Solution set of an LMI is convex

The affine function is $T(x) = B - x_1 A_1 - \cdots - x_n A_n $.

The solution set to your LMI can be described as \begin{equation} \{ x \mid T(x) \succeq 0 \} = T^{-1}(S^m_+), \end{equation} where $S^m_+$ is the positive semidefinite cone in $\mathbb R^{m\times m}$.

Further details:

If we view $A_1,\ldots,A_n$ and $B$ as column vectors in $\mathbb R^{m^2}$, then \begin{equation} T(x) = \underset{\substack{\Bigg \uparrow \\m^2 \times 1}}{B} - \underset{\substack{\Bigg \uparrow \\ m^2 \times n}}{A} \underset{\substack{\uparrow \\ n \times 1}}{x} \end{equation} where \begin{equation} A = \begin{bmatrix} A_1 & A_2 & \cdots & A_n \end{bmatrix}. \end{equation} In this equation, the $A$ is multiplied by $x$ using ordinary matrix multiplication.

Given a linear matrix polynomial $\mathrm A : \mathbb{R}^{n} \to \mbox{Sym}_k (\mathbb{R})$ defined by

$$\mathrm A (x) = \mathrm A_0 + x_1 \mathrm A_1 + \cdots + x_n \mathrm A_n$$

where $\mbox{Sym}_k (\mathbb{R})$ is the set of $k \times k$ real symmetric matrices, and matrices $\mathrm A_0, \mathrm A_1, \dots, \mathrm A_n$ are symmetric, we form the following linear matrix inequality (LMI)

$$\mathrm A (\mathrm x) \succeq \mathrm O_k$$

whose solution set is the spectrahedron

$$\mathcal S := \{ \mathrm x \in \mathbb{R}^n \mid \mathrm A (\mathrm x) \succeq \mathrm O_k\}$$

Let $\mathrm x, \mathrm y \in \mathcal S$ and $\lambda \in [0,1]$. Hence,

$$\mathrm A (\lambda \mathrm x + (1-\lambda) \mathrm y) = \cdots = \lambda \, \underbrace{\mathrm A (\mathrm x)}_{\succeq \mathrm O_k} + (1-\lambda) \, \underbrace{\mathrm A (\mathrm y)}_{\succeq \mathrm O_k} \succeq \mathrm O_k$$

Since a convex combination of two positive semidefinite matrices is also positive semidefinite, we conclude that set $\mathcal S$ is convex.

---

convex-analysis linear-matrix-inequality spectrahedra