# Linear Transforamtion proof?

• Feb 2nd 2013, 07:34 PM
jackGee
Linear Transformation proof?
Let T: V-> W be a bijective linear transformation
Prove that if {v1,v2,...,vn} is a basis for V,
then {T(v1),T(v2),...,T(vn)} is a basis for W
• Feb 2nd 2013, 07:55 PM
jakncoke
Re: Linear Transforamtion proof?
Assuming $\{v_1,..,v_n\}$ is a basis for V.
To show the set $\{T(v_1),...,T(v_n)\}$ 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 $w \in W$ we can find a coordinate in $V$ such that
$L(c_1v_1 + ... + c_nv_n) = w$ so that equals $c_1L(v_1)+...+c_nL(v_n) = w$. There we picked any arbitray vector in W and wrote it as the linear combination of $L(v_1)..L(v_n)$

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