my notes say (I'm self studying) that the Rayleigh quotient minimizes $Aq=pq$
I have no idea what that means. I did a quick search online and I found out that this quotient is a complex number $p$, $q^HAq = q^Hqp $
I have no idea how I shoud prove this. how are we supposed to show that p minimizes $Aq-pq$??
Not entirely sure I understand your question but this is my guess about what you're asking.
For a matrix $A$ and a vector $q$, the rayleigh quotient $R_{A}(q) := \frac{q^{H}Aq}{q^{H}q}$ is bounded between the eigenvalues of $A$. Therefore, its minimum is the smallest eigenvalue $\lambda_{min}$, which is the smallest possible $p$ in the eigenvalue equation $Aq = pq$. So the minimum Rayleigh quotient is the smallest $p$ in the eigenvalue equation (which also corresponds to the eigenvector $q$ that minimises the Rayleigh quotient).
Additional clarification:
If $q$ is a linear combination of the eigenvalues of $A$ (i.e. $q = a_{1}u_{1} + ... a_{n}u_{n}$ where $\sum a_{i} = 1$ and the $u_{i}$ are the eigenvectors) and has norm of 1, clearly $R(q) = \frac{q^{H}Aq}{1} = q^{H}(\lambda_{1}a_{1}u_{1} + ... \lambda_{n}a_{n}u_{n}) = \sum_{i} \lambda_{i}a_{i}^{2}$.
To minimise this expession we just set the $a_{i}$ corresponding to the minimum eigenvalue to 1. Scaling of $q$ by some constant does not affect this.
Therefore the minimum of $R_{A}(q)$ is the minimum eigenvalue, which by definition is the minimum $p$ in $Aq = pq$.
