If $A = L+D+U $ is symmetry positive definite (s.p.d) then so is $(D+L)D^{-1}(D+U)$ 
Let $A $ be symmetry positive definite (s.p.d). Matrix $A= L+D+U$, where L,D,U denote lower, diagonal, upper matrix of A, respectively. Show that matrix $M= (D+L)D^{-1}(D+U)$ is s.p.d

A is s.p.d then $U = L^T$, hence $M^T=M$. But I don't know how to show that M is p.d. Can anyone help me? Thank you so much.
 A: If $A$ is positive definite and $B$ is invertible then $B^TAB$ is positive definite because
$$ \left< B^TABx, x \right> = x^TB^TABx = (Bx)^T A (Bx) = \left< A(Bx), Bx \right> \geq 0 $$
and the expression is equal to zero if and only if $Bx = 0$ which happens if and only if $x = 0$ (because $B$ is invertible).
The matrix $A$ is positive definite and so the diagonal entries $a_{ii}$ of $A$ must be positive (because $a_{ii} = \left<Ae_i, e_i \right>$). This implies that $D$ is invertible and $D^{-1}$ is also positive definite. Hence, by the previous observation
$$ M = (D + L)D^{-1}(D + U) = (D + U)^T D^{-1} (D + U) $$
is also positive definite.
A: First note that
$$\begin{align}
M&=(D+L)D^{-1}(D+U)\\
&=(D+L)(D^{-1}D+D^{-1}U) \\
&=(D+L)(I+D^{-1}U) \\
&=D+DD^{-1}U+LI+LD^{-1}U\\
&=D+IU+L+LD^{-1}U \\
&=(L+D+U)+LD^{-1}U \\
&=(L+D+U)+U^{T}D^{-1}U \\
&= A+B
\end{align}$$
But by your assumptions $A$ is S.P.D. Also you can easily show that $B$ is S.P.D too. So the sum of two S.P.D matrices is a S.P.D matrix.
