Question

Find ker (T)
a) T: R^2-->R^2; where T(x,y) = (y,x)
b) T : R^2-->T^2; where T(x,y)= (0, 2x+3y)
c) T: R^2---> R^3; where T(x,y)= (x-y,y-x,2x-2y)
Please, help

Answer 1

set it up as a solution to \(c_1v_1+c_2v_2=0\)

Answer 2

where v1 and v2 .. vn are the vectors presented

Answer 3

more, please, still not get

Answer 4

T(x,y) = (y,x) is notation for a setup such that

0x + 1y = 0
1x + 0y = 0

Answer 5

got the step

Answer 6

T(x,y)= (0, 2x+3y) is notation for

0x + 0y = 0
2x + 3y = 0

Answer 7

got it, too. but , please, each case at a time, don't mix them up

Answer 8

T(x,y)= (x-y,y-x,2x-2y)

1x- 1y =0
-1x +1y =0
2x - 2y =0

the setup is then augmented to determine the free variables

Answer 9

rref each coefficient matrix

Answer 10

ok, then, teh rref tell me what? for case a) I got (0,0) b) I got (1,-3/2) so??

Answer 11

the rref tells us the free variables and the ... not free? .... variables. forgot the names of the others :)

so for the first one:

0 1 1 1 1 1 1 0
1 0 1 0 0 -1 0 1

has no free variables so the basis of the kernel is just the 0 vector

Answer 12

got it

Answer 13

for the second one

0 0 2 3 1 3/2
2 3 0 0 0 0

1 free variable such that x = -3/2 y
y = 1 y

the basis of our kernel is the vector <-3/2,1> , or to proper it up ... <3,-2>

Answer 14

ok, got it

Answer 15

1 -1
-1 1
2 -2



1 -1
0 0
0 0

in this case we have one free variable again, such that:

x = 1y
y = 1y
z = 0y

Answer 16

stuck here, 1 equation, 3 variables, we must have at least 2 free variables???

Answer 17

what is the dim of the 3rd setup?

Answer 18

1

Answer 19

might have forgotten a specific along the way; the basis of the kernel on the last one is simply (1,1)

Answer 20

so, I just take rref and consider how many free variables it has, and base on the result to conclude about the ker(T)?

Answer 21

yes

Answer 22

what if the matrix give out linear dependent . I mean 1 free variable and 2 independent . how to conclude about ker(T)?

Answer 23

A = 1 -1
-1 1
2 -2

Ax = 0 when x = <1,1>

Answer 24

define the 2 nonfree variables in terms of the free variable.

Answer 25

the next step is " base on the ker(T) consider whether it is one to one or not." so I need how to find it, and how to consider ker(t) =0 or not

Answer 26

cant say a recall the details of that

Answer 27

in case a) matrix gives out linear independent, as you state, its ker (T) = (0,0) . got it

Answer 28

case b) matrix gives out linear dependent, since it has free variable there, ker (T) = (-3,2) got it, too

Answer 29

stuck at case c)

Answer 30

hey, explanation please

Answer 31

A linear transformation T is one-to-one ⇔ kernelT = {0} .... from the internet

Answer 32

from my book, too

Answer 33

if a system has a unique solution for Ax=b, then it is one to one

Answer 34

if Ax = 0 has only the trivial solution, it is one to one

Answer 35

I know, but the problem asks for find ker(T) and then consider.... Cannot come up by another way

Answer 36

another property is such that

T:R^n -> R^m is 1-1 if m>=n

Answer 37

you mean we can conclude right there by the last property? it's not one to one.

Answer 38

http://algebra.math.ust.hk/matrix_linear_trans/04_oneone/lecture2.shtml

according to this, yes

Answer 39

got it, thanks a lot, friend.

Answer 40

hope it helps

Answer 41

thanks for the link, I read it now