I have seen in multiple sources that if we start with writing a time series as follows (assuming $n$ is odd):
$$x_t = a_o + \sum_{j=1}^{(n-1)/2}{[a_jcos(2\pi tj/n) + b_jsin(2\pi tj/n)]}$$
Then we will be able to estimate the $a_j$'s and $b_j$'s by treating it as a linear regression problem. Specifically, (letting $z_{tj}=cos(2\pi tj/n)$ or $z_{tj}=sin(2\pi tj/n)$, we can write
$$a_j = \frac{\sum_{t=1}^nx_tz_{t,j}}{\sum_{t=1}^nz^2_{t,j}}$$
and $$b_j = \frac{\sum_{t=1}^nx_tz_{t,j}}{\sum_{t=1}^nz^2_{t,j}}$$
This is where I get stuck. I understand that the above formulations come from estimating the coefficients of the linear regression, however, the denominator of each coefficient confuses me.
Given that this is a multiple linear regression problem, I've been trying to work through it using vectors/matrices. That is, using
$$\hat{\beta}=(\sum^n_{t=1}z_tz_t')^{-1}\sum^n_{t=1}z_tx_t$$
where $\hat{\beta}$ is the vector of coefficients (in our case this will represent the $a_j$'s and $b_j$'s) and $z_t$ is the vector of $cos(2\pi tj/n)$'s and $sin(2\pi tj/n)$'s, which we are regressing on.
By letting $q=n-1$ arbitrarily and $z_t = [1, z_{t1}, ... z_{tq}]'$, we can work out the numerator of $a_j$ and $b_j$ as follows
$$\sum_{t=1}^nz_tx_t=\sum_{t=1}^n[1, z_{t1}, ... z_{t1}]'x_t$$ $$=[\sum_{t=1}^nx_t, \sum_{t=1}^nz_{t1}x_t, ...\sum_{t=1}^nz_{tq}x_t]$$
The first element is $a_o$ and every other element is the numerator of the corresponding coefficient it is estimating.
Then, to calculate the denominator, we write
$$\sum^n_{t=1}z_tz_t'= \begin{bmatrix} 1 & z_{t1} & ... & z_{tq} \\ z_{t1} & z_{t1}^2 & ... & z_{t1}z_{tq} \\ ... & ... & & ...\\ z_{tq} & z_{t1}z_{tq} & ... & z_{tq}^2 \\ \end{bmatrix} $$
From here we take the inverse of this matrix, but I fail to see how this would equate to $\sum_{t=1}^nz^2_{t,j}$ being the denominator of each coefficient. Unless there is some serious math involved, I don't see how we would be able to generalise the results of this inverse matrix so succinctly.
If there is more advanced math that I am missing, I imagine the matrix being symmetric has something to do with, as this would mean the inverse is also symmetric, but that's as far as I can get. The only thing I can think of is that I am overlooking an important piece of information before I got to this point, which is probably more likely.
