Thread: Linear Transforamtion proof?

Let T: V-> W be a bijective linear transformation
Prove that if {v₁,v₂,...,v_n} is a basis for V,
then {T(v₁),T(v₂),...,T(v_n)} is a basis for W

Assuming $\displaystyle \{v_1,..,v_n\} $ is a basis for V.
To show the set $\displaystyle \{T(v_1),...,T(v_n)\} $ is a basis. We need to show 2 tihings
1)Every vector in W can be written as a linear combination of this set (or this set spans W)
2)This set is linearly independent

The first one is pretty straightforward, since T is bijective it means that for every $\displaystyle w \in W $ we can find a coordinate in $\displaystyle V $ such that
$\displaystyle L(c_1v_1 + ... + c_nv_n) = w $ so that equals $\displaystyle c_1L(v_1)+...+c_nL(v_n) = w $. There we picked any arbitray vector in W and wrote it as the linear combination of $\displaystyle L(v_1)..L(v_n) $

The second one is also pretty straight forward. if $\displaystyle \{L(v_1),...,L(v_n) \} $ was linearly dependent, it would have atleast one non zero solution to the homogenous equation, so there is a coordinate $\displaystyle x_1 \in V $ such that $\displaystyle x_1 \not = 0 $ and $\displaystyle L(x_1) = 0 $ Now Pick another non zero vector $\displaystyle x_2 \in V $ now observe that $\displaystyle x_3 = x_1 + x_2 $ and $\displaystyle x_3 \not = x_1 $ but $\displaystyle L(x_1 + x_2) = L(x_1) + L(x_2) = 0 + L(x_2) = L(x_3) $ how can 2 different vectors in $\displaystyle V $ go to the same vector in W? it was supposed to be one-to-one, thus only the zero solution must exist for the homogenous equation, thus $\displaystyle \{L(v_1),...,L(v_n) \} $ is linearly independent and thus a basis for W.