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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.

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  • $\begingroup$ What is $A_n$? Are you interested to show $C$ is positive definite? $\endgroup$
    – G_0_pi_i_e
    Mar 18, 2016 at 2:54
  • $\begingroup$ Hint: Use interlacing property. Note $2$ is always an eigenvalue. It's multiplicity is $2$ if $n$ is even, otherwise it is $1$. $\endgroup$
    – G_0_pi_i_e
    Mar 18, 2016 at 4:00
  • $\begingroup$ I refer to the notion of classified lie-algebras link so $C=(2 Id - A)$ where $A$ is the adjacency matrix of the corresponding Dynkin diagram. I know that $\det C=4$ but I either need all eigenvalues to be positiv or is need all the determinants of the minors both are not easy to calculate beause the matrix doesnt have a block diagonal form. Knowing that 2 is an eigenvalue and its multiplicity isnt enought I still have $n-1$ resp. $n-2$ to check weather they are positiv. $\endgroup$
    – milkpirate
    Mar 18, 2016 at 13:49
  • $\begingroup$ You need only to show the smallest eigenvalue is positive. Observe that the components of eigenvector corresponding to the smallest eigenvalue are positive. $\endgroup$
    – G_0_pi_i_e
    Mar 19, 2016 at 3:30
  • $\begingroup$ Another Hint: A matrix is positive definite if it is symmetric and all its pivots are positive. $\endgroup$
    – G_0_pi_i_e
    Mar 19, 2016 at 3:34

1 Answer 1

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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.

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