symmetric matrix and eigenvalues Consider A as an nxn symmetric positive definite matrix. Suppose there are $L, L^T$ such that:
$$A'=L A L^T$$
Sylvester's law of inertia says, that A' has the same positive eigenvalues.
How can I conclude, that A' is also positive definite and symmetric?
 A: Well, first we must assume that $L$ is nonsingular otherwise this will not work. Taking $(LAL^T)^T$ we obtain again $LAL^T$.
Now suppose we have a nonzero vector $x$. Then $x^T(LAL^T)x = (x^TL)A(L^Tx)$ and since $A$ is positive definite, we know this will be greater than $0$ and so $A'$ is also positive definite. 
A: All matrices involved will be $n \times n.$First, $$A'^T=(LAL^T)^T$$
$$=(L^T)^TA^TL^T$$
$$=LAL^T$$
$$=A'$$ so $A'$ is symmetric. [Your question is about eigenvalues. Before I answer that, let's look at some issues with Sylvester's law.]Second, Sylvester's law of inertia is about diagonalization by congruence, $i.e.$ given a symmetric matrix $M$ over the reals, it is always possible to find a real invertible matrix $P$ such that
$P^TAP=D$ where $D$ is diagonal. We say that we have diagonalized the matrix $M$ by congruence. The matrices $P$ and $D$ not at all unique. Sylvester's law of inertia says that no matter which $P$ and $D$ are used, the number of positive, negative and 0 terms on the diagonal will be constant. In particular if all the terms on the diagonal are positive in one diagonalization by congruence, then they will all be positive in every diagonalization by congruence. Third, the definition of a matrix $M$ being positive-definite [We may as well assume that $M$ is symmetric.] is that $v^TMv>0$ for every non-zero column-vector $v.$ Fourth, for every real symmetric matrix $M$ there is an orthogonal matrix $B$ such that $$B^TMB=E$$ where $E$ is diagonal. Since an orthogonal matrix is invertible, orthogonal diagonalization is a special case of digonalization by congruence. However, the calculations are much harder. In the case of orthogonal digonalization of a real symmetric matrix $M$, the elements on the diagonal are the eigenvalues of $M.$ Thus the following conditions are equivalent for a real symmetric matrix $M$ (i) in some diagonalization by congruence, all the diagonal elements are positive. (ii) in every diagonalization by congruence, all the diagonal elements are positive. (iii) the matrix $M$ is positive-definite. (iv) all the eigenvalues are positive. Going back to your notation $$A'=LAL^T.$$ Since $A$ is positive-definite, there exists an invertible matrix $P$ such that $P^TAP=D$ where $D$ is diagonal and every element on the diagonal is positive. Then $$A=L^{-1}A'(L^{-1})^T$$ and thus
$$((L^{-1})^TP)^TA'((L^{-1})^TP)=D,$$ so $A$ is positive-definite.
