Positive Semidefinite Matrices form a Convex Cone

Last updated Nov 1, 2022

Suppose $n \in \mathbb{N}$. Then $\mathbb{S}_{n}^{+}$ is a Convex Cone over $\mathbb{R}$.

Suppose $X, Y \in \mathbb{S}_{n}^{+}$ and $a, b \geq 0$. Consider $Z = aX + bY$. For any $x \in \mathbb{R}^{n}$, we have that

$$\begin{align*} x^{T}Z x &= x^{T}(aX + bY) x\\ &=a x^{T}X x + b x^{T} Y x\\ &\geq 0 \end{align*}$$

Since $X, Y \in \mathbb{S}_{n}^{+}$ and $a, b \geq 0$. $\blacksquare$

• Positive Semidefinite Matrix
• Real Numbers