# Existence of a linear map resulting in the determinant being an elementary symmetric polynomial

Let $$1 \leq k \leq n$$ be fixed integers. Let $$\mathcal{M}_n^{\mathrm{H}}$$ be the set of $$n \times n$$ complex Hermitian matrices (if it makes it easier to answer this question, you may instead use the set of real symmetric matrices). The $$k$$-th elementary symmetric polynomial is \begin{align} S_k(x_1,x_2,\ldots,x_n) = \sum_{1 \leq i_1 < \cdots < i_k \leq n} \left(\prod_{j=1}^k x_{i_j}\right). \end{align}

Question: Does there exist a linear map $$\Phi : \mathcal{M}_n^{\mathrm{H}} \rightarrow \mathcal{M}_{k}^{\mathrm{H}}$$ with the property that, for all diagonal $$X \in \mathcal{M}_n^{\mathrm{H}}$$, we have $$\det(\Phi(X)) = S_k(\lambda_1, \lambda_2, \ldots, \lambda_n)$$, where the $$\lambda$$s are the eigenvalues (i.e., diagonal entries) of $$X$$?

Partial results: The answer is trivially true if $$k = 1$$. In this case, you can choose $$\Phi$$ to be the trace map, so that $$\det(\Phi(X)) = \det(\mathrm{tr}(X)) = \mathrm{tr}(X) = S_1(\lambda_1, \lambda_2, \ldots, \lambda_n)$$ (here I have been a bit loose and freely interpreted a $$1 \times 1$$ matrix as a scalar). The answer is similarly trivially true if $$k = n$$, by choosing $$\Phi$$ to be the identity map.

The result is also (less trivially) true if $$k = n-1$$. In this case, let $$U$$ be the $$n \times n$$ complex Fourier matrix, but with one column removed (so $$U$$ is $$n \times (n-1)$$). Then the linear map $$\Phi$$ defined by $$\Phi(X) = U^*XU$$ has the property that \begin{align*} \det(\Phi(X)) = S_{n-1}(\lambda_1, \lambda_2, \ldots, \lambda_n) \end{align*} whenever $$X$$ is diagonal (well, this equation might be off by some scalar multiple, but that's unimportant since we can just scale $$\Phi$$ appropriately to make the equality true).

I haven't been able to find a $$\Phi$$ that works for any other values of $$k$$.

• $\mathcal{M}_n^{\mathrm{H}}$ is not a linear space. So, how can a linear map be defined on it? Commented Jun 17 at 13:00
• $\mathcal{M}_n^{\mathrm{H}}$ is a real vector space (not a complex one), and I mean it in that sense. But if it makes anything easier, just work with $\mathcal{M}_n$ itself and require that the output is Hermitian whenever the input is diagonal. Commented Jun 17 at 13:09
• If your $\Phi$ exists for $n=4$ and $k=2$, then the elementary symmetric polynomial $e_2\left(x,y,z,w\right) = xy+xz+xw+yz+yw+zw$ can be written as a determinant of a $2\times 2$-matrix of linear forms in $x,y,z,w$. Thus, there must be a $2$-dimensional subspace of $\mathbb{C}^4$ on which it completely vanishes (just set two entries of that matrix to $0$). Something tells me that this is impossible, though I don't quite see why (it certainly can't work for $\mathbb{R}$ instead of $\mathbb{C}$). Commented Jun 18 at 18:07
• Ah, we need not look at subspaces. Just observe that $e_2$ is a quadratic form of full rank (i.e., its representing matrix $\dfrac{1}{2}\begin{pmatrix}0&1&1&1\\ 1&0&1&1\\ 1&1&0&1\\ 1&1&1&0\end{pmatrix}$ is invertible), and thus it cannot be written as $fg-hj$ for four linear maps $f,g,h,j$. So $\Phi$ cannot exist for $n=4$ and $k=2$. Please check! Commented Jun 18 at 18:11
• Okay, I'm not quite sure about the above, since quadratic forms don't uniquely determine the respective matrices: the quadratic form might be $fg-hj$ while the representing matrix will only be $\dfrac{1}{2}\left(fg-hj + \left(fg-hj\right)^T\right)$ (in the appropriate sense), which can well have rank $4$. But my argument still works for $n=6$ and $k=2$ (here the representing matrix will have rank $6$). Commented Jun 18 at 18:14

Such a map $$\Phi$$ does not exist for $$n=5$$ and $$k=2$$.

Let me in fact show a stronger claim: There exists no $$\mathbb{R}$$-linear map $$\Phi : \mathbb{R}^5 \to \mathbb{C}^{2\times 2}$$ such that every five reals $$x_{1},x_{2},\ldots,x_{5}\in\mathbb{R}$$ satisfy $$$$\det\left( \Phi\left( x_{1},x_{2},\ldots,x_{5}\right) \right) =e_{2}\left( x_{1},x_{2},\ldots,x_{5}\right) . \label{eq.darij1.1} \tag{1}$$$$

Proof. Assume the contrary. Thus, such a map $$\Phi$$ exists. Consider it. Write it as \begin{align} \Phi\left( x_{1},x_{2},\ldots,x_{5}\right) = \begin{pmatrix} f_{11}\left( x\right) & f_{12}\left( x\right) \\ f_{21}\left( x\right) & f_{22}\left( x\right) \end{pmatrix} , \label{eq.darij1.2} \tag{2} \end{align} where the $$f_{ij}\left( x\right)$$ are four linear forms in $$x_{1},x_{2},\ldots,x_{5}$$ with complex coefficients (that is, four linear maps from $$\mathbb{R}^5$$ to $$\mathbb{C}$$). Thus, by our assumption, any five reals $$x_{1},x_{2},\ldots,x_{5}$$ satisfy \begin{align} & e_{2}\left( x_{1},x_{2},\ldots,x_{5}\right) \nonumber\\ & =\det\left( \Phi\left( x_{1},x_{2},\ldots ,x_{5}\right) \right) \qquad\left( \text{by \eqref{eq.darij1.1}} \right) \nonumber\\ & =\det \begin{pmatrix} f_{11}\left( x\right) & f_{12}\left( x\right) \\ f_{21}\left( x\right) & f_{22}\left( x\right) \end{pmatrix} \qquad\left( \text{by \eqref{eq.darij1.2}}\right) . \label{eq.darij1.3} \tag{3} \end{align} Let us extend the four linear forms $$f_{ij}$$ from $$\mathbb{R}^5$$ to $$\mathbb{C}^5$$ in the obvious way (i.e., preserving their coefficients). Thus, the $$f_{ij}$$ are now four $$\mathbb{C}$$-linear forms from $$\mathbb{C}^5$$ to $$\mathbb{C}$$.

Then, \eqref{eq.darij1.3} is an equality between two polynomials in $$x_{1},x_{2},\ldots,x_{5}$$. Thus, since we know that it holds for all reals $$x_{1},x_{2},\ldots,x_{5}$$, we conclude that it holds for all complex numbers $$x_{1},x_{2},\ldots,x_{5}$$.

Now, recall that the $$f_{ij}$$ are linear forms on $$\mathbb{C}^{5}$$. Let $$W$$ be the $$\mathbb{C}$$-vector subspace $$\operatorname*{Ker}\left( f_{21}\right) \cap\operatorname*{Ker}\left( f_{22}\right)$$ of $$\mathbb{C}^{5}$$. The dimension of this subspace $$W$$ is at least $$3$$ (since it is defined by two linear equations in $$\mathbb{C}^{5}$$), thus larger than $$5/2$$.

But $$e_{2}\left( x_{1},x_{2},\ldots,x_{5}\right) =x_{1}x_{2}+x_{1} x_{3}+\cdots+x_{4}x_{5}$$ is a symmetric bilinear form on $$\mathbb{C}^{5}$$. This form is easily seen to be nondegenerate, since it is represented by the invertible symmetric matrix $$\dfrac{1}{2} \begin{pmatrix} 0 & 1 & 1 & 1 & 1\\ 1 & 0 & 1 & 1 & 1\\ 1 & 1 & 0 & 1 & 1\\ 1 & 1 & 1 & 0 & 1\\ 1 & 1 & 1 & 1 & 0 \end{pmatrix}$$. Hence, a known fact (see, e.g., https://math.stackexchange.com/questions/4303680/ ) says that it cannot become identically zero when restricted to a subspace of $$\mathbb{C}^{5}$$ whose dimension is larger than $$5/2$$. In particular, it cannot become identically zero when restricted to $$W$$ (since $$W$$ has dimension larger than $$5/2$$). Hence, the corresponding quadratic form also cannot become identically zero when restricted to $$W$$ (because if a symmetric bilinear form is nonzero, then the the corresponding quadratic form is also nonzero). In other words, there exists some $$x=\left( x_{1},x_{2},\ldots,x_{5}\right) \in W$$ such that $$e_{2}\left( x_{1},x_{2},\ldots,x_{5}\right) \neq0$$. However, $$x\in W$$ entails $$f_{21}\left( x\right) =0$$ and $$f_{22}\left( x\right) =0$$ and thus \begin{align*} \det \begin{pmatrix} f_{11}\left( x\right) & f_{12}\left( x\right) \\ f_{21}\left( x\right) & f_{22}\left( x\right) \end{pmatrix} =\det \begin{pmatrix} f_{11}\left( x\right) & f_{12}\left( x\right) \\ 0 & 0 \end{pmatrix} =0, \end{align*} so that \eqref{eq.darij1.3} rewrites as $$e_{2}\left( x_{1},x_{2},\ldots ,x_{5}\right) =0$$. This contradicts $$e_{2}\left( x_{1},x_{2},\ldots ,x_{5}\right) \neq0$$. This contradiction completes our proof. $$\blacksquare$$

• Thank you. I don't think it affects the actual argument in the 2nd half of this answer, but is (1) really stronger than my claim? A general map $\Phi$ is a sum of maps of the form $X \mapsto UXV$, not just a single map of that form. Commented Jun 21 at 10:54
• Oh, you're right! I was a lazy reader. Commented Jun 21 at 11:58
• Fixed, I believe. Can you cofnirm? Commented Jun 21 at 12:09
• Looks good to me now, thanks! Commented Jun 21 at 12:11