Cofactor matrices and positive semi-definiteness If $A$ is an $n\times n$ matrix with real entries, let me write $\widehat A$ its cofactor matrix. Since the map $A\mapsto\widehat A$ is polynomial, homogeneous of degree $n-1$, it can be multi-linearized:
$$\widehat A=P_{n-1}(A,\ldots,A)$$
where $P_{n-1}:{\bf M}_n({\mathbb R})^{n-1}\to{\bf M}_n({\mathbb R})$ is multilinear and symmetric. For instance if $n=3$, then
$$P_2(A,B)=\frac12\left(\widehat{A+B}-\widehat A-\widehat B\right),\qquad A,B\in{\bf M}_3({\mathbb R}).$$
If $A$ is symmetric, then so is $\widehat A$. Moreover, if $A\in{\bf Sym}_n^+$ (symmetric positive semi-definite), then so is $\widehat A$.

Is it true that if $A_1,\ldots,A_{n-1}\in{\bf Sym}_n^+$, then $P_{n-1}(A_1,\ldots,A_{n-1})\in{\bf Sym}_n^+$ ?

I know that the answer is positive for $n=3$. Proof: by density we may assume that $B$ is positive definite. With $H=\sqrt B$, we have $P_2(A,B)=\widehat HP_2(C,I_3)\widehat H$ for $C=H^{-1}AH^{-1}$. Thus it is sufficient to consider the case where $B=I_3$. Then diagonalizing $A$, we may restrict to the diagonal case. Then
$$P_2(A,I_3)=\frac12{\rm diag}\left(\prod_{j\ne i}(1+a_j)-\prod_{j\ne i}a_j-1\right)=\frac12{\rm diag}(a_2+a_3,a_1+a_3,a_1+a_2),$$
which is obviously positive. $\blacksquare$
Remark. There is no trivial answer to this question, because the map $A\mapsto A^2$ satisfies similar assumptions (it sends ${\bf Sym}_n^+$ into itself), but the bilinear function $(A,B)\mapsto\frac12(AB+BA)$ does not.
 A: After further thinking, I believe that the answer is positive for every $n\ge2$. Here is a tantative proof.

Given an arbitrary vector $v$, we want to show that $q(v):=v^TP_{n-1}(A_1,,\ldots,A_{n-1})v$ is non-negative. Because the map $A\mapsto\widehat A$ is unitary equivariant, it is enough to prove this when $v=e_n$ is the $n$-th element of the canonical basis. Because the $(n,n)$-entry of $\widehat A$ is the determinant of $A'$, the matrix obtained from $A$ by deleting the $n$-th row and column, $q(n)$ equals the mixed determinant of the $(n-1)$-uplet $(A_1',\ldots,A_{n-1}')$. On the one hand, every $A_j'$ is positive semi-definite. On the other hand the determinant is a hyperbolic polynomial over ${\bf Sym}_{n-1}$ in the direction of $I_{n-1}$ (in the sense of Garding). Its cone of future is precisely ${\bf Sym}_{n-1}^+$. At last, one of Garding's results is that if $Q$ is a homogenenous polynomial of degree $k$, hyperbolic with future cone $\Gamma$, and if $\Phi$ is the corresponding $k$-linear symmetric map, then $\Phi\ge0$ over $\Gamma\times\cdots\times\Gamma$. Here we obtain that the mixed determinant of positive semi-definite symmetric matrices is non-negative. Hence $q(e_n)\ge0$.

