Every convex polyhedron is a spectrahedron I'm trying to show that convex polyhedra are special cases of spectrahedra. This was left as an exercise to the reader in a convex optimization text that I'm reading. I'm not sure how standard the definitions and notation in this book are, so I'll include them below.

Notation: Letting $A, B \in \mathbb{R}^{m \times m}$ be symmetric matrices, write $B \succeq A$ if $B - A \succeq 0$, i.e., $B-A$ is positive semidefinite. For vectors $a, b \in \mathbb{R}^n$, write $a \geq b$ if $ a_i \geq b_i \forall i \in \{1, 2, \dots, n\}$. 

Definition: Given symmetric matrices $A_1, A_2, \dots, A_n, B \in \mathbb{R}^{m \times m}$, the following convex set $$\left\{ x \in \mathbb{R}^n \,\,\middle|\,\, \sum_{i = 1}^n x_i A_i \preceq B \right\}$$ is called a spectrahedron.

Definition: Given $A \in \mathbb{R}^{m \times n}$ and $b \in \mathbb{R}^m$, $\left\{ x \in \mathbb{R}^n \mid Ax \leq b \right\}$
is called a convex polyhedron.

My attempt so far:
I've been trying to come up with a representation of a spectrahedron, probably with $B = bb^T$. I was thinking of $AA^T = A_i$ for $i = 1, \dots, n$. But I haven't made much progress with this approach. I figure making a solid guess on the form and then showing "$\subseteq$" and "$\supseteq$" is the best way to prove this.
 A: Given $\mathrm A \in \mathbb{R}^{m \times n}$ and $\mathrm b \in \mathbb{R}^m$, let
$$\mathcal P := \{ \mathrm x \in \mathbb{R}^n \mid \rm A x \leq b \}$$
which is a (convex) $\mathcal H$-polytope (i.e., defined by the intersection of closed half-spaces). Let
$$p_i (\mathrm x) := b_i - a_{i1} x_1 - a_{i2} x_2 - \dots - a_{in} x_n$$
where $i \in \{1, 2, \dots, m\}$, be $n$-variate polynomials of degree $1$. Hence,
$$\mathcal P = \{ \mathrm x \in \mathbb{R}^n \mid p_1 (\mathrm x) \geq 0 \land p_2 (\mathrm x) \geq 0 \land \cdots \land p_m (\mathrm x) \geq 0 \}$$
The conjunction of the $m$ non-strict linear inequalities is equivalent to the following linear matrix inequality (LMI)
$$\mathrm M (\mathrm x) := \mathrm M_0 + x_1 \mathrm M_1 + x_2 \mathrm M_2 + \dots + x_n \mathrm M_n = \begin{bmatrix} p_1 (\mathrm x) & 0 & \ldots & 0\\ 0 & p_2 (\mathrm x) & \ldots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \ldots & p_m (\mathrm x)\end{bmatrix} \succeq \mathrm O_m$$
where the $m \times m$ diagonal matrices $\mathrm M_0, \mathrm M_1, \dots, \mathrm M_n$ are given by
$$\mathrm M_0 := \begin{bmatrix} b_1 & 0 & \ldots & 0\\ 0 & b_2 & \ldots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \ldots & b_m\end{bmatrix} \qquad \qquad \mathrm M_j := \begin{bmatrix} -a_{1j} & 0 & \ldots & 0\\ 0 & -a_{2j} & \ldots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \ldots & -a_{m j}\end{bmatrix}$$
for $j \in \{1, 2,\dots, n\}$. Thus, convex polytope $\mathcal P$ can be defined using an LMI
$$\mathcal P = \{ \mathrm x \in \mathbb{R}^n \mid \mathrm M (\mathrm x) \succeq \mathrm O_m \}$$
and, thus, is a (convex) spectrahedron.

convex-analysis discrete-geometry polyhedra polytopes linear-matrix-inequality spectrahedra
A: I was able to solve this using Johan's hint. Rodrigo's answer is sleeker than mine (albeit the same idea), but I thought that I'd post this as well, since it may be easier to understand for someone with my background or less in this area of math.

Let $P$ be a convex polyhedron, defined as
$$
P := \left\{ x \in \mathbb{R}^n \middle| Ax \leq b \right\},
$$
where $A =
\begin{pmatrix}
\vec{a}_1^T \\ \vdots \\ \vec{a}_m^T
\end{pmatrix}
=
\begin{pmatrix} 
a_{11} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{m1} & \cdots & a_{mn}
\end{pmatrix}
\in \mathbb{R}^{m \times n}$ and $b = \begin{pmatrix} b_1 \\ \vdots \\ b_m \end{pmatrix}$. 
Define 
$B := 
\begin{pmatrix}
b_1 & \cdots & 0 \\ 
\vdots & \ddots & \vdots \\ 
0 & \cdots & b_m
\end{pmatrix}$
and 
$Q_i := 
\begin{pmatrix}
a_{1i} & \cdots & 0 \\ 
\vdots & \ddots & \vdots \\ 
0 & \cdots & a_{mi}
\end{pmatrix}$, $i = 1, \cdots, n$.
Note:
\begin{align*}
\sum_{i=1}^n x_i Q_i = \sum_{i=1}^n x_i \begin{pmatrix}
a_{1i} & \cdots & 0 \\ 
\vdots & \ddots & \vdots \\ 
0 & \cdots & a_{mi}
\end{pmatrix}
&= \begin{pmatrix}
\sum_{i=1}^n a_{1i} x_i & \cdots & 0 \\ 
\vdots & \ddots & \vdots \\ 
0 & \cdots & \sum_{i=1}^n a_{mi} x_i
\end{pmatrix} \\
&= 
\begin{pmatrix}
\vec{a}_1^T x & \cdots & 0 \\ 
\vdots & \ddots & \vdots \\ 
0 & \cdots & \vec{a}_m^T x
\end{pmatrix} \\
&=
\begin{pmatrix}
\left( Ax \right)_1 & \cdots & 0 \\ 
\vdots & \ddots & \vdots \\ 
0 & \cdots & \left( Ax \right)_m
\end{pmatrix}.
\end{align*}
Now define the spectrahedron
$$
P' := \left\{ x \in \mathbb{R}^n \middle| \sum_{i = 1}^n x_i Q_i \preceq B \right\}.
$$
Claim: $P = P'$. 
Proof:
"$\subseteq$": 
Let $x \in P$, i.e. $Ax \leq b \iff b_i - (Ax)_i \geq 0$ for all $i \in \{1, \cdots, n\}$.
Let $z \in \mathbb{R}^m$.
$$
z^T \left(B - \sum_{i=1}^n x_i Q_i \right)z = \sum_{j=1}^m z_j^2 \left( b_j - \left( \sum_{i=1}^n x_i Q_i \right)_{jj} \right),
$$
since $B - \sum_{i=1}^n x_i Q_i$ is a diagonal matrix.
From the the note above, we have
$$
z^T \left(B - \sum_{i=1}^n x_i Q_i \right)z = \sum_{j=1}^m z_j^2 \underbrace{\left( b_j - (Ax)_j \right)}_{\geq 0 \text{ since } x \in P} \geq 0.
$$
Hence $B - \sum_{i=1}^n x_i Q_i \succeq 0$ and $x \in P'$.
"$\supseteq$":
Let $x \in P'$, i.e. $z^T \left( B - \sum_{i=1}^n x_i Q_i \right) z \geq 0$ $\forall z \in \mathbb{R}^m$. 
Suppose that $Ax \nleq b$.
Define the index set $J := \left\{ j \in \{1, \cdots, m\} \middle| (Ax)_j > b_j \right\}$, which is nonempty by hypothesis.
Define $z \in \mathbb{R}^m$ by $z_j = 1$ if $j \in J$, $z_j = 0$ otherwise. Then
\begin{align*}
z^T \left( B - \sum_{i=1}^n x_i Q_i \right) z &= \sum_{j \in J} z_j^2 \left( b_j - \left( \sum_{i=1}^n x_i Q_i \right)_{jj} \right) \\
&= \sum_{j \in J} \underbrace{z_j^2}_{=1} \underbrace{ \left( b_j - \left( Ax \right)_j \right) }_{ < 0 \text{ since } j \in J} < 0,
\end{align*}
which contradicts $x \in P'$. Hence $Ax \leq b$, so $x \in P$.
