Problem with this question(lenear algebra)

  • Thread starter transgalactic
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As for finding the polynomial g(y), you can use the same process as in B to find the basis for Im(T).
  • #1
transgalactic
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in A i am asked to find the matrix of T

in B i need to find ker T and I am T

in C i need to find polinomial g(y) so T(y)=y+2

i tried to solve each one and i got problematic points

http://img165.imageshack.us/img165/5331/unt.th.gif
 
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  • #2
A) Your solution looks correct.

B) You found correctly that Ker(T) consists of vectors of the form (x,0,0,0). This can can be written as Ker(T)=span{(1,0,0,0)}.
Regarding the image of T: Your solution is correct, but one can also find a simpler basis consisting of standard basis vectors.

C) Look at the system of linear equations you extracted from the matrix equation, there is a small error. Also, since T has a one-dimensional kernel there will be infinitely many solutions.
 
  • #3
regarding B i get 0*x=0
on what basis i can state that its
(x,0,0,0)
??

where is the error in C
??
 
  • #4
transgalactic said:
regarding B i get 0*x=0
on what basis i can state that its
(x,0,0,0)
??

Well, you are trying to find all possible values of x,y,z,t so that T((x,y,z,t))=0. You found that this means that 0=x*0, y=0, z=0, t=0. So x can be any number (0=x*0 is true for all x) and y,z,t have to be zero. So the solutions of the equation T((x,y,z,t))=0 are exactly the vectors of the form (x,0,0,0), where x is any number.

where is the error in C
??

The first equation, x+2y=2, should be y+2z=2.
 

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