Let $V$ and $W$ finite vector spaces with dimension $n$ and $r$ respectively and $T: V \rightarrow W$ linear transformation, $\{ v_1,v_2,\ldots,v_n \} \subset \ker T$ and $\{ u_1,u_2,\ldots,u_s \} \subset V$.
How can i prove that any of the following statements ( $a)$ and $b)$ ) imply the third, i.e, $c)$ :
$a)$ $\{ v_1,v_2,\ldots,v_r \}$ is a basis of $ \ker T$;
$b)$ $\{ T(u_1),T(u_2),\ldots,T(u_s) \}$ is a basis of $ \operatorname{im} T$;
$c)$ $\{ v_1,v_2,\ldots,v_n \}\cup \{ u_1,u_2,\ldots,u_s \}$ is a basis of $V$.
I really stuck in this problem, please i really need help for this, thanx for your time.
$$\dim(\ker T) + \dim(\operatorname{im}(T) = \dim V$$
So if $\{ v_1,v_1,\ldots,v_n \}$ is a basis of $ \ker T$ and $\{ T(u_1), \ldots, T(u_s) \}$ is a basis of $\operatorname{im}T$ we have that $$\dim V = n+s. $$
Moreover $\lbrace v_1 \ldots v_n , u_1 \ldots u_s \rbrace $ are linearly independent due to the following $$\sum_{i=1}^n \alpha_i v_i \ + \sum_{j=1}^s \beta_j u_j = 0$$ $$T\left(\sum_{i=1}^n\alpha_i v_i \ + \sum_{j=1}^s \beta_j u_j\right) = 0$$ $$\sum_{j=1}^s \beta_j T(u_j) = 0$$
But $\{ T(u_1), \ldots T(u_s) \}$ is a basis of $\text{im}T$ so $\beta_j = 0 \ \ \forall j$ and then $\alpha_i = 0 \ \ \forall i $ because $\{ v_1,v_1,\ldots,v_n \}$ is a basis of $ \ker T$.
Thus $\lbrace v_1 \ldots v_n , u_1 \ldots u_s \rbrace $ are $n+s$ linearly independent vectors $\Longrightarrow $ they are a basis.