Made an edit to the above, which you already said was incorrect -- I'm now sure how though...
If f(x,y) = xy

and i substitute the components of vector u (u1,u2) and vector v (v1,v2) into this function, then I get the answer I got.

 

[Mark44]

f(x,y)=xy

f(u+v) = u1*v1+u2*v2

f(cu) = f(c(u1*u2)

Both of the above are wrong. You are skipping steps like crazy, and writing stuff without thinking about it.

You already told me what u + v was; namely u + v = $<u_1 + v_1, u_2 + v_2>$, and what cu was; namely, cu = $<cu_1, cu_2>$
What is f($<u_1 + v_1, u_2 + v_2>$)?
What is f($<cu_1, cu_2>$)?

 

[says]

f(x,y) = xy

f(u1+v1,u2+v2) = (u1+v1)*(u2+v2)
f(cu1,cu2) = cu1*cu2

 

[Ray Vickson]

f(x,y)=xy

f(u+v) = u1*u2+v1*v2

f(cu) = f(c(u1*u2)

The first one is wrong; the second one just displays the vector cu explicitly, but without actually saying anything.

You are told that if you have a vector $\vec{v} = (\text{thing 1}, \text{thing 2})$ then $f(\vec{v}) = \text{thing 1} \times \text{thing 2}$. OK so far?

Now, what would you get if $\text{thing 1} = u_1 + v_2$ and $\text{thing 2} = u_2 + v_2$? What would you get if $\text{thing 1} = c u_1$ and $\text{thing 2} = c u_2$?

 

[Mark44]

f(x,y) = xy

f(u1+v1,u2+v2) = (u1+v1)*(u2+v2)
f(cu1,cu2) = cu1*cu2

YES!

So f(u + v) = f($<u_1 + v_1, u_2 + v_2>) = (u_1 + v_1) \cdot (u_2 + v_2)$
Is this equal to f(u) + f(v)?

And f(cu) = f($<cu_1, cu_2>) = cu_1 \cdot cu_2$
Is this equal to cf(u)?

 

[Mark44]

The first one is wrong; the second one just displays the vector cu explicitly, but without actually saying anything.

You are told that if you have a vector $\vec{v} = (\text{thing 1}, \text{thing 2})$ then $f(\vec{v}) = \text{thing 1} \times \text{thing 2}$. OK so far?

Now, what would you get if $\text{thing 1} = u_1 + v_2$ and $\text{thing 2} = u_2 + v_2$? What would you get if $\text{thing 1} = c u_1$ and $\text{thing 2} = c u_2$?

Ray, he corrected these in post #21.

 

[says]

f(u + v) = f(<u1+v1,u2+v2>)=(u1+v1)⋅(u2+v2)

f(u) = u1*u2
f(v) = v1*v2

No, this is not equal.

f(cu) = f(<cu1,cu2>)=cu1⋅cu2

c(u1*u2)

This is not equal either. Therefore it is not a linear transformation.

 

[Ray Vickson]

Ray, he corrected these in post #21.

OK. That seems to be another one of those cases where some things appear on my screen only after I press the 'enter' key.

 

[Mark44]

f(u + v) = f(<u1+v1,u2+v2>)=(u1+v1)⋅(u2+v2)

f(u) = u1*u2
f(v) = v1*v2

No, this is not equal.

I believe you, but you're not very convincing here. Show me why (u1+v1)⋅(u2+v2) is not equal to u1*u2 + v1*v2.

f(cu) = f(<cu1,cu2>)=cu1⋅cu2

c(u1*u2)

This is not equal either. Therefore it is not a linear transformation.

Again, show me why cu1*cu2 is not equal to c(u1 * u2).

 

[says]

I gave each variable values and subbed them in.
u1=1 u2=2 v1=3 v2=4 c=5

(u1+v1)⋅(u2+v2) ≠ (u1*u2) + (v1*v2)
(1+3)*(2+4) ≠ (1*2)+(3*4)
4*6 ≠ 2+12
24 ≠ 12
cu1⋅cu2 ≠ c(u1*u2)
5(1)*5(2) ≠ 5(1*2)
50 ≠ 10

 

[Mark44]

I gave each variable values and subbed them in.
u1=1 u2=2 v1=3 v2=4 c=5

(u1+v1)⋅(u2+v2) ≠ (u1*u2) + (v1*v2)
(1+3)*(2+4) ≠ (1*2)+(3*4)
4*6 ≠ 2+12
24 ≠ 12
cu1⋅cu2 ≠ c(u1*u2)
5(1)*5(2) ≠ 5(1*2)
50 ≠ 10

I would be much more convinced (and your instructor would as well) if you could show my why (u1+v1)⋅(u2+v2) ≠ (u1*u2) + (v1*v2), and why cu1⋅cu2 ≠ c(u1*u2), without resorting to specific number examples.

For the first one, expand (u1+v1)⋅(u2+v2) and show that this is not equal to u1u2 + v1v2.
The second one is very easy to show.

 

[Mark44]

You're almost done with this problem, and you have made a lot of improvement along the way. When you're done here, you might want to revisit the other question you posted yesterday. It might make more sense now.

 

[says]

Yep! I'm going to do exactly that! Thanks a lot for your help @Mark44

 

[says]

f(u + v) = f(<u1+v1,u2+v2>)=
f(u) = u1*u2
f(v) = v1*v2
f(cu) = f(<cu1,cu2>)=cu1⋅cu2(u1+v1)⋅(u2+v2) = u1u2+u1v2+v1u2+v1v2 ≠ u1*u2 + v1*v2

cu1⋅cu2 ≠ c*u1*u2

 

[Mark44]

f(u + v) = f(<u1+v1,u2+v2>)=
f(u) = u1*u2
f(v) = v1*v2
f(cu) = f(<cu1,cu2>)=cu1⋅cu2(u1+v1)⋅(u2+v2) = u1u2+u1v2+v1u2+v1v2 ≠ u1*u2 + v1*v2

This looks good. Even better would be:
f(u + v) = f(<u1+v1,u2+v2>) = (u1+v1)⋅(u2+v2) = u1u2+u1v2+v1u2+v1v2 ≠ u1*u2 + v1*v2 = f(u) + f(v)

cu1⋅cu2 ≠ c*u1*u2

because c²u1⋅u2 ≠ c*u1*u2 unless c = 0 or c = 1.