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I have this exercise

 Define and find the expression of the linear transformation  

Information given:

$ T : \Bbb R^5\to \Bbb R^4$

Im T = ⟨(3, -1, 0, 1), (1, 0, 0, 2), (0, 0, 1 -1)⟩

⟨(3,3 0, 0,0)⟩⊂ Ker T

Solve T(⟨(1; 1; 0; 0); (0; 0; 1; 0)⟩) and give a basis.

I have already define the transformation and found the expresion that is T(x,y,z,w,u)=(x-y+z,-x+y,w,x-y+z-w) And T(⟨(1, 1, 0, 0), (0, 0, 1, 0)⟩) that is T(1,1,0,0,0)=(0,0,0,0)

T(0,0,1,0,0)=(1,0,0,1)

But I have a problem giving the basis, is the transformation basis the imagen basis,I don't really undestand what is the problem asking in that part?

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  • $\begingroup$ Welcome to Maths SX! What do you denote SubGen? $\endgroup$ – Bernard Jan 30 at 23:55
  • $\begingroup$ Thanks,generated subspace. $\endgroup$ – Emmaaaaa Jan 30 at 23:57
  • $\begingroup$ The usual name is the Span. And I suppose c Ker T means contained in Ker T? $\endgroup$ – Bernard Jan 31 at 0:09
  • $\begingroup$ Yes,i suppose it is that. $\endgroup$ – Emmaaaaa Jan 31 at 0:27
  • $\begingroup$ (SubGen(1, 1, 0, 0), (0, 0, 1, 0)) could also be written like this <(1, 1, 0, 0), (0, 0, 1, 0)> $\endgroup$ – Emmaaaaa Jan 31 at 12:57

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