Question

Let T be a linearly transformation defines by
T(x,y,z)=(x+y+z;x-y-3z;-x+y+3z)


find a basis for image(T) and a kernel(T)

Answer 1

ok lets start with the representation of the transformation as a matrix

Answer 2

1 1 1
1 -1 -3
-1 1 3

Answer 3

agree?

Answer 4

yes

Answer 5

now for the kernel we have to solve
1 1 1 | 0
1 -1 -3 | 0
-1 1 3 | 0

Answer 6

yes

Answer 7

so what vectors did you get ?

Answer 8

x=z and y=-2z

Answer 9

1,-2,1

Answer 10

for example right ?

Answer 11

how did u do that

Answer 12

ok if you got x=z and y = -2z

lets call z = t
so we have
x = t
y = -2t
z = t

so the vector representation of this is : t(1,-2,1)

taking t = 1 (we can take any value we want) will give us (1,-2,1)

Answer 13

and it is a valid basis for our kernel

Answer 14

ok

Answer 15

now for the Image what ive learned to do is to apply the transformation on the
vectors of the standard basis

so for example :
apply T on (1,0,0)
we will get : (1,1,-1)
apply T on (0,1,0)
we will get ( 1,-1,1)

since we know that the basis of the image should have only 2 vectors
and we got here two independent vectors
they are a valid basis for the image

Answer 16

is the part i want to know of standard basis are we always using them

Answer 17

well i guess.. soon you will learn about representation of linear transformations on different basis

but for finding the image i guess you do it this way

Answer 18

i always did it this way

Answer 19

wat if is in R4 or any Rn where n is any number but not 3

Answer 20

so your standard basis would look like (1,0,0,0)
(0,1,0,0) ... (for R4)

Answer 21

ok u can continue with question

Answer 22

i think we done ?
we found a basis for kernel and image

Answer 23

wat about (0,0,1)

Answer 24

we can check for it but it will be FOR SURE linear dependent on the two others

Answer 25

so are u sure about (1,0,0) and (0,1,0)

Answer 26

i know it using the rank–nullity theorem .. dim(Im) + dim(Ker) = dim(V)

Answer 27

(1,0,0) and (0,1,0) -> it is not basis for ker nor for im

Answer 28

image basis : (1,1,-1),( 1,-1,1) ker basis : (1,-2,1)

Answer 29

ok

Answer 30

i used (1,0,0) and (0,1,0) in order to find the basis of the image

Answer 31

oh thanx now i understand