How to show that this matrix is symmetric definite positive I need help.
We want to show that the matrix $A$ is symmetric definite positive for all $n$ where $n$ is the size of the matrix $A$.
Here is $A$ :
$$
A = \begin{bmatrix}
 1&  -1&  0& ...& 0\\ 
 -1&  2&  -1& \ddots & \vdots\\ 
 0&  -1&  2& \ddots &0\\ 
 \vdots &  \ddots &  \ddots&  \ddots &-1 \\
0 & ...  &0 & -1& 2
\end{bmatrix} 
$$
I tried brute forcing by calculating for $x$ a vector of size $n$, $x^TAx$, but it seems tricky.
Thanks for your help
 A: HINT:
The matrix is symmetric and diagonally dominant, so positive semidefinite. To show that it is actually positive definite, you need to check that the kernel is null. Take a vector in the kernel. You notice that the first two components are equal, and then, the components are in arithmetic progression, but then notice the relation between the last two components. Thus the vector is $0$.
$\bf{Added:}$. Place $-\epsilon < 0$ instead of $-1$, and call the matrix $M_{\epsilon}$.  For every $\epsilon \in [0,1)$ the matrix is strictly diagonally dominant, so the eigenvalues are not $0$. The eigenvalues of $M_{\epsilon}$ vary continuously with $\epsilon$. Now the eigenvalues of $M_{0}$ are $>0$, so no eigenvalue of $M= M_1$ can be negative.
A: This follows directly from the Cholesky decomposition $A=(I-J)^T(I-J)$, where $J$ is the $n\times n$ upper triangular nilpotent Jordan block.
A: Let $f_n$ be the determinant of the $n$th leading principal minor.
We can quickly check that $f_1=1, f_2= 1$ and
$f_n = 2f_{n-1}-f_{n-2}$. It is not hard to check that the solution is $f_n =1$.
Hence $A$ is positive definite by Sylvester's criterion.
A: A symmetric matrix  is positive definite iff its leading principal minors are positive. It should be easy to show it by induction (maybe you can expand the determinants by Laplace on the last column).
