# Thread: [SOLVED] Linear transform - need a hand

1. ## [SOLVED] Linear transform - need a hand

Let T:V->W be a linear transformation with NS(T)={0}. Suppose {u1, u2} is a linearly independent subset of V. Show that {T(u1), T(u2)} is linearly independent.

My attempt:
Suppose {T(u1), T(u2)} is linearly dependent, then there exsists c1, c2 =/ 0 (can't find the "is not equal" symbol, sorry) such that:
c1T(u1) + c2T(u2) = 0
c1u1 + c2u2 = 0 (as T(0)=0 ) **
c1=c2=0, as {u1, u2} are linearly indepedent
Then {T(u1), T(u2)} is also linearly independent as the only solution of c1T(u1) + c2T(u2) = 0 is the trivial solution.

The part I'm wondering if I've done wrong is at the **. Given NS(T)={0}, instead of having c1u1 + c2u2 = 0, I could have c1u1 + c2u2 = x where x $\in$ V : T(x) = 0 , couldn't I? Then I haven't proved anything at all...

2. Originally Posted by Cloud_pin
Let T:V->W be a linear transformation with NS(T)={0}. Suppose {u1, u2} is a linearly independent subset of V. Show that {T(u1), T(u2)} is linearly independent.

My attempt:
Suppose {T(u1), T(u2)} is linearly dependent, then there exsists c1, c2 =/ 0 (can't find the "is not equal" symbol, sorry) such that:
c1T(u1) + c2T(u2) = 0
c1u1 + c2u2 = 0 (as T(0)=0 ) ** your reasoning here is not correct!
$T(c_1u_1 + c_2u_2)=c_1T(u_1)+c_2T(u_2)=0,$ which means $c_1u_1 + c_2u_2 \in NS(T)=\{0\}.$ thus $c_1u_1 + c_2u_2 = 0.$

3. Originally Posted by NonCommAlg
$T(c_1u_1 + c_2u_2)=c_1T(u_1)+c_2T(u_2)=0,$ which means $c_1u_1 + c_2u_2 \in NS(T)=\{0\}.$ thus $c_1u_1 + c_2u_2 = 0.$
Aaaaah. Thanks oodles!!