Currently all i know is:
All hermitian matrices are diagonlizable but not vice versa.
The eigenvectors of a diagonlizable matrix can form a Basis for the vector space it operates on.
The eigenvectors for a Hermitian matrix can form an orthogonal or orthonormal Basis for the vector space it operates on. And its eigenvalues have to be real.
Is what I have said correct? Is there anything im missing? Thank you in advance.
Your statements are all correct.
A Hermitian matrix $A$ can be diagonalized with a unitary matrix, so $A=UDU^H$, with $D$ diagonal and $U^H=U^{-1}$. In particular, the columns of $U$ are eigenvectors for $A$ and form an orthonormal basis.
A matrix can be diagonalizable without being Hermitian; simple example $$ \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} $$
If a matrix $A$ is diagonalizable, then $A=SDS^{-1}$, with $D$ diagonal, and the columns of $S$ form a basis consisting of eigenvectors of $A$.
The eigenvalues of a Hermitian matrix are real. Indeed, if $A=A^H$ and $Av=\lambda v$, with $v\ne0$, then $$ \lambda v^Hv=v^H(\lambda v)=v^H Av=v^HA^Hv=(Av)^Hv=(\lambda v)^Hv= \bar{\lambda}v^Hv $$ Since $v^Hv\ne0$, we conclude $\lambda=\bar{\lambda}$ (complex conjugate), hence $\lambda$ is real.