No linear matrix inequality describes the cone

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

This lesson proves the second half of Theorem A of Hyperbolic polynomials and spectrahedral cones. Suppose some linear matrix inequality \(L(X,Z,y)\succeq0\) of finite size describes the hyperbolicity cone \(K\). The slices of A positive map and a hyperbolic polynomial force \(L\), after a normalization and a limit, into a Schur-complement form. Letting \(X^{-1}\) and \(Z^{-1}\) tend to rank-one projections turns it into an identity between the Choi–Lam form and the norms of certain matrices. Differentiating that identity produces a nonzero bilinear form whose square lies below \(b\), which The Choi–Lam form excludes. The proof follows OpenAI [OpenAI-NS, Sections 4 to 6]. It never compares the determinant of the pencil with \(p\).

Throughout, \(\|\cdot\|_{\mathrm{op}}\) is the operator norm for the Euclidean norms, and for \(u\in\mathbf R^4\) we write \(u=(u_0,u')\) with \(u'\in\mathbf R^3\).

1. A pencil gives a Schur-complement inequality

Proposition 4.1 (OpenAI). Suppose \(K=\{x:L(x)\succeq0\}\) for a linear map \(L:\mathbb S^4\times\mathbb S^4\times\mathbf R^3\to\mathbb S^N\). Then there are integers \(a,c\ge1\), unital positive linear maps \(D:\mathbb S^4\to\mathbb S^a\) and \(E:\mathbb S^4\to\mathbb S^c\), and a linear map \(B:\mathbf R^3\to\mathbf R^{a\times c}\) such that for all \(X\succ0\), \(Z\in\mathbb S^4\) and \(y\in\mathbf R^3\), \[ \begin{aligned} &Z\succeq\Phi_y(X^{-1})\\ &\quad\iff E(Z)\succeq B(y)^{\mathsf T}D(X)^{-1}B(y). \end{aligned} \]

Proof. Write \(L(X,Z,y)=L_1(X)+L_2(Z)+L_3(y)\).

Removing a common kernel. Since \(e\) is an interior point of \(K\) (Proposition 3.3), for every \(x\) there is \(\varepsilon>0\) with \(e\pm\varepsilon x\in K\), so \(L(e)\pm\varepsilon L(x)\succeq0\), and \(\ker L(e)\subseteq\ker L(x)\) by Lemma 1.3(6) of the first lesson. A symmetric matrix maps into the orthogonal complement of its kernel. So in an orthonormal basis adapted to \(V\oplus\ker L(e)\), \(V=\ker L(e)^\perp\), every \(L(x)\) is block diagonal with lower block \(0\), and replacing \(L(x)\) by its upper block \(L_V(x)\) does not change the set \(\{x:L(x)\succeq0\}\). Moreover \(L_V(e)\succ0\). The space \(V\) is nonzero: otherwise \(L=0\) and \(K\) would be the whole space, while \((-I_4,I_4,0)\notin K\) by Proposition 3.3. We may therefore assume \(L(e)\succ0\).

Splitting off a block. By the second slice of Proposition 3.3, \((X,0,0)\in K\) and \((0,Z,0)\in K\) whenever \(X,Z\succeq0\); so \(L_1\) and \(L_2\) are positive maps. Let \(U=\ker L_1(I_4)\). For every \(X\in\mathbb S^4\) and small \(\varepsilon>0\), \(I_4\pm\varepsilon X\succeq0\), so \(L_1(I_4)\pm\varepsilon L_1(X)\succeq0\), and \(U\subseteq\ker L_1(X)\). In an orthonormal basis adapted to \(U^\perp\oplus U\), \[ L_1(X)=\begin{pmatrix}D(X)&0\\0&0\end{pmatrix}. \] Both summands are nonzero. If \(U^\perp=0\), then \(L_1=0\) and \(L(-I_4,I_4,0)=L_2(I_4)\succeq0\), so \((-I_4,I_4,0)\in K\), which is false. If \(U=0\), then \(L_1(I_4)\succ0\), so \(L_1(I_4)-\varepsilon L_2(I_4)\succ0\) for small \(\varepsilon>0\), and \((I_4,-\varepsilon I_4,0)\in K\), which is false. Let \(a=\dim U^\perp\), \(c=\dim U\), and let \(E(Z)\) be the lower diagonal block of \(L_2(Z)\). The maps \(D\) and \(E\) are positive, as compressions of positive maps. \(D(I_4)\succ0\) by the choice of \(U\), and \(E(I_4)\succ0\) because it is the compression of \(L(e)=L_1(I_4)+L_2(I_4)\succ0\) to \(U\). Replacing \(L\) by \(S^{\mathsf T}LS\) with \(S=\operatorname{diag}(D(I_4)^{-1/2},E(I_4)^{-1/2})\) does not change the described set (Lemma 1.3(4) of the first lesson) and makes \(D(I_4)=I_a\) and \(E(I_4)=I_c\). Write \[ \begin{aligned} L_2(Z)&=\begin{pmatrix}F(Z)&C(Z)\\ C(Z)^{\mathsf T}&E(Z)\end{pmatrix},\\ L_3(y)&=\begin{pmatrix}A(y)&B(y)\\ B(y)^{\mathsf T}&G(y)\end{pmatrix}. \end{aligned} \]

The block \(G\) vanishes. Fix \(y\) and put \(Z_0=\Phi_y(I_4)\). For every real \(s\), \(\Phi_{sy}(I_4)=s^2Z_0\), so \((I_4,s^2Z_0,sy)\in K\) by the first slice of Proposition 3.3. The lower block of \(L(I_4,s^2Z_0,sy)\) is \(s^2E(Z_0)+sG(y)\), and it is positive semidefinite, as a compression of a positive semidefinite matrix. Dividing by \(|s|\) and letting \(s\to0\) through positive and through negative values gives \(G(y)\succeq0\) and \(-G(y)\succeq0\), so \(G(y)=0\).

Rescaling. Let \(X\succ0\), \(Z\in\mathbb S^4\), \(y\in\mathbf R^3\) and \(s\ne0\). Conjugating \(L(X,s^2Z,sy)\) by \(\operatorname{diag}(I_a,s^{-1}I_c)\) gives \[ \begin{aligned} M_s&=\begin{pmatrix}D_s&B_s\\ B_s^{\mathsf T}&E(Z)\end{pmatrix},\\ D_s&=D(X)+sA(y)+s^2F(Z),\\ B_s&=B(y)+sC(Z). \end{aligned} \] As \(\Phi_{sy}=s^2\Phi_y\), the first slice gives \((X,s^2Z,sy)\in K\) if and only if \(Z\succeq\Phi_y(X^{-1})\). Hence, for every \(s\ne0\), \[ M_s\succeq0\iff Z\succeq\Phi_y(X^{-1}). \] As \(s\to0\), \(M_s\) tends to \[ M=\begin{pmatrix}D(X)&B(y)\\B(y)^{\mathsf T}&E(Z)\end{pmatrix}. \] If \(Z\succeq\Phi_y(X^{-1})\), all \(M_s\) are positive semidefinite, and so is their limit \(M\).

The converse. Suppose \(M\succeq0\). Since \(X\succeq\lambda_{\min}(X)I_4\) and \(D\) is positive and unital, \(D(X)\succeq\lambda_{\min}(X)I_a\succ0\). For \(\delta>0\), the matrix \(M+\operatorname{diag}(0,\delta I_c)\), which is the limit matrix for \(Z+\delta I_4\), is positive definite: on a vector \((w_1,w_2)\ne0\) its quadratic form is at least \(\delta\|w_2\|^2\), and equals \(w_1^{\mathsf T}D(X)w_1>0\) when \(w_2=0\). Positive definite matrices form an open set, so the matrices \(M_s\) formed with \(Z+\delta I_4\) are positive definite for small \(s\ne0\), and the equivalence for \(M_s\) gives \(Z+\delta I_4\succeq\Phi_y(X^{-1})\). Letting \(\delta\to0\) gives \(Z\succeq\Phi_y(X^{-1})\).

So \(Z\succeq\Phi_y(X^{-1})\) if and only if \(M\succeq0\), and by the Schur complement, Lemma 1.3(5) of the first lesson with \(P=D(X)\succ0\), if and only if \(E(Z)\succeq B(y)^{\mathsf T}D(X)^{-1}B(y)\). \(\square\)

2. Rank-one limits and a norm identity

For unit vectors \(u,v\in\mathbf R^4\) let \(P(v)\) be the orthogonal projection of \(\mathbf R^a\) onto \(\ker D(I_4-vv^{\mathsf T})\), and \(R(u)\) that of \(\mathbf R^c\) onto \(\ker E(I_4-uu^{\mathsf T})\).

Lemma 4.2. If \(H\in\mathbb S^n\) is positive semidefinite, then \([I+(t-1)H]^{-1/2}\) tends, as \(t\to\infty\), in operator norm to the orthogonal projection onto \(\ker H\).

Proof. In an orthonormal basis of eigenvectors of \(H\), with eigenvalues \(\mu\ge0\), the matrix is diagonal with entries \((1+(t-1)\mu)^{-1/2}\), which equal \(1\) for \(\mu=0\) and tend to \(0\) for \(\mu>0\). \(\square\)

Proposition 4.3 (OpenAI). Under the conclusion of Proposition 4.1, for all unit vectors \(u,v\in\mathbf R^4\) and all \(y\in\mathbf R^3\), \[ \begin{aligned} &b(v_0u'-u_0v',\,y)\\ &\quad=\bigl\|P(v)B(y)R(u)\bigr\|_{\mathrm{op}}^2 . \end{aligned} \tag{4.3} \]

Proof. For \(t\ge1\) put \(X_t=vv^{\mathsf T}+t(I-vv^{\mathsf T})\) and \(Z_t=uu^{\mathsf T}+t(I-uu^{\mathsf T})\). Their eigenvalues are \(1\) and \(t\), so \(X_t,Z_t\succeq I\), and positivity and unitality give \(D(X_t)\succeq I_a\) and \(E(Z_t)\succeq I_c\). Put \[ \begin{aligned} A_t&=Z_t^{-1/2}\,\Phi_y(X_t^{-1})\,Z_t^{-1/2}\succeq0,\\ C_t&=D(X_t)^{-1/2}\,B(y)\,E(Z_t)^{-1/2}. \end{aligned} \] For real \(r\), congruence by \(Z_t^{-1/2}\) and Lemma 1.3(3) of the first lesson give: \(rZ_t\succeq\Phi_y(X_t^{-1})\) if and only if \(r\ge\lambda_{\max}(A_t)\). Congruence by \(E(Z_t)^{-1/2}\) and Lemma 1.3(7) give: \(E(rZ_t)\succeq B(y)^{\mathsf T}D(X_t)^{-1}B(y)\) if and only if \(r\ge\|C_t\|_{\mathrm{op}}^2\). By the equivalence of Proposition 4.1 for \((X_t,rZ_t,y)\), these two sets of \(r\) coincide, so \[ \lambda_{\max}(A_t)=\|C_t\|_{\mathrm{op}}^2 . \] Let \(t\to\infty\). Since \(X_t^{-1}=vv^{\mathsf T}+t^{-1}(I-vv^{\mathsf T})\) and \(Z_t^{-1/2}=uu^{\mathsf T}+t^{-1/2}(I-uu^{\mathsf T})\), \[ A_t\to uu^{\mathsf T}\,\Phi_y(vv^{\mathsf T})\,uu^{\mathsf T}=\alpha\,uu^{\mathsf T} \] with \(\alpha=u^{\mathsf T}\Phi_y(vv^{\mathsf T})u\ge0\), and \(\lambda_{\max}(A_t)\to\alpha\) by the continuity in Lemma 1.3(3). On the other side, unitality gives \(D(X_t)=I_a+(t-1)D(I-vv^{\mathsf T})\) and \(E(Z_t)=I_c+(t-1)E(I-uu^{\mathsf T})\), with \(D(I-vv^{\mathsf T})\succeq0\) and \(E(I-uu^{\mathsf T})\succeq0\). By Lemma 4.2, \(D(X_t)^{-1/2}\to P(v)\) and \(E(Z_t)^{-1/2}\to R(u)\), so \(C_t\to P(v)B(y)R(u)\), and \(\|C_t\|_{\mathrm{op}}^2\) tends to the right side of (4.3). By Lemma 3.1 of the previous lesson, \(\alpha=b(v_0u'-u_0v',y)\). \(\square\)

At \(u=v\) the left side of (4.3) is \(b(0,y)=0\), so for every \(y\in\mathbf R^3\) \[ P(v)B(y)R(v)=0 . \tag{4.4} \]

3. The tangent obstruction

Proof of Theorem A. Proposition 3.2 of the previous lesson proves that \(p\) is homogeneous of degree \(20\), with \(p(e)=1\), and hyperbolic with respect to \(e\). Suppose \(K\) is described by a linear matrix inequality of some finite size, and let \(D,E,B\) be given by Proposition 4.1.

A point where \(R\) is smooth. Let \(S^3\) be the unit sphere of \(\mathbf R^4\) and \(\Omega=\{u\in S^3:u_0\ne0\}\). The matrices \(A(u)=E(I-uu^{\mathsf T})\), \(u\in\Omega\), have finitely many possible ranks; let \(\rho\) be the largest, attained at some \(v\in\Omega\). If \(\rho=0\), then \(R(u)=I\) on \(\Omega\). If \(\rho>0\), choose \(\rho\) linearly independent columns of \(A(v)\), and let \(W(u)\) be the matrix formed by the same columns of \(A(u)\). A \(\rho\times\rho\) minor of \(W\) is nonzero at \(v\), hence on a neighbourhood \(\mathcal U\) of \(v\) in \(\mathbf R^4\). For \(u\in\mathcal U\cap\Omega\), the columns of \(W(u)\) are independent and lie in the range of \(A(u)\), which has dimension at most \(\rho\); so they span it. As \(A(u)\) is symmetric, its kernel is the orthogonal complement of its range, and for \(u\in\mathcal U\cap\Omega\) \[ \begin{aligned} R(u)&=I-W(W^{\mathsf T}W)^{-1}W^{\mathsf T},\\ W&=W(u). \end{aligned} \] The right side is a smooth function of \(u\in\mathcal U\); write \(dR_v\) for its differential at \(v\), a linear map from \(\mathbf R^4\) to matrices.

Differentiating the identity. For \(h=(h_0,h')\in v^\perp\) put \[ \begin{aligned} Jh&=v_0h'-h_0v',\\ u_s&=\frac{v+sh}{\sqrt{1+s^2\|h\|^2}} . \end{aligned} \] For small \(s\), \(u_s\in\mathcal U\cap\Omega\), and \[ v_0u_s'-(u_s)_0v'=\frac{s\,Jh}{\sqrt{1+s^2\|h\|^2}} . \] By (4.3) at the pair \((u_s,v)\), (4.4) and the homogeneity of \(b\) of degree \(2\) in its first argument, for \(s\ne0\), \[ \begin{aligned} &\frac{b(Jh,y)}{1+s^2\|h\|^2}\\ &\quad=\Bigl\|P(v)B(y)\,\frac{R(u_s)-R(v)}{s}\Bigr\|_{\mathrm{op}}^2 . \end{aligned} \] As \(s\to0\), the difference quotient tends to \(dR_v(h)\), since \(u_s\) has velocity \(h\) at \(s=0\). Therefore, for all \(h\in v^\perp\) and \(y\in\mathbf R^3\), \[ b(Jh,y)=\bigl\|P(v)B(y)\,dR_v(h)\bigr\|_{\mathrm{op}}^2 . \] We call this the tangent identity.

The contradiction. The linear map \(J:v^\perp\to\mathbf R^3\) is an isomorphism: if \(Jh=0\), then \(h'=(h_0/v_0)v'\), because \(v_0\ne0\), so \(h=(h_0/v_0)v\), and \(h\perp v\) forces \(h=0\); both spaces have dimension \(3\). The matrix \[ F(z,y)=P(v)B(y)\,dR_v(J^{-1}z), \] defined for \(z,y\in\mathbf R^3\), depends bilinearly on \((z,y)\). In orthonormal bases each entry \(\ell_{ij}(z,y)\) is a bilinear form, and by the tangent identity with \(z=Jh\), \[ \ell_{ij}(z,y)^2\le\|F(z,y)\|_{\mathrm{op}}^2=b(z,y). \] By Lemma 2.3 of The Choi–Lam form, every \(\ell_{ij}\) vanishes, so \(F=0\), and the tangent identity gives \(b=0\), contradicting \(b(e_1,e_1)=1\). Hence no linear matrix inequality describes \(K\). \(\square\)

The proof uses four properties of the construction: the two slices of the cone, the positivity of \(\Phi_y\), the rank-one identity of Lemma 3.1, and the absence of dominated bilinear squares below \(b\). Replacing \(b\) by a form that dominates a nonzero bilinear square breaks only the last step (Exercise 4.3).

4. Exercises

4.1 (easy). Where does the proof use that \(v_0\ne0\)? What goes wrong at a point \(v\) with \(v_0=0\)?

4.2 (medium). For \(H\succeq0\) with least positive eigenvalue \(\mu\), show that the distance in operator norm between \([I+(t-1)H]^{-1/2}\) and the projection onto \(\ker H\) is \((1+(t-1)\mu)^{-1/2}\).

4.3 (medium). Replace \(Q(y)\) by \(\|y\|^2I_3\), so that \(b\) becomes \(\|z\|^2\|y\|^2\). Show that \(\ell(z,y)=z_1y_1\) satisfies \(\ell^2\le\|z\|^2\|y\|^2\), so that the last step of the proof fails for the modified polynomial.

4.4 (easy). Show that conjugating a pencil \(L\) by a fixed invertible matrix \(S\), \(x\mapsto S^{\mathsf T}L(x)S\), does not change the set \(\{x:L(x)\succeq0\}\).

5. Solutions

4.1. Only in showing that \(J\) is injective. If \(v_0=0\), then \(\|v'\|=1\) and \(Jh=-h_0v'\) for \(h\in v^\perp\). So \(J\) maps \(v^\perp\) onto the line \(\mathbf Rv'\), its kernel consists of the vectors \((0,w)\) with \(w\perp v'\), and the tangent identity controls \(b(z,y)\) only for \(z\) in the line \(\mathbf Rv'\).

4.2. In an eigenbasis the difference is diagonal with entries \(0\) on \(\ker H\) and \((1+(t-1)\mu')^{-1/2}\) for the positive eigenvalues \(\mu'\); the largest is at \(\mu'=\mu\).

4.3. By the Cauchy–Schwarz inequality, \(z_1^2y_1^2\le\|z\|^2\|y\|^2\). In the proof, the entries \(\ell_{ij}\) need not vanish, so no contradiction arises.

4.4. \(S^{\mathsf T}L(x)S\succeq0\) if and only if \(L(x)\succeq0\), by Lemma 1.3(4) of the first lesson.

References