Prove a specific Cartan matrix is positive definite

Question

I am trying to prove that the following matrix is positive definite, but I am stuck in the last step of my proof... Any help would be really appreciated. Thanks!

Question

Let $A$ be a matrix with entries as follows: $$A_{ii}=2, A_{i,i+1}=A_{i+1,i}=-1, i=1,...,n,$$ and all the remaining entries equal to zero.

Answer

I am going to prove this by induction.

Let us first analyse the basic cases:

• $k=1$: $M_1(A)=2>0$
• $k=2$: $M_2(A)=det\left( \begin{array}{ccc} 2 & -1 \\ -1 & 2 \end{array} \right)=3>0$
• $k=3$: $M_3(A)=det\left( \begin{array}{ccc} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{array} \right)=4>0$

And we see that the principal minors for $k=1,2,$ and $3$ are positive. Now let us assume that the same happens when $k=n-1$, $M_{n-1}(A)=n>0$. Then we have that $$M_n(A)=2M_{n-1}(A)-(-1)(\text{some matrix})??$$

So there is where I get lost... How do I show that $det(A)=M_n(A)=n+1$?

Thanks a lot!

Answer 1

The previous answer had a computational mistake, so here is another (hopefully correct approach). We want to find a recurrence relation between det($M_{n}$) for different $n$. I will denote $D_{n} = det (M_{n})$.

Let us consider $D_{n+1}$ and we will compute it along the last row.

Step1: $$D_{n+1} = 2D_{n} + (-1)^{2n+1}. (-1). det(X)$$ Here $X$ is a $n \times n$ with $M_{n-1}$ in the principal diagonal and the last row is $(0,0,0,...-1, -1)$ and everything else is zero. But then, expanding along the last column of $X$ we get det$(X)= (-1)*D_{n-1}$.

Step2: Assuming $n \geq 1$ we get $D_{n+1} = 2D_{n} - D_{n-1}$ or $$ D_{n+1} - D_{n} = D_{n} - D_{n-1} $$.

Hopefully there are no embarrassing mistakes this time!