[proofplan]
We rewrite the semidefinite programs as a primal-dual pair over the cone $\mathbb{S}^n_+$ and compute the adjoint of the affine constraint map. [Weak duality](/theorems/2549) follows directly from the trace [inner product](/page/Inner%20Product) and [self-duality of the positive semidefinite cone](/theorems/6709). The attainment and zero-gap conclusions are exactly the finite-dimensional conic strong duality theorem under Slater's condition, applied first to the primal problem and then to the dual problem with the roles interchanged.
[/proofplan]
[step:Express the programs through the linear constraint map]
Define the [linear map](/page/Linear%20Map)
\begin{align*}
\mathcal{A}: \mathbb{S}^n &\to \mathbb{R}^m
\end{align*}
by
\begin{align*}
\mathcal{A}(X) := (\langle A_1,X\rangle,\dots,\langle A_m,X\rangle).
\end{align*}
Then the primal problem is
\begin{align*}
p^*=\inf\{\langle C,X\rangle : X \in \mathbb{S}^n_+,\ \mathcal{A}(X)=b\}.
\end{align*}
The adjoint map
\begin{align*}
\mathcal{A}^*: \mathbb{R}^m &\to \mathbb{S}^n
\end{align*}
is determined by the identity
\begin{align*}
\mathcal{A}(X)\cdot y=\langle X,\mathcal{A}^*y\rangle
\end{align*}
for every $X\in\mathbb{S}^n$ and every $y\in\mathbb{R}^m$. Since
\begin{align*}
\mathcal{A}(X)\cdot y=\sum_{i=1}^m y_i\langle A_i,X\rangle=\left\langle \sum_{i=1}^m y_iA_i,X\right\rangle,
\end{align*}
we have
\begin{align*}
\mathcal{A}^*y=\sum_{i=1}^m y_iA_i.
\end{align*}
Thus the dual cone condition $C-\mathcal{A}^*y\in\mathbb{S}^n_+$ is precisely
\begin{align*}
C-\sum_{i=1}^m y_iA_i\in\mathbb{S}^n_+.
\end{align*}
[guided]
The purpose of introducing $\mathcal{A}$ is to put the problem into standard conic form. Define
\begin{align*}
\mathcal{A}: \mathbb{S}^n &\to \mathbb{R}^m
\end{align*}
by
\begin{align*}
\mathcal{A}(X):=(\langle A_1,X\rangle,\dots,\langle A_m,X\rangle).
\end{align*}
Then the constraints $\langle A_i,X\rangle=b_i$ for all $i=1,\dots,m$ are exactly the single vector equation $\mathcal{A}(X)=b$, so the primal problem becomes
\begin{align*}
p^*=\inf\{\langle C,X\rangle : X\in\mathbb{S}^n_+,\ \mathcal{A}(X)=b\}.
\end{align*}
Now compute the adjoint. The adjoint
\begin{align*}
\mathcal{A}^*: \mathbb{R}^m &\to \mathbb{S}^n
\end{align*}
is the unique linear map satisfying
\begin{align*}
\mathcal{A}(X)\cdot y=\langle X,\mathcal{A}^*y\rangle
\end{align*}
for every $X\in\mathbb{S}^n$ and $y\in\mathbb{R}^m$. Expanding the left-hand side gives
\begin{align*}
\mathcal{A}(X)\cdot y=\sum_{i=1}^m y_i\langle A_i,X\rangle.
\end{align*}
By bilinearity and symmetry of the trace inner product,
\begin{align*}
\sum_{i=1}^m y_i\langle A_i,X\rangle=\left\langle \sum_{i=1}^m y_iA_i,X\right\rangle.
\end{align*}
Therefore
\begin{align*}
\mathcal{A}^*y=\sum_{i=1}^m y_iA_i.
\end{align*}
Consequently the abstract conic dual constraint $C-\mathcal{A}^*y\in\mathbb{S}^n_+$ is exactly the semidefinite inequality
\begin{align*}
C-\sum_{i=1}^m y_iA_i\in\mathbb{S}^n_+.
\end{align*}
[/guided]
[/step]
[step:Verify weak duality for every primal-dual feasible pair]
Let $X\in\mathbb{S}^n_+$ satisfy $\mathcal{A}(X)=b$, and let $y\in\mathbb{R}^m$ satisfy $C-\mathcal{A}^*y\in\mathbb{S}^n_+$. Since $\mathbb{S}^n_+$ is self-dual with respect to the trace inner product, the matrices $X\in\mathbb{S}^n_+$ and $C-\mathcal{A}^*y\in\mathbb{S}^n_+$ satisfy
\begin{align*}
\langle C-\mathcal{A}^*y,X\rangle \ge 0.
\end{align*}
Using the definition of the adjoint and the equality constraint,
\begin{align*}
\langle C,X\rangle-b\cdot y
=
\langle C,X\rangle-\mathcal{A}(X)\cdot y
=
\langle C,X\rangle-\langle \mathcal{A}^*y,X\rangle
=
\langle C-\mathcal{A}^*y,X\rangle.
\end{align*}
Hence
\begin{align*}
b\cdot y\le \langle C,X\rangle.
\end{align*}
Taking the supremum over all dual feasible $y$ and the infimum over all primal feasible $X$ gives $d^*\le p^*$.
[/step]
[step:Apply conic Slater duality to primal strict feasibility]
Assume that the primal feasible set is nonempty, that $p^*\in\mathbb{R}$, and that there exists $X_0\in\mathbb{S}^n$ with $X_0\succ 0$ and $\mathcal{A}(X_0)=b$. The condition $X_0\succ 0$ says precisely that the affine space $\{X\in\mathbb{S}^n:\mathcal{A}(X)=b\}$ meets the interior of the cone $\mathbb{S}^n_+$.
By the finite-dimensional conic strong duality theorem under Slater's condition, applied to the cone program
\begin{align*}
\inf\{\langle C,X\rangle : X\in\mathbb{S}^n_+,\ \mathcal{A}(X)=b\},
\end{align*}
the dual optimum is attained and the dual optimal value equals the primal optimal value. This theorem is being invoked here as the standard Slater strong-duality theorem for finite-dimensional conic programs; citing a result not yet in the wiki: finite-dimensional conic strong duality under Slater's condition.
Using the adjoint computation from the first step, the dual problem supplied by that conic theorem is exactly
\begin{align*}
\sup\left\{b\cdot y : y\in\mathbb{R}^m,\ C-\sum_{i=1}^m y_iA_i\in\mathbb{S}^n_+\right\}.
\end{align*}
Therefore $d^*=p^*$ and there exists $y^*\in\mathbb{R}^m$ such that
\begin{align*}
C-\sum_{i=1}^m y_i^*A_i\in\mathbb{S}^n_+
\end{align*}
and
\begin{align*}
b\cdot y^*=d^*=p^*.
\end{align*}
[/step]
[step:Apply the same conic duality theorem to dual strict feasibility]
Assume that the dual feasible set is nonempty, that $d^*\in\mathbb{R}$, and that there exists $y_0\in\mathbb{R}^m$ such that
\begin{align*}
S_0:=C-\sum_{i=1}^m (y_0)_iA_i \succ 0.
\end{align*}
The dual can be written as the conic program
\begin{align*}
\sup\{b\cdot y : y\in\mathbb{R}^m,\ C-\mathcal{A}^*y\in\mathbb{S}^n_+\}.
\end{align*}
The condition $S_0\succ 0$ says that this feasible affine set meets the interior of $\mathbb{S}^n_+$.
Applying the finite-dimensional conic strong duality theorem under Slater's condition to this dual conic program, with the roles of primal and dual interchanged, gives attainment of the opposite program and equality of optimal values. The opposite program is the original primal program because $\mathbb{S}^n_+$ is self-dual and because the adjoint of $\mathcal{A}^*$ is $\mathcal{A}$ under the trace and Euclidean inner products.
Hence there exists $X^*\in\mathbb{S}^n_+$ such that
\begin{align*}
\mathcal{A}(X^*)=b
\end{align*}
and
\begin{align*}
\langle C,X^*\rangle=p^*=d^*.
\end{align*}
Equivalently,
\begin{align*}
\langle A_i,X^*\rangle=b_i
\end{align*}
for every $i=1,\dots,m$, so the primal infimum is attained and the primal and dual optimal values are equal.
[/step]
