[proofplan] We prove both inclusions directly. If $Y$ is positive semidefinite, then for each positive semidefinite $X$ we diagonalize $X$ and express $\operatorname{tr}(XY)$ as a nonnegative weighted sum of quadratic forms of $Y$. Conversely, if $Y$ is not positive semidefinite, a vector detecting the negative quadratic form produces the rank-one positive semidefinite matrix $vv^\top$, whose trace pairing with $Y$ is negative. [/proofplan] [step:Pair a positive semidefinite matrix with an arbitrary positive semidefinite test matrix] Fix $Y \in \mathbb{S}_+^n$. We prove that $Y \in (\mathbb{S}_+^n)^*$. Let $X \in \mathbb{S}_+^n$ be arbitrary. By the finite-dimensional spectral theorem for real symmetric matrices (citing a result not yet in the wiki: finite-dimensional spectral theorem for real symmetric matrices), there exist an [orthonormal basis](/page/Orthonormal%20Basis) $u_1,\dots,u_n \in \mathbb{R}^n$ and real eigenvalues $\lambda_1,\dots,\lambda_n \in \mathbb{R}$ such that \begin{align*} X = \sum_{i=1}^n \lambda_i u_i u_i^\top. \end{align*} Since $X$ is positive semidefinite, each eigenvalue is nonnegative: for each $i \in \{1,\dots,n\}$, \begin{align*} \lambda_i = u_i^\top X u_i \geq 0. \end{align*} Using linearity of the trace and the identity $\operatorname{tr}(ab^\top M) = b^\top M a$ for vectors $a,b \in \mathbb{R}^n$ and a matrix $M \in \mathbb{R}^{n \times n}$, we obtain \begin{align*} \langle X,Y\rangle = \operatorname{tr}(XY) = \sum_{i=1}^n \lambda_i \operatorname{tr}(u_i u_i^\top Y). \end{align*} For each $i$, \begin{align*} \operatorname{tr}(u_i u_i^\top Y) = u_i^\top Y u_i. \end{align*} Because $Y \in \mathbb{S}_+^n$, each quadratic form value $u_i^\top Y u_i$ is nonnegative. Therefore \begin{align*} \langle X,Y\rangle = \sum_{i=1}^n \lambda_i u_i^\top Y u_i \geq 0. \end{align*} Since $X \in \mathbb{S}_+^n$ was arbitrary, $Y \in (\mathbb{S}_+^n)^*$. [guided] Fix $Y \in \mathbb{S}_+^n$. To show that $Y$ belongs to the dual cone, we must prove that its trace pairing with every positive semidefinite matrix is nonnegative. Let $X \in \mathbb{S}_+^n$ be arbitrary. The useful move is to diagonalize $X$, because positive semidefiniteness of $X$ then appears as nonnegativity of its eigenvalues. By the finite-dimensional spectral theorem for real symmetric matrices (citing a result not yet in the wiki: finite-dimensional spectral theorem for real symmetric matrices), there are an orthonormal basis $u_1,\dots,u_n \in \mathbb{R}^n$ and eigenvalues $\lambda_1,\dots,\lambda_n \in \mathbb{R}$ such that \begin{align*} X = \sum_{i=1}^n \lambda_i u_i u_i^\top. \end{align*} For each $i$, the vector $u_i$ has Euclidean norm $1$, and the eigenvalue relation gives $Xu_i = \lambda_i u_i$. Hence \begin{align*} u_i^\top X u_i = u_i^\top \lambda_i u_i = \lambda_i |u_i|^2 = \lambda_i. \end{align*} Since $X$ is positive semidefinite, the left-hand side is nonnegative, so $\lambda_i \geq 0$ for every $i$. Now compute the trace pairing. Linearity of the trace gives \begin{align*} \operatorname{tr}(XY) = \operatorname{tr}\left(\sum_{i=1}^n \lambda_i u_i u_i^\top Y\right) = \sum_{i=1}^n \lambda_i \operatorname{tr}(u_i u_i^\top Y). \end{align*} For vectors $a,b \in \mathbb{R}^n$ and a matrix $M \in \mathbb{R}^{n \times n}$, the identity $\operatorname{tr}(ab^\top M) = b^\top M a$ follows by expanding matrix entries. Applying it with $a = u_i$, $b = u_i$, and $M = Y$, we get \begin{align*} \operatorname{tr}(u_i u_i^\top Y) = u_i^\top Y u_i. \end{align*} Because $Y$ is positive semidefinite, every quadratic form value $u_i^\top Y u_i$ is nonnegative. Thus each summand $\lambda_i u_i^\top Y u_i$ is a product of two nonnegative [real numbers](/page/Real%20Numbers). Therefore \begin{align*} \langle X,Y\rangle = \operatorname{tr}(XY) = \sum_{i=1}^n \lambda_i u_i^\top Y u_i \geq 0. \end{align*} Since this holds for every $X \in \mathbb{S}_+^n$, the definition of the dual cone gives $Y \in (\mathbb{S}_+^n)^*$. [/guided] [/step] [step:Detect a nonpositive semidefinite matrix by a rank-one test matrix] We prove the reverse inclusion by contraposition. Let $Y \in \mathbb{S}^n$ and suppose $Y \notin \mathbb{S}_+^n$. By the definition of $\mathbb{S}_+^n$, there exists a vector $v \in \mathbb{R}^n$ such that \begin{align*} v^\top Yv < 0. \end{align*} In particular, $v \neq 0$. Define the rank-one symmetric matrix $X \in \mathbb{S}^n$ by \begin{align*} X := vv^\top. \end{align*} For every $z \in \mathbb{R}^n$, \begin{align*} z^\top Xz = z^\top vv^\top z = (v^\top z)^2 \geq 0. \end{align*} Thus $X \in \mathbb{S}_+^n$. Using the trace identity $\operatorname{tr}(ab^\top M) = b^\top M a$ with $a = v$, $b = v$, and $M = Y$, we obtain \begin{align*} \langle X,Y\rangle = \operatorname{tr}(vv^\top Y) = v^\top Yv < 0. \end{align*} Therefore $Y$ fails the defining inequality for membership in $(\mathbb{S}_+^n)^*$, because $X \in \mathbb{S}_+^n$ but $\langle X,Y\rangle < 0$. Hence $Y \notin (\mathbb{S}_+^n)^*$. [/step] [step:Combine the two inclusions] The first step proves $\mathbb{S}_+^n \subseteq (\mathbb{S}_+^n)^*$. The second step proves the contrapositive of $(\mathbb{S}_+^n)^* \subseteq \mathbb{S}_+^n$. Therefore the two sets are equal: \begin{align*} (\mathbb{S}_+^n)^* = \mathbb{S}_+^n. \end{align*} This proves the [self-duality of the positive semidefinite cone](/theorems/6709) with respect to the trace [inner product](/page/Inner%20Product). [/step]