Given $A^{n\times n}$ a complex matrix and $z=\begin{pmatrix} z_{1} \\ \vdots \\ z_{n} \end{pmatrix}\in\mathbb{C}$. Show that $\overline {A\cdot z}=\overline {A}\cdot \overline {z}$. Note that $\overline {A}, \overline{z}$ means that conjugate of $A$ and $z$, respectively.
I showed for $2\times2$, $3\times3$ but how should I show for $n\times n$ because there are too many entires for this, can you help, can you give me a hint?
I see that you're fine working through the particular cases, but you have trouble generalizing to the $n \times n$ cases.
When you hit an obstacle like that, it's often a good idea to use summation notation. In this particular case, we note that the $j$th entry of $Az$ is $$ \sum_{k=1}^n A_{jk}z_k $$ That being said, we can rewrite the equality you're trying to prove as $$ \overline{\sum_{k=1}^n A_{jk}z_k} = \sum_{k=1}^n \bar A_{jk}\bar z_k, \quad j=1,\dots,n $$ I suspect that you'll be able to figure out the trick from there.