Question

Let T(x1, x2, x3) = (x1+ x2, x2+ x3 , x1— x3) be
a linear operator on R³. Find its kernel. Show
that T is not onto. Show that (1, 1, 0) is in the
image of T. Also, find two distinct vectors
u1and u2such that T(u1 ) = (2, 2, 0) = T(u2).

Answer 1

Kernel (T)=

$$\{(x_{1},x_{2},x_{3})|T(x_{1},x_{2},x_{3})=0\}=\{(x_{1},x_{2},x_{3})|x_{1}+x_{2}=0,x_{2}+x_{3}=0,x_{1}-x_{3}=0\}$$

=$\{c*(1,-1,1)|c\in R\}$

(1,1,2) has no preimage with respect to linear operator T. Hence T is not onto.

T($\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2})=(1,1,0)$

$u_{1}=(\dfrac{1}{2},$ $\dfrac{3}{2},\dfrac{1}{2}), u_{2}=(1,1,1), T(u_{1})=T(u_{2})=(2,2,0)$