Proof - Conjugate of the conjugate of a matrix equals the original matrix I have speared much time looking for the proof of $\overline{\overline A} = A$ where $A$ is any matrix of order $(m  \times n)$.
There are other questions of conjugation that I need help with but I just require  a clue on how to prove something that is so general.
 A: If you want a long winded and general proof where you can see what is happening to the elements of $A$, here it is.
If $A \in \mathbb{C}^{m\times n}$, then its general form will be
$$A=\begin{pmatrix}a_{11} & a_{12} & \dots & a_{1n} \\
a_{21} & a_{22} & \dots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m1} & a_{m2} & \dots & a_{mn}
\end{pmatrix}.$$ 
Taking the conjugate gives
$$\overline{A}=\begin{pmatrix}\overline{a_{11}} & \overline{a_{12}} & \dots & \overline{a_{1n}} \\
\overline{a_{21}} & \overline{a_{22}} & \dots & \overline{a_{2n}} \\
\vdots & \vdots & \ddots & \vdots \\
\overline{a_{m1}} & \overline{a_{m2}} & \dots & \overline{a_{mn}}
\end{pmatrix}.$$
Hence,
$$\overline{\overline{A}}=\begin{pmatrix}\overline{\overline{a_{11}}} & \overline{\overline{a_{12}}} & \dots & \overline{\overline{a_{1n}}} \\
\overline{\overline{a_{21}}} & \overline{\overline{a_{22}}} & \dots & \overline{\overline{a_{2n}}} \\
\vdots & \vdots & \ddots & \vdots \\
\overline{\overline{a_{m1}}} & \overline{\overline{a_{m2}}} & \dots & \overline{\overline{a_{mn}}}
\end{pmatrix}=\begin{pmatrix}a_{11} & a_{12} & \dots & a_{1n} \\
a_{21} & a_{22} & \dots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m1} & a_{m2} & \dots & a_{mn}
\end{pmatrix}=A.$$
But as @RideTheWavelet commented, you can just as easily say $$\overline{\overline{a_{ij}}}=a_{ij},$$ $\forall i=1,2,\dots,m$ and $j=1,2,\dots,n$. Hence, $\overline{\overline{A}}=A$.
