Thread: proving if such transformation exists

is there a transformation which follows these rules:
T: ->
T(-1,1,0,1)=(1,-1,2,1)
T(1,1,1,0)=(2,3,1,-1)
T(-1,5,2,3)=(3,7,0,-3)

ImT is defined as span of the base of the images
so i found that this group is dependant
{(1,-1,2,1),(2,3,1,-1),(3,7,0,-3)}

so ImT=Sp{(1,-1,2,1),(2,3,1,-1)}
so dim ImT=2 and dim kerT=2

so i took basis
and defined T as:
T(-1,1,0,1)=(1,-1,2,1)
T(0,2,1,1)=(0,5,-3,-3)
T(0,0,1,0)=(0,0,0,0)
T(0,0,0,1)=(0,0,0,0)

is it ok?

Originally Posted by transgalactic

is there a transformation which follows these rules:
T: ->
T(-1,1,0,1)=(1,-1,2,1)
T(1,1,1,0)=(2,3,1,-1)
T(-1,5,2,3)=(3,7,0,-3)

You mean "linear transformation"- you make a general transformation that does anything!

ImT is defined as span of the base of the images
so i found that this group is dependant

You mean "set". In "mathematical" English, "group" has a very specific meaning that is not correct here.

{(1,-1,2,1),(2,3,1,-1),(3,7,0,-3)}

Note that {(-1, 1, 0, 1), (1, 1, 1, 0), (-1, 5, 2, 3)} are also dependent. We can write
-3(-1, 1, 0, 1)+ 2(1, 1, 1, 0)= (-1, 5, 2, 3).
Is T(-1, 5, 2, 3)= -3T(-1, 1, 0, 1)+ 2T(1, 1, 1, 0)?

so ImT=Sp{(1,-1,2,1),(2,3,1,-1)}
so dim ImT=2 and dim kerT=2

so i took basis

I don't understand what you mean by this. What did you take as the basis? Are you referring to the standard basis?

[tex]and defined T as:
T(-1,1,0,1)=(1,-1,2,1)
T(0,2,1,1)=(0,5,-3,-3)
T(0,0,1,0)=(0,0,0,0)
T(0,0,0,1)=(0,0,0,0)

is it ok?

Well, you haven't answered the question! What is your answer?

a transformation of a basis is single.
so if we write the transformation of a each vector in the R^4 basis
then it defines a single transformation
T(x,y,z,t)=xT(1,0,0,0)+yT(0,1,0,0)+zT(0,0,1,0)+tT( 0,0,0,1)

ok there is no problem for me to find the value of
T(0,0,0,1) etc..

i ask if it ok to constract a transformation in such a way?

how to know if it follows tle laws given in the question

how do you see a definition of transformation?