Let V be a vector space with basis $u_1, u_2, u_3$ and let $T : V → V$ be the linear transformation such that
$T(u_1)=u_1 −u_2 −u_3$
$T(u_2)=2u_2 −6u_3$
$T(u_3)=−u_1 +3u_2 −5u_3$
Find bases for the null space and the image of T and determine the rank and nullity of T .
I think I solved for the null space and nullity first.
After solving the system of equations with the solutions equal to $0$, I came up with
$u_3=k$ which gave $u_1=4k$ and $u_2=3k$
which gives $(4k,3k,k)$ nullity 3. Was this the correct method (gaussian elimination)?
And for solving for the basis of the image would I use the same method but instead of having $0$ in the right-most column for the solutions I would have 1,2,-5 (in descending order) corresponding to the coefficient in the given equation.
*After a second look I'm almost sure this is incorrect since the coefficients aren't the same as the T($u_n$)
Best Answer
$imT=\{(T(u_1),T(u_2),T(u_3)):u_1,u_2,u_3 \in V \}=\{(u_1-u_2-u_3,2u_2-6u_3,-u_1+3u_2-5u_3):u_1,u_2,u_3 \in V\}=\{u_1(1,0,-1)+u_2(-1,2,3)+u_3(-1,-6,-5):u_1,u_2,u_3 \in V \}$
$=<(1,0,-1),(-1,2,3),(-1,-6,-5)>$,but $-4(1,0,-1)=3(-1,2,3)+(-1,-6,-5) $ so
$imT=<(1,0,-1),(-1,2,3)>\Rightarrow dim(imT)=2$