Prove that the set of every PSD symmetric matrices are convex (4)

P={A∣A∈Sn,A≥0}P = \{A|A \in \mathbb{S}^n, A \ge 0\}

We know that

∀x⃗∈Rn,x⃗⊤Ax⃗≥0\forall \vec{x} \in \mathbb{R}^n, \vec{x}^\top A \vec{x} \ge 0

Then let any A1,A2∈PA_1, A_2 \in P

B=θA1+(1−θ)A2B = \theta A_1+(1-\theta) A_2

Let’s check if BB is PSD

x⃗⊤Bx⃗=θx⃗⊤A1x⃗undefined≥0+(1−θ)x⃗⊤A2x⃗undefined≥0≥0\vec{x}^\top B\vec{x} = \theta \underbrace{\vec{x}^\top A_1 \vec{x}}_{\ge 0} + (1-\theta) \underbrace{\vec{x}^\top A_2 \vec{x}}_{\ge 0} \ge 0

Therefore

B≥0B \ge 0

The symmetry can be checked easily without multiplying vectors