# Homework Help: Linear transformation

1. Dec 1, 2015

### says

1. The problem statement, all variables and given/known data
R2 to R1 f(x,y)=xy
Determine if the transformation is linear or not

2. Relevant equations
T(V1+V2) = T(V1) + T(V2)
T(cV1) = Tc(V1)
3. 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!

2. Dec 1, 2015

### 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.

3. Dec 1, 2015

### Staff: Mentor

Why are you doing this? You have not defined another function. All you have done is use different input arguments.
?? 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.
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 $f(a\vec{u} + b\vec{v})$?

4. Dec 1, 2015

### 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.

5. Dec 1, 2015

### 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.

6. Dec 1, 2015

### Staff: Mentor

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

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: $\begin{bmatrix} x \\ y\end{bmatrix}$
cumulative?
Are you talking about multiplication of a vector by a scalar?
See what I wrote at the end of post #3.

7. Dec 1, 2015

### 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

?

8. Dec 1, 2015

### 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...

9. Dec 1, 2015

### Staff: Mentor

Yes, assuming that u = <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.
Scalars
f is a function. xy is a number.
None of these makes any sense.

10. Dec 1, 2015

### says

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

11. Dec 1, 2015

### Staff: Mentor

No, you only have the one function, f.
Good.
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.
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)?

12. Dec 1, 2015

### says

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

13. Dec 1, 2015

### Staff: Mentor

NO! This is just plain old vector addition!
NO! This is nothing more than multiplication of a vector by a scalar.
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...

14. Dec 1, 2015

### 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...

15. Dec 1, 2015

### Staff: Mentor

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.

16. Dec 1, 2015

### says

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

17. Dec 1, 2015

### Staff: Mentor

Yes
No. Look at how f is defined.
Yes
No. Look at how f is defined.

18. Dec 1, 2015

### says

f(x,y)=xy

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

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

19. Dec 1, 2015

### 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.

20. Dec 1, 2015

### Staff: Mentor

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($<cu_1, cu_2>$)?

21. Dec 1, 2015

### says

f(x,y) = xy

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

22. Dec 1, 2015

### Ray Vickson

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 $\text{thing 2} = c u_2$?

23. Dec 1, 2015

### Staff: Mentor

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)?

24. Dec 1, 2015

### Staff: Mentor

Ray, he corrected these in post #21.

25. Dec 1, 2015

### 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.