Let $T:V\rightarrow W$ be a linear mapping, with $V$ finite-dimensional. We wish to show that $$\dim V = \dim \Ker T + \dim\Img T$$

The images of a basis of $V$ will span $\Img T$ and hence $\Img T$ is finite-dimensional. Choose then a basis $w_1,\ldots,w_n$ of $\Img T$ and choose preimages $v_1,\ldots,v_n\in U$ such that $$w_i = T(v_i),\quad i=1\ldots n$$ Choose a basis $u_1,\ldots,u_k$ of $\Ker T$ The result will follow once we show that $u_1,\ldots,u_k,v_1,\ldots,v_n$ is a basis of $V$

Let $v\in V$ be given. Since $T(v)\in \Img T$ by definition, we can choose scalars $b_1,\ldots,b_n$ such that $$T(v)= b_1 w_1 + \ldots b_n w_n.$$ Linearity of $T$ now implies that $T(b_1 v_1 + \ldots + b_n v_n-v) = 0,$ and hence we can choose scalars $a_1,\ldots, a_k$ such that $$b_1 v_1 + \ldots + b_n v_n-v = a_1 u_1 + \ldots a_k u_k.$$ Therefore $u_1,\ldots,u_k,v_1,\ldots,v_n$ span $V$

Next, let $a_1,\ldots, a_k,b_1,\ldots,b_n$ be scalars such that $$a_1 u_1+\ldots +a_k u_k + b_1 v_1 + \ldots + b_n v_n = 0.$$ By applying $T$ to both sides of this equation it follows that $$b_1 w_1 + \ldots + b_n w_n =0,$$ and since $w_1,\ldots, w_n$ are linearly independent that $$b_1= b_2 = \ldots = b_n = 0.$$ Consequently $$a_1 u_1+\ldots +a_k u_k = 0$$ as well, and since $u_1,\ldots, u_k$ are also assumed to be linearly independent we conclude that $$a_1= a_2 = \ldots = a_k = 0$$ also. Therefore $u_1,\ldots,u_k,v_1,\ldots,v_n$ are linearly independent, and are therefore a basis. Q.E.D.