let $\mathbb F$ be a field. let $A\in M_n(\mathbb F)$.Then we can see $A$ as a linear map over $\mathbb F^n$.（sent $x$ to $Ax$） Denote the $A^T=(\beta_1^T,…,\beta_n^T)$,where $\beta_i$ is the row vector of $A$.$Ker A=\{x\in \mathbb F^n\mid Ax=0\}$.this is equivalent to $\beta_ix=0,\forall i$.
First,let $\mathbb F=\mathbb R$.denote $Im A^T=\{A^Tx\mid x\in \mathbb R^n\}$.that is $Im A^T$ is generated by $\beta_i^T$.it is interesting that $KerA\cap ImA^T=0$,since if $A^Tx\in Ker A$,then $AA^Tx=0$,hence $ (A^Tx)^TA^Tx=0 \in \mathbb R$.So $A^Tx=0$.by the formula of dimension in vector space,we know $\mathbb R^n=KerA \oplus Im A^T$.
The same way,$if \mathbb F=\mathbb C$, $Im \overline {A^T}=\{\overline{A^T}\mid x\in \mathbb C^n\}$.and we get $\mathbb C^n=Ker A\oplus Im\overline {A^T}$.
My question is how to calculate the decomposition of vector space as above way in more field？
Thanks.
Over a field $\mathbb F$, a vector space $V$ over $\mathbb F$, and a map $A:V\to V$, we can define the adjoint of $A$ as the map $A^*:V^*\to V^*$ by $A^*(f)=f\circ A$. ($V^*=\{f:V\to \mathbb {F}\,|\,f\text{ linear}\}$). But because $V$ is finite-dimensional, we can pick a basis for it, $\{e_i\}_{i=1}^n$, and then a basis for $V^*$ is $\{\delta_i\}_{i=1}^n$, where $\delta_i(\sum \alpha_je_j)=\alpha_i$. Then the mapping $\phi: V^*\to V$ by $e_i\mapsto \delta_i$ is an isomorphism of vector spaces, so we can view the adjoint $A^*$ as the map $\phi\circ A^*:V\to V$. Note that this depended on our choice of basis.
In particular, when $\mathbb F=\mathbb R$, the adjoint map so defined is exactly the transpose, and when $\mathbb F=\mathbb C$, the adjoint map is the conjugate transpose as above.