conjugate of Ax, where x is and eigenvector Learning linear algebra on my own time. 
I must be missing something really simple. Say I have a matrix $A$ and its eigenvector $x$. Then $Ax = \lambda x$, where $\lambda$ is an eigenvalue. I can't figure out why the following is true,
$$
\bar A \bar x = \bar \lambda \bar x
$$
where $\bar A, \bar x, $ and $\bar \lambda$ are complex conjugates of $A, x, $and $\lambda$ respectively. 
 A: Well certainly you would have to have $$Ax = \lambda x \implies \overline{Ax} = \overline{\lambda x}$$
Complex conjugation satisfies $$\overline{z_1z_2} = \overline{z_1} *\overline{z_2} \quad \text{ and }\quad \overline{z_1+ z_2} =\overline{z_1} + \overline{z_2}  $$
And so looking at how matrix multiplication is defined its pretty routine to show that $$\overline{AB}=
\overline{A} * \overline{B}$$
The same also holds for addition of matrices. Let me know if that doesn't clear it up and I'll go into more detail. I would also add, if you haven't done a lot of work with complex numbers, it might be easier just to learn linear algebra with real matrices only. In fact, if you were to take an introductory linear algebra class, you would only use real matrices. (at least I'm not aware of any schools that use complex matrices in intro to linear algebra). Then when you want to add in complex numbers, it just requires tweaking a couple of things ( mainly just replacing any appearances of
 $A^{T}$ with $\overline{A}^T$ in all of the theorems) but everything else stays pretty much the same. A really great book to use if you wanted to learn linear algebra on your own is the one by Lay. 
