So I'm trying to prove the classic dimension theorem for linear transformations. Here is what I have so far:
Given a linear transformation $T:V\rightarrow W$
Let the basis of $\mathrm{Im}T$ be $\{u_1,…,u_m\}$ where $\forall u\in V:T(u)$.
Let the basis of $\ker T$ be $\{v_1,..,v_k\}$ where $\forall v:T(v)=0$.
Lets prove $B=\{u_1,…,u_m,v_1,..,v_k\}$ is a basis of $V$:
(1) $B$ is linearly independent: Assume (falsely) that $B$ isn't linearly independent. Since $\{u_1,…,u_m\}$ and $\{v_1,..,v_k\}$ are both the basis of $\ker T$ and $\mathrm{Im}T$ repectively, they must each be linearly independent.If this is the case, $\exists b\in V:b\in \{v_1,..,v_k\}\cap \{v_1,..,v_k\}$ however this contradicts $\mathrm{Im}T\cap \ker T=\emptyset$ so $B$ must be linearly independent.
(2)$\mathrm{span}(B)=V$: Assume (falsely) that $b_2\in V| b_2\notin \mathrm{span}(B)$. If this is the case, then $b_2\notin \{\forall u\in V:T(u)\}\cap\{\forall v:T(v)=0\}$. This is impossible because $b_2\in V$ so $\mathrm{span}(B)=V$
Since (1) and (2) are true, according to basis identities, $B$ is a basis of $V$. This means that $\dim V=|B|=\dim\ker T+\dim\mathrm{Im}T$ QED
Best Answer
Your proof doesn't work. For instance, consider the transformation $$ T = \pmatrix{0&1\\0&0} $$ Both the image and kernel are spanned by $(1,0)^T$, so there is no pair of bases for your initial step whose union will form a linearly independent set.
If you look up a correct proof, you'll see that the usual idea is to take the basis of the kernel and extend it to a basis for $V$, without regard for what happens with the image.