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 $k$.

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    $\begingroup$ $\mathcal{M}_n^{\mathrm{H}}$ is not a linear space. So, how can a linear map be defined on it? $\endgroup$ Commented Jun 17 at 13:00
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    $\begingroup$ $\mathcal{M}_n^{\mathrm{H}}$ 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. $\endgroup$ Commented Jun 17 at 13:09
  • $\begingroup$ 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 $\mathbb{C}$). $\endgroup$ Commented Jun 18 at 18:07
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    $\begingroup$ 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! $\endgroup$ Commented Jun 18 at 18:11
  • $\begingroup$ 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 $6$). $\endgroup$ Commented Jun 18 at 18:14

1 Answer 1


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 \begin{equation} \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} \end{equation}

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. $\blacksquare$

  • $\begingroup$ 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. $\endgroup$ Commented Jun 21 at 10:54
  • $\begingroup$ Oh, you're right! I was a lazy reader. $\endgroup$ Commented Jun 21 at 11:58
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    $\begingroup$ Fixed, I believe. Can you cofnirm? $\endgroup$ Commented Jun 21 at 12:09
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    $\begingroup$ Looks good to me now, thanks! $\endgroup$ Commented Jun 21 at 12:11

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