The question requires me to diagonalize the following circulant matrix where h is a real value. I know I am supposed to use discrete fourier transforms to diagonalize the matrix based on online research, but I am having trouble understanding how to do that. Currently the only way I know how to diagonalize matricies are to find the eigenvalues and corresponding eigenvectors to find the Jordan normal form. I can not apply this method to the arbitrary n x n matrix so how do I diagonalize this matrix?
$$ A= \begin{bmatrix} 2&-h&0&0&...&...&0&-h\\ -h&2&-h&0&...&...&0&0\\ 0&-h&2&-h&...&...&...&...\\ 0&0&-h&2&...&...&...&...\\ ...&...&...&...&...&...&0&0\\ ...&...&...&...&...&...&-h&0\\ 0&0&...&...&0&-h&2&-h\\ -h&0&...&...&0&0&-h&2 \end{bmatrix} $$
Hint: Try finding another matrix $S$ that has the same eigenvectors as this matrix. This matrix $S$ will be easier to diagonalize than $A$, and once we find the eigenvectors that it shares with $A$, we can easily compute the eigenvalues of these eigenvectors in $A$.
To find $S$, note that a necessary condition for $A$ and $S$ to have simultaneous eigenvectors is for $A$ and $S$ to commute. (Try showing this). Then, you can show that if $X$ is an eigenvector of $A$, then so is $SX$. Thus $S^2 X$ must be an eigenvector of A. Thus $S^3 X$ must be an eigenvector of A. Find $S$ such that $S^n X = X$, that is, find $S$ such that $S^n = I$, the identity matrix. Then the eigenvalues are the $n^{th}$ roots of unity, and one can easily write down the eigenvectors. I'll leave it up to you to find the $S$, but it will help to think of it in terms of some symmetry of the system- observe the action of the matrix $A$ on some $n \times 1$ column vector.