Problem with this question(lenear algebra)

1. Mar 2, 2009

transgalactic

Last edited by a moderator: May 4, 2017
2. Mar 2, 2009

yyat

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. Mar 2, 2009

transgalactic

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. Mar 2, 2009

yyat

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.

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