Question

How to prove a liner map/transformation?
Question in comments.

Answer 1

what defines linearity?

Answer 2

T(x+y) = T(x) + T(y)

T(kx) = kT(x), for some constant k

Answer 3

are these conditions sufficient?

Answer 4

I know those conditions, I don't know how to apply them.

Answer 5

I did this for part a) and now I'm stuck at part b), same question

Answer 6

These are the questions

Answer 7

does (x1+x2,y1+y2) =((x1+x2)(y1+y2),0)

Answer 8

or rather, does the process work out?

Answer 9

Yes,
so if I multiply them out, that I am supposed just to go backwards again?

Answer 10

yes, the idea is to show that the transform of the sum of 2 vectors is equal to the sum of the transforms

Answer 11

or that it fails ...

Answer 12

So, is this it?

Answer 13

I don't really see the point of doing this :|

Answer 14

And I think it fails the second test?

Answer 15

the point is to give you practice :) make you familiar with the concepts

Answer 16

Ok, I am willing to practice, I am just not sure if I am doing the right thing, or if I am writing nonsense

Answer 17

T((2,1)+(0,3))= T(2,4) = T(2,1) +T(0,3)

T(2,4) = T(2,1) +T(0,3) now we see if the transforms are equal

(2(4),0) = (2(1),0) +(0(3),0)

(8,0) = (2,0) +(0,0)

well, it fails in that we can find an example at least; (8,0) is nto the same as (2,0) is it?

Answer 18

if T(a+b)=T(a)+T(b), then T(x) is a linear tranformation.

and T(x) is defined so that we can compare results

Answer 19

Ok, give me few mins to figure this out. Thank you

Answer 20

Ok, I understand. I'll write that down neatly now. Would you mind taking a look at question 5, if that is not too much to ask? Just part a, if I get a), I will be able to solve b) on my own.

Answer 21

as for a, and assuming ive read your notations correctly

T((x,xn)+(y,yn))= T(x,xn) + T(y,yn)

T((x+y),(xn+yn))= T(x,xn) + T(y,yn)

T(a,b)= 3a+2b

3(x+y)+ 2(xn+yn) = 3x +2xn + 3y+2yn

3(x+y)+ 2(xn+yn) = 3(x+y) +2(xn+yn), works out fine
--------------------------

T(kx,ky) = k T(x,y)

3kx+2ky = k (3x+2y)

k(3x+2y) = k (3x+2y) works fine too

so a is linear

Answer 22

ill look at it

Answer 23

hmmm, it gives you a hint ... my idea was dealing with matrixes.

Answer 24

im still thinking on a good approach, the hint doesnt do much in my mind ....

Answer 25

If I multiply f(0, 1) = (3, 2, 1) with 2 as the hits says, and
f(1, 1) = (−1, 0, 1) with 3.
And combine them together (subtract them) I will get:
f(3, 1) = (-9, -4, 1)

Answer 26

And now, I should use 3 and 1 to somehow get 5 and -9, and 0 and -4, and -2 and 1. Which seems very confusing.
f(3, 1) = (5, 0, −2)
f(3, 1) = (-9, -4, 1)

Answer 27

\[\begin{pmatrix}v_1&u_1\\v_2&u_2\\v_3&u_3 \end{pmatrix}

\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}b_1\\b_2\\b_3 \end{pmatrix}\]

Answer 28

we can setup 3 equations for each row and try to solve the system. was my idea

Answer 29

Can you give me an example how would you construct an equation from one of those? (because they are vectors)

Answer 30

Or if (3, 1) = 3(1, 1) − 2(0, 1)
Than (5, 0, −2) should be equal to 3 (−1, 0, 1) - 2 (3, 2, 1) ?
And if it is, I proved it! I guess? :D

Answer 31

0a+1b=3
0c+1d=2
0e+1f=1

1a+1b=-1
1c+1d=0
1e+1f=1

3a+1b=5
3c+1d=0
3e+1f=-2


solving for a b

rref{{0,1,3},{1,1,-1},{3,1,5}}

Answer 32

Okay, I see where you're going, I'll try to solve them.

Answer 33

it was a thought, but i dont think itll pan out that well.

Answer 34

the transform will look like some contrived thing as:

T(x,y)=(x^2+2, xy^2+2y, 2x)

something like this but this is just a random jotting for descriptive purposes

Answer 35

The solutions I got to the equations above, does not match. As we had 9 equations and 6 unknowns. Does that mean the map is not linear?

Answer 36

Because I got
a=-4 b=3 c=-2 d=2 e=0 f=1
And when I use those values in third set of equations I'm not getting 5,0,-2, instead, I am getting -9,-4,1 which is actually in part b of this question. So there is a linear map for part b, but not for part a.

Answer 37

It doesn't look like the best way to do it, but it perfectly makes sense (and does the job I guess)?

Answer 38

it means that the mapping we chose is not linear for 'a' ... which is not the same as not having a linear map

Answer 39

but at least it works in b :)

Answer 40

How is that? Shouldn't it be linear for every value?

Answer 41

f and g are 2 different objects, the mapping on g does not imply that it has to perform the same for f

Answer 42

Yeah, but the values for first two equations are the same, so I will get same solutions for a,b,c,d,e and f for both function f and g.

Answer 43

all that means is the f and g converge for part of their existence.