Sum of positive definite matrices still positive definite?

Solution 1:

A real matrix $M$ is positive-definite if and only if it is symmetric and $u^TMu > 0$ for all nonzero vectors $u$.

Now if $A$ and $B$ are positive-definite, $u^T(A+B)u = \ldots?$

This is an more detailed answer.

Now let $ A $ and $B$ be positive definite matrices, that is for all $ h \; \in \mathbb{R} ^{n}$ we must have $ h^{T}Ah > 0 $ and $ h^{T}Bh > 0 $.

From properties of real numbers

$0 < h^{T}Bh + h^{T}Ah $

Now from the distributive laws of matrix multiplication we must have

$0< h^{T}Bh + h^{T}Ah =h^{T} (B + A)h $

This implies that $0<h^{T} (B + A)h $ meaning $ B+A \succ 0$.

That is $ B+A $ is positive definite.