Condition for positive definite symmetric matrix in linear equations Given the equation
$B\cdot s = y$
for the positive definite symmetric matrix $B$ and the vectors $s$ and $y$, I read in http://perso.unifr.ch/ales.janka/numeroptim/09_newton1.pdf that the matrix B can only be positive definite symmetric if
$s^T\cdot y > 0$
Why is that?
 A: The assertion in your question is not true. A simple counter-example:
$$B = \begin{bmatrix}
    1 & .3 \\
    .3 & 1
  \end{bmatrix}$$
The eigen values of $B$ are 1.3 and 0.7, so it is positive definite and also symmetric.
Also let $s=\begin{bmatrix}1 & 1\end{bmatrix}$
We get $y = Bs = \begin{bmatrix}1.3 & 1.3\end{bmatrix}$
Obviously we can see $s^Ty \neq 0$
As far as I can tell, the only matrix for which your assertion would be true would be the matrix that rotates by $\pi/2$.
$$B = \begin{bmatrix}
    0 & -1 \\
    1 & 0
  \end{bmatrix}$$

EDIT: Just noticed equation (8) in your paper and you simultaneously corrected your question. If the matrix $B$ is positive definite, then it can be written as (assuming it's diagonalizable):
$$B = P \Lambda P^{-1}$$
Here, $\Lambda$ is a diagonal matrix with eigen values along the diagonals.
For a symmetric matrix, $P$ is a rotation matrix and so $P^{-1}=P^T$. So we can write:
$$s^Ty = s^TBs = s^TP \Lambda P^Ts$$
Now let $P^Ts = u$
The equation above becomes:
$$s^Ty = u^T \Lambda u = \sum \lambda_i u_i^2$$
Where $u_i$ are the elements of the vector $u$ and $\lambda_i$ are the eigen values of $B$ that make up the diagonals of $\Lambda$. Since they are all positive, $\sum \lambda_i u_i^2$ must be positive as well.
A: I found the answer in a different book (of course after asking, not during the last couple of days...):
Simply multiply both sides with $s^T$. Combined this with the definition of positive definiteness (https://en.wikipedia.org/wiki/Definiteness_of_a_matrix):

$B$ is positive definite if and only if $s^T B s > 0$ for any vector $s$.

So if $B$ is supposed to be positive definite, then $s^T B s > 0$ and with $B s = y$ we get $s^T y > 0$.
A: We also need to assume
$s \ne 0; \tag 0$
then for positive definite symmetric $B$ we have an orthogonal matrix $O$,
$O^TO = OO^T = I, \tag 1$
with
$O^TBO = D = \text{diag}(b_1, b_2, \ldots, b_n), \tag 2$
where the
$b_i > 0, 1 \le i \le n, \tag 3$
are the eigenvalues of $B$; then
$Bs = y \tag 4$
yields
$s^TBs = s^Ty; \tag 5$
but
$s^TBs = s^TIBIs = s^T(OO^T)B(OO^T)s =  (s^TO)(O^TBO)(O^Ts)$
$= (s^TO)D(O^Ts) = (O^Ts)^TD(O^Ts) = \displaystyle \sum_1^n (O^Ts)_i d_i (O^Ts)_i = \sum_1^n d_i (O^Ts)_i^2 > 0, \tag 6$
provided
$s \ne 0, \tag 7$
where the $(O^Ts)_i$, $1 \le n$, are the components of $O^Ts$.  Combining (5) and (6) we find
$s^Ty = s^TBs > 0, \tag 8$
the desired inequality.
