1. ## Linear Transformation

Let T: V -> W be a linear transformation, and let the dimension of V be n, Let { $v_1, v_2, ..., v_m$} be a basis for the kernel of T.

1) Let V_(m+1), v_(m+2), ..., $v_n$ be vectors in V such that { $v_1$, $v_2$, ..., $v_m$, v_(m+1), ..., $v_n$} in a basis for V. Prove that {Tv_(m+1), Tv_(m+2), ..., $Tv_n$} is a basis for the image of T.

2) Using 1), prove that the rank of T plus the nullity of T is n.

Thanks!

2. Originally Posted by cindy
Let T: V -> W be a linear transformation, and let the dimension of V be n, Let { $v_1, v_2, ..., v_m$} be a basis for the kernel of T.
It is safe to assume $m\geq 1$.

1) Let V_(m+1), v_(m+2), ..., $v_n$ be vectors in V such that { $v_1$, $v_2$, ..., $v_m$, v_(m+1), ..., $v_n$} in a basis for V. Prove that {Tv_(m+1), Tv_(m+2), ..., $Tv_n$} is a basis for the image of T.
Let $\bold{a}$ be in the image of $T$.
Then $\bold{a} = k_1 \bold{v}_1 + ... + k_m \bold{v}_m + k_{m+1} \bold{v}_{m+1} + ... + k_n \bold{v}_n$.
Thus, $T\bold{a} = k_1 T\bold{v}_1 + ... + k_m T\bold{v}_m + k_{m+1}T \bold{v}_{m+1} + ... + k_n T\bold{v}_n = k_{m+1}T \bold{v}_{m+1} + ... + k_n T\bold{v}_n$.
Thus, $T\bold{v}_{m+1}, ... T\bold{v}_n$ spams the image of $T$.

You need to show $T\bold{v}_{m+1}, ... T\bold{v}_n$ is linearly independent.
If $a_{m+1}T\bold{v}_{m+1} + ... + a_n T\bold{v}_n = \bold{0}$
Then $T(a_{m+1}\bold{v}_{m+1} + ... + a_n \bold{v}_n) = \bold{0}$.
Thus, $a_{m+1}\bold{v}_{m+1} + ... + a_n \bold{v}_n \in \ker T$.

We can therefore write, $a_{m+1}\bold{v}_{m+1} + ... + a_n \bold{v}_n = a_1\bold{v}_1 + ... + a_2 \bold{v}_m$.
This implies $a_1=a_2= ... 0$ because those vectors form a basis.
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Then #2 follows because $m + (n-m) = n$.