please consider the following Cartan matrix (it corresponds to $D_n$ $-$ zeros are replaced by "."'s for better view)
$ C=C_{D_n}=\begin{bmatrix} % dd 2 & . & -1 & . & \cdots & . & . & . \\ . & 2 & -1 & . & \cdots & . & . & . \\ -1 & -1 & 2 & -1 & \cdots & . & . & . \\ . & . & -1 & 2 & \cdots & . & . & . \\ \vdots & \vdots & \vdots & \vdots& \ddots &\vdots & \vdots & \vdots \\ . & . & . & . & \cdots & 2 & -1 & . \\ . & . & . & . & \cdots & -1 & 2 & -1 \\ . & . & . & . & \cdots & . & -1 & 2 \end{bmatrix} $
I am interested in the eingenvalues of the matrix to show that a bilinear form made up by
$ Q(x):=x^t C x $
is positiv definite. The Eigenvalues of $A_n$ are easy to get because its Cartan matrix is also a Teoplitz matrix (i.e. a special tridiagonal one) and there exists a formula for the eigenvalues.
Since my $C$ as above doesnt have this convinent shape I am hoping that you can give me a hint where I can find the proof for that.
Thank you very much
EDIT: Thank you. I used the Schur Complement to calculate the determinatens. Since $B$ is really simple and $C=B^t$ it breaks down to get the $(1,1)$ entry of the inverse of the really simple tridiagonal matrix $D$ (schur notion). This can be explicitly given (see: p. 16 and 11 of Explicit inverses of some tridiagonal matrices). And since the principal minors are sub matrices of one another (as you stated) the result turns out well.
A symmetric matrix is positive definite if and only if all its leading principal minors are positive. Let $MC_i$ denote the leading principal minor of order $i$. Then $MC_1=2,~~ MC_2= MC_3= \ldots = MC_n=4$. Hence the result.