# ILinear Transformations / Isomorphs

1. Nov 26, 2015

### says

1. The problem statement, all variables and given/known data
Determine if the following is a linear transformation or not

f : IR2 → IR, f(x, y) = x + y

2. Relevant equations
T(x+y) = T(x) + T(y)
T(ca) = Tc(a)

3. The attempt at a solution
I can't tell you how much I've read, how many youtube videos I've watched over the last couple of hours. I'm so confused with linear transformations.
They are basically functions, only we want to think of them slightly differently and visualise them as moving, squishing, stretching a vector.
Can somebody eli5 (explain like I'm 5) and help me understand linear transformations / isomorphs. I added the example above as a reference.

2. Nov 26, 2015

### Samy_A

Linearity is a property of functions between vector spaces. Of course some functions are linear, others not.
Linearity is defined by the relevant equations you stated. Although you didn't state it, in order for a function to be linear the equations have to be true for all vectors x, y, a and for all (real) numbers c.

Take two functions, f and g
$f: \mathbb R² \to \mathbb R$: $(x,y) \mapsto x+y$
$g: \mathbb R² \to \mathbb R$: $(x,y) \mapsto x²+y$

$g$ is not linear because it doesn't satisfy the relevant equations you stated, and which define what a linear function is.

Let's check it:
$g(1,1)+g(2,3)=1²+1+2²+3=9$
$g((1,1)+(2,3))=g(1+2,1+3)=g(3,4)=3²+4=13$
$9 \neq 13$ , so g doesn't satisfy your first relevant equation.

$g(3(2,4))=g(3*2,3*4)=g(6,12)=36+12=48$
$3g(2,4)=3(2²+4)=3*8=24$
$48 \neq 24$ , so g doesn't satisfy your second relevant equation.

Now try to prove that f does satisfy the two conditions for linearity.

Last edited: Nov 26, 2015
3. Nov 26, 2015

### jk22

Basically you have to check if f ((x,y)+(a,b))=f (x,y)+f (a,b)

And f (a (x,y))=af (x,y)

We suppose that the operations are (x,y)+(a,b)=(x+a,y+b) and a (x,y)=(ax,ay)

4. Nov 26, 2015

### says

so f(x,y) = x+y

We write this as a column vector
x
y
Then we create out own corresponding column vector a+b, which gives us two column vectors:
x a
y b
for it to be a linear transformation
x+a
y+b
has to =
x+y
a+b

Is this correct for the first rule of linear transformations?

5. Nov 27, 2015

### jk22

The result of your function is in R so it is not a column vector it is the sum of the components.

6. Nov 27, 2015

### says

I dont understand.

7. Nov 27, 2015

### jk22

x+a
y+b

Is x+a+y+b which indeed equals to x+y+a+b

8. Nov 27, 2015

### says

So it is a linear transformation? If it wanted the answer in R2 though it wouldn't be a linear transformation? Was my answer in R2?

9. Nov 27, 2015

### jk22

A function from R2->R2 takes a 2-vector and gives as output a 2-vector for exemple

$$\begin{array}{c}x\\y\end {array}\mapsto\begin {array}{c} 2x+y \\x-y\end {array}$$

We are not supposed to give direct answers but to give hints to the problems.

10. Nov 27, 2015

### says

I'm conceptually having problems though -- my question isn't homework, it's about understanding the theory behind what I'm doing without loading myself with too many new terms.

11. Nov 27, 2015

### says

Isn't x+y one vector though?

12. Nov 27, 2015

### Samy_A

You should try to use clear and consistent notation.

For example, write vectors in $\mathbb R²$ as $(x,y)$.
That way, you can easily see the difference between a vector, $(x,y)$, and a number, $x+y$.

13. Nov 27, 2015

### says

A transformation from R2 to R1 would take 2 vectors V1: x+y and V2: a+b and give an output as one vector.

so T:x+y= T(x+y+a+b)
This is equal to T(V1) + T(V2), which = T(x+y) + T(a+b) = T(x+y+a+b)
T(ca) = Tc(a)

T(c*x+y) = T(cx+y)
Tc(x+y) = Tcx+Tcy

T(cx+y) does not equal Tcx+Tcy

I think this is better. The first condition is met, but the second is not, which means it is not a linear transformation.

14. Nov 27, 2015

### Samy_A

Let's rewrite the first part of your post:
$V1=(x,y), V2=(a,b)$
We have to prove that $T(V1)+T(V2)=T(V1+V2)$
$T(V1)+T(V2)=x+y + a+b$
$T(V1+V2)=T((x+a,y+b))=x+a+y+b$
As $x+y + a+b=x+a+y+b$, the first condition is met.

Now try to check the second condition, using a clear notation.

15. Nov 27, 2015

### says

T(cx) = Tc(x), where x=V1

T(cV1) = Tc(V1)
T(c(x+y)) = Tc(x+y)

T(cx+cy) = Tcx+Tcy
Tcx+Tcy = Tcx+Tcy

So it is a linear transformation because the second condition is met. I made a little mistake in the post above.

16. Nov 27, 2015

### Samy_A

It would help to use consistent notation. For example, in your post $x$ represents a vector in $\mathbb R²$ but also the first coordinate of that vector: very confusing.

Moreover "Tcx" and "Tcy" are meaningless, as T is an operator on $\mathbb R²$, while cx and cy are (I assume) numbers.

Your conclusion is correct, T is linear, but try to write the proof in a correct way.

17. Nov 27, 2015