How to minimize $|z_1 - z_2|^2 + |z_1 - z_4|^2 + |z_2 - z_3|^2 + |z_3 - z_4|^2$?

Question

If $z_1,z_2,z_3,z_4 \in \mathbb{C}$ satisfy $z_1 + z_2 + z_3 + z_4 = 0$ and $|z_1|^2 + |z_2|^2 + |z_3|^2 + |z_4|^2 = 1$, then the least value of $|z_1 - z_2|^2 + |z_1 - z_4|^2 + |z_2 - z_3|^2 + |z_3 - z_4|^2$ is $2$.

I solved this by taking two of the values to be $\frac{1}{2}$ and other two as $-\frac{1}{2}$. But I couldn't prove this in an efficient manner.

Answer 1

Let $\vec w$ be related to $\vec z$ by the self-inverse Hadamard transform:

$$H = \frac{1}{\sqrt4} \begin{pmatrix} 1&1&1&1\\ 1&1&-1&-1\\ 1&-1&1&-1\\ 1&-1&-1&1 \end{pmatrix}$$

Then $w_1 = 0$, i.e. we have reduced the problem to three dimensions. $H$ is unitary so $|\vec w|^2 = |\vec z|^2 = 1$.

Let us now look at the terms of the objective function.

$$z_1-z_2 = w_3+w_4$$ $$z_1-z_4 = w_2+w_3$$ $$z_2-z_3 = w_2-w_3$$ $$z_3-z_4 = w_3-w_4$$

Then:

$$|z_1-z_2|^2+|z_1-z_4|^2 = 2(|w_3|^2+|w_4|^2)$$ $$|z_2-z_3|^2+|z_3-z_4|^2 = 2(|w_3|^2+|w_2|^2)$$

So the objective function is equal to $2+2|w_3|^2$.

Answer 2

Definitely not more efficient, but here goes anyway:

Let $z \in \mathbb{C}^4$, $e=(1,...,1)^T$, $A= \begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & -1 \\ -1 & 0 & 0 & 1 \end{bmatrix}$.

Then the problem is $\min \{ \|Az\|^2 \ | \ e^T z = 0, \|z\|^2 = 1 \}$. Some tedious multiplication shows that $A^* A = A A^*$, so $A$ is unitarily diagonalizable, and $\ker A = \operatorname{sp} \{ e \}$.

Since $A$ is unitarily diagonalizable, we can write $A z = \sum_{k=1}^3 \lambda_k (u_k^* z ) u_k$, where the $\lambda_k$ are non-zero, the $u_k$ are orthonormal, and $u_k^* e = 0$ for all $k$. Furthermore, $\|A z\|^2 = \sum_{k=1}^3 |\lambda_k|^2 |u_k^* z|^2$. If we let $\sigma = \min_k |\lambda_k|^2$, we have $\|A z\|^2 \ge \sigma\sum_{k=1}^3 |u_k^* z|^2$. If $e^T z = 0$, we have $z \in \operatorname{sp} \{ u_1, u_2, u_3\}$, and so $\sum_{k=1}^3 |u_k^* z|^2 = \|z\|^2$.

Hence if $e^T z = 0$, we have $\|A z\|^2 \ge \sigma \|z\|^2$, and if $u$ is the eigenvector that corresponds to $\sigma$, then we have $\|Au\|^2 = \sigma \|u\|^2$.

Hence the minimum value is the square of the modulus of the smallest non-zero eigenvalue of $A$, and the corresponding unit eigenvector is a minimizer.

We have $\chi_A(s) = x(x-2)(x^2-2x+2)$, hence the non-zero eigenvalues are $2, 1\pm i$, and so we have $\sigma = 2 = |1\pm i|^2$. The corresponding eigenvector is $(i,-1,-i,1)^T$, hence a minimizer is $z = \frac{1}{2}(i,-1,-i,1)^T$.

Answer 3

Derive respect $z_{i}^{*}$ with the conditions $\sum_{i}z_{i}^{*} = 0$ and $\sum_{i}z_{i}^{*}z_{i} - 1 = 0$ by using Lagrange multipliers. In this way,we get the values of $\left\lbrace z_{i}\right\rbrace$. Plug these values in the expression to get the minimum.