# For a unitary matrix $U$, what's the minimal value of the real part of $\det(U^*)\prod_i U_{ii}$?

For an $$n$$-by-$$n$$ unitary matrix $$U$$, what's the minimal value of the real part of $$\Delta(U)=\det(U^*)\prod_i U_{ii}$$?

Let $$V$$ be the orthogonal matrix with diagonal entries equal to $$1-2/n$$ and all other entries equal to $$-2/n$$. This achieves $$\Delta(V)=-(1-2/n)^n$$, which computer experiments suggest is optimal. Interestingly this would mean that the large $$n$$ limit is $$-e^{-2}$$.

For $$n=2$$ the minimum is $$0$$, which can be proven by writing $$U$$ in the form $$\begin{pmatrix}\alpha & \beta \\ -e^{-i\theta}\bar\beta & e^{-i\theta}\bar\alpha\end{pmatrix}.$$

The average value of $$\Delta(U)$$ across the unitary group is $$1/n!$$. Indeed, for any permutation $$\sigma$$ with permutation matrix $$P_\sigma$$, $$\Delta_\sigma(U)=(-1)^\sigma\det(U^*)\prod_i U_{i,\sigma(i)}$$ equals $$\Delta(UP_\sigma)$$. The sum $$\sum_\sigma \Delta_\sigma(U)$$ equals $$\det(U^*)\det(U)=1$$, and each $$\int_{U(n)}\Delta_\sigma(U)dU$$ is equal because multiplication by $$P_\sigma$$ preserves the Haar measure.

Let $$n>2$$. Define $$f:M_n(\mathbb{C})\to\mathbb{R}$$ with $$f(X)=\operatorname{Re}\left(\det(X^*)\prod_{k=1}^nx_{kk}\right)$$.

Note that $$f(X) = f(X^T) = f(\overline{X}) = f(X^*) = f(PXP^T) = f(XD)$$ for every permutation matrix $$P$$ and diagonal unitary matrix $$D$$.

Function $$f$$ is continuous and therefore obtains its minimum on the set of unitary matrices. Let $$U$$ be a matrix for which minimum $$m=f(U)<0$$ is attained. We may assume $$u_{kk} > 0$$, otherwise we note that $$f(U)<0$$ implies diagonal elements are non-zero and we can multiply each column $$k$$ with $$|u_{kk}|/u_{kk}$$ to achieve our assumption. Let $$\zeta=-\det{U^*}$$ and note that $$f(U)<0$$ implies $$\operatorname{Re}(\zeta)>0$$.

We are going to show that $$U$$ is also a Hermitian matrix. To obtain relation between its off-diagonal elements we will exploits the fact that $$f(UQ)\geq f(U)$$ for every unitary $$Q$$.

Let $$i,j\in\{1,\ldots,n\}$$, $$i\neq j$$ be arbitrary. For these $$i$$ and $$j$$ we define a unitary matrix $$Q(\varphi)$$ as a matrix obtained from identity by replacing submatrix at the intersection of rows and columns $$i$$ and $$j$$ with $$\begin{bmatrix}1&0\\0&\xi\end{bmatrix}\begin{bmatrix}\cos\varphi & -\sin\varphi\\\sin\varphi & \cos\varphi\end{bmatrix}\begin{bmatrix}1&0\\0&\overline{\xi}\end{bmatrix}\,,$$ where $$\xi$$ is unimodular number such that $$u_{ij}\xi=|u_{ij}|$$. We note that $$\det(UQ(\varphi))=\det(U)$$ and that diagonal of $$U$$ and $$UQ(\varphi)$$ differ only at positions $$(i,i)$$ and $$(j,j)$$.

Now, we define function $$g:\mathbb{R}\to\mathbb{R}$$ with $$g(\varphi)=f(UQ(\varphi))$$. Using previous results, we have \begin{align} g(\varphi) &=f(UQ(\varphi)) = \operatorname{Re}\left(\det(Q(\varphi)^*U^*)\prod_{k=1}^n[UQ(\varphi)]_{kk}\right)\\ &= \operatorname{Re}\left(\det(U^*)\prod_{k=1}^nu_{kk}\cdot\frac{1}{u_{ii}u_{jj}}(u_{ii}\cos\varphi+\xi u_{ij}\sin\varphi)(u_{jj}\cos\varphi-\overline{\xi}u_{ji}\sin\varphi)\right)\\ &= -\left(\prod_{k=1}^nu_{kk}\right)\operatorname{Re}\left(\frac{\zeta}{u_{ii}u_{jj}}(u_{ii}\cos\varphi+\xi u_{ij}\sin\varphi)(u_{jj}\cos\varphi-\overline{\xi}u_{ji}\sin\varphi)\right)\\ &= -\left(\prod_{k=1}^nu_{kk}\right)\left(\cos\varphi+\frac{|u_{ij}|}{u_{ii}}\sin\varphi\right)\left(\operatorname{Re}(\zeta)\cos\varphi-\frac{\operatorname{Re}(\zeta\overline{\xi}u_{ji})}{u_{jj}}\sin\varphi\right)\,.\tag{1} \end{align}

Global minimum of function $$g$$ is obtain whenever the product of the last two factors in $$(1)$$ is maximized. Using trigonometric addition formulas, we can show this happens for every $$\varphi$$ satisfying $$2\varphi-\arctan\left(\frac{|u_{ij}|}{u_{ii}}\right)+\arctan\left(\frac{\operatorname{Re}(\zeta\overline{\xi}u_{ji})}{u_{jj}\operatorname{Re}(\zeta)}\right)\in 2\pi\mathbb{Z}\,.\tag{2}$$ On the other hand, global minimum is obtained for $$\varphi=0$$, because $$g(0)=f(U)$$. This and $$(2)$$ implies that $$\operatorname{Re}\big((|u_{ij}|u_{jj}-\overline{\xi}u_{ii}u_{ji})\zeta\big)=0\,.$$ Multiplying the last result with $$|u_{ij}|u_{jj}$$, we obtain $$\operatorname{Re}\big((|u_{ij}|^2u_{jj}^2-u_{ii}u_{ij}u_{ji}u_{jj})\zeta\big)=0\,.\tag{3}$$

Repeating this procedure with matrix $$PUP^T$$ obtained from $$U$$ by exchange of rows $$i$$ and $$j$$ and columns $$i$$ and $$j$$ gives $$\operatorname{Re}\big((|u_{ji}|^2u_{ii}^2-u_{ii}u_{ij}u_{ji}u_{jj})\zeta\big)=0\,.\tag{4}$$ Repeating the same procedure with matrix $$U^*$$ gives $$\operatorname{Re}\big((|u_{ji}|^2u_{jj}^2-u_{ii}u_{ij}u_{ji}u_{jj})\overline{\zeta}\big)=0\,.\tag{5}$$

We subtract $$(3)$$ from $$(4)$$ to obtain $$|u_{ij}|u_{jj}=|u_{ji}|u_{ii}$$. From here, it follows that $$u_{ji}=\overline{u_{ij}}u_{jj}u_{ii}^{-1}\rho_{ij}\,,\tag{6}$$ for some $$\rho_{ij}$$ such that $$|\rho_{ij}|=1$$.

Using $$(6)$$ we show that all diagonal elements of $$U$$ are equal. If necessary, we may symmetrically permute rows and columns of $$U$$ so that $$u_{11}$$ is the largest diagonal element. Now, $$1 = \sum_{k=1}^n|u_{1k}|^2 = \sum_{k=1}^n|u_{k1}|^2\frac{|u_{11}|^2}{|u_{kk}|^2} \geq\sum_{k=1}^n|u_{k1}|^2=1\,,$$ from where our claim follows.

Replacing $$u_{ji}$$ in $$(3)$$ with $$(6)$$ implies $$u_{ij}=0$$ or $$\operatorname{Re}((1-\rho_{ij})\zeta)=0$$. The same replacement in $$(5)$$ implies $$u_{ij}=0$$ or $$\operatorname{Re}((1-\rho_{ij})\overline{\zeta})=0$$. If $$u_{ij}\neq0$$, solving this two equations for $$\rho_{ij}$$ shows that $$\rho_{ij}\in\{1,\zeta^2\}\cap\{1,(\overline{\zeta})^2\}=\{1\}\,.$$ The last equality is true because only solution of $$\zeta^2=(\overline{\zeta})^2$$ satisfying $$\operatorname{Re}(\zeta)>0$$ is $$\zeta=1$$. If $$u_{ij}=0$$, we are free to take $$\rho_{ij}=1$$. In any case, $$u_{ji}=\overline{u_{ij}}u_{jj}u_{ii}^{-1}\tag{7}\,.$$ Since all diagonal elements are equal, $$(7)$$ implies $$U$$ is a Hermitian matrix.

We now know matrix $$U$$ is unitary and Hermitian. Therefore, its only eigenvalues are $$\pm1$$. From $$\operatorname{Re}(\zeta)>0$$ and $$\det(U)\in\{-1,1\}$$ we conclude that $$\det(U)=-1$$, and in turn, that at least one eigenvalue of $$U$$ is $$-1$$. Now, $$m=f(U)=-\prod_{k=1}^nu_{kk}=-u_{11}^n = -\left(\frac{1}{n}\operatorname{tr}(U)\right)^n \geq -\left(\frac{n-2}{n}\right)^n\,,$$ shows the conjecture was correct.

We consider the real case.

$$\textbf{Proposition}$$. Let $$n>2$$ and

$$f:U=[u_{i,j}]\in O(n)\mapsto \det(U^T)\Pi_{i=1}^n u_{i,i}$$.

Then the minimum of $$f$$ is $$m=-(1-2/n)^n$$.

$$\textbf{Proof}$$. Since $$O(n)$$ is compact, the lower bound of $$f$$ is reached in at least a matrix $$A=[a_{i,j}]\in O(n)$$. Note that if we change a column of $$U\in O(n)$$ into its opposite, then the obtained matrix $$U'\in O(n)$$ satisfies $$f(U)=f(U')$$.

Consequently, we may assume that, for every $$i$$, $$a_{i,i}\geq 0$$. Since we know, from the OP, that $$m\leq -(1-2/n)^n< 0$$, we deduce that, for every $$i$$, $$a_{i,i}>0$$ and $$\det(A)<0$$.

Then $$-1\in spectrum(A)$$ and $$\sum_i a_{i,i}=trace(A)\leq n-2$$.

Consequently, $$0<\Pi_i a_{i,i}\leq (\dfrac{n-2}{n})^n$$ (fixing the sum of the $$(a_i)$$ in $$n-2$$, the max of the product is reached when the $$(a_{i,i})$$ are equal)

and we are done. $$\square$$

• Nice! By multiplying columns by $e^{i\theta}$'s, the argument works in the unitary case up to the point of $a_{ii}>0$ and $\mathrm{Re}(\det A)<0$. I wonder if the next step can be modified. – MTyson May 11 at 21:16
• Yes, I saw that; yet I am unable to finish because I find only the bound $n-1$ for the trace. – loup blanc May 11 at 21:19