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> [!theorem] Theorem. ([[Rank-Nullity theorem]])
>Let $k$ be a [[field]] and $0 \to U \to V \to W \to 0$ be a [[short exact sequence]] of finite-dimensional $k$-[[vector space|vector spaces]]. Then
>
>- $\text{dim }V=\text{dim }U + \text{dim }W$;
>- $\text{dim }V / U = \text{dim }V - \text{dim }U$;
>- For any $T \in \text{Hom}_{k\text{-}\mathsf{Vect}}(V,W)$, $\text{dim }V=\text{dim ker }T + \text{dim im } T.$
>
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> [!proof] Proof 2.
> [[every vector space has a basis|Let]] $\lb u_i \rb_{i=1}^m$ be a [[basis]] of $\null T$. The [[linearly independent]] list $u_1, \dots, u_m$ [[every linearly independent list extends to a basis|can be extended]] to a [[basis]] $B :=\lb u_i, v_j \rb_{i=1,j=1}^{m,n}$ of $V$ such that $\dim V = m+n$. We need to show that $\im T$ is [[vector space#Finite-Dimensional Vector Space|finite-dimensional]] with [[dimension]] $n$.
>
> Let $v \in V$. Because $B$ [[spans]] $V$, we can write $v=\sum_{i=1}^m a_iu_i + \sum_{i=1}^n c_iv_i, $where $a_i \and b_i \in \ff$ for $1 \leq i \leq m$ and $1 \leq i \leq n$ respectively.
>
> Applying $T$ to both sides of this equation yields $T(v) = T\left(\sum_{i=1}^m a_iu_i + \sum_{i=1}^n c_iv_i\right) = \cancel{\sum_{i=1}^m a_iT(u_i)}^{u_i \in \ker T} + \sum_{i=1}^n c_i T(v_i).$
> Since we've written an arbitrary $T(v)$ as a [[linear combination]] of $Tv_1, \dots, Tv_n$, we see that $\lb v_i \rb_{i=1}^n$ [[spans]] $\im T$, hence $\im T$ is [[vector space#Finite-Dimensional Vector Space|finite-dimensional]].
>
> We're done if we can show that the $T\left(\lb v_i \rb_{i=1}^n\right)$ are [[linearly independent]], since then they would form a [[basis]] for $\im T$. So suppose $d_1, \dots, d_n \in \ff$ and $\sum_{i=1}^n d_i Tv_i=0.$
> Then by $T