Proving the distillability problem of two-copy $4\times 4$ Werner states for monomial matrices

Abstract

The distillability conjecture of two-copy $4\times 4$ Werner states is one of the main open problems in quantum information (https://arxiv.org/abs/2002.03233, P. Horodecki, L. Rudnicki, and K. Zyczkowski). We prove three special cases of the conjecture in terms of the $4\times 4$ non-normal matrices A, B involved in the conjecture. The first case, namely the main result of this paper, occurs when A, B are monomial matrices. Then, we apply it to the remaining two cases. One case occurs when A, B both have at most two nonzero entries. The other case works for rank-one A and some rank-two B. Our results present the latest progress on the conjecture.

Appendices

Appendix

The proof of Proposition 4

We have

$$\begin{aligned} X=A\otimes I+I\otimes B= \left( \begin{array}{cccc}B&{}a_1I&{}0&{}0\\ a_2I&{}B &{}0&{}0\\ 0&{}0&{}B&{}a_3I\\ 0&{}0&{}a_4I&{}B\end{array}\right) , \end{aligned}$$

and

$$\begin{aligned} XX^\dag= & {} \left( \begin{array}{cccc}B&{}a_1I&{}0&{}0\\ a_2I&{}B &{}0&{}0\\ 0&{}0&{}B&{}a_3I\\ 0&{}0&{}a_4I&{}B\end{array}\right) \left( \begin{array}{cccc}B^\dag &{}a_2^*I&{}0&{}0\\ a_1^*I&{}B^\dag &{}0&{}0\\ 0&{}0&{}B^\dag &{}a_4^*I\\ 0&{}0&{}a_3^*I&{}B^\dag \end{array}\right) \\= & {} \left( \begin{array}{cccc}BB^\dag +|a_1|^2I&{}a_2^*B+a_1B^\dag &{}0&{}0\\ a_2B^\dag +a_1^*B&{}BB^\dag +|a_2|^2I&{}0&{}0\\ 0&{}0&{}BB^\dag +|a_3|^2I &{} a_3B^\dag +a_4^*B\\ 0&{}0&{}a_4B^t+a_3^*B&{}|a_4|^2I+BB^\dag \end{array}\right) \\= & {} \left( \begin{array}{cc}C&{}0\\ 0&{}D\end{array}\right) =C\oplus D. \end{aligned}$$

where

$$\begin{aligned}&C=\left( \begin{array}{cc}BB^\dag +|a_1|^2&{}a_2^*B+a_1B^\dag \\ a_2B^\dag +a_1^*B&{}|a_2|^2I+BB^\dag \end{array}\right) \nonumber \\&\quad =\left( \begin{array}{cccccccc}|b_1|^2+|a_1|^2&{}0&{}0&{}0&{}0&{}a_2^*b_1&{}0&{}a_1b_4^*\\ 0&{}|b_2|^2+|a_1|^2&{}0&{}0&{}a_1b_1^*&{}0&{}a_2^*b_2&{}0\\ 0&{}0&{}|b_3|^2+|a_1|^2&{}0&{}0&{}a_1b_2^*&{}0&{}a_2^*b_3\\ 0&{}0&{}0&{}|b_4|^2+|a_1|^2&{}a_2^*b_4&{}0&{}a_1b_3^*&{}0\\ 0&{}a_1^*b_1&{}0&{}a_2b_4^*&{}|b_1|^2 +|a_2|^2&{}0&{}0&{}0\\ a_2b_1^*&{}0&{}a_1^*b_2&{}0&{}0&{}|b_2|^2+|a_2|^2&{}0&{}0\\ 0&{}a_2b_2^*&{}0&{}a_1^*b_3&{}0&{}0&{}|b_3|^2 +|a_2|^2&{}\\ a_1^*b_4&{}0&{}a_2b_3^*&{}0&{}0&{}0&{}0&{}|b_4|^2 + |a_2|^2\end{array}\right) \nonumber \\&\quad \sim \left( \begin{array}{cccccccc}|b_1|^2+|a_1|^2&{}0&{}a_2^*b_1&{}a_1b_4^*&{}0&{}0&{}0&{}0\\ 0&{}|b_3|^2+|a_1|^2&{}a_1b_2^*&{}a_2^*b_3&{}0&{}0&{}0&{}0\\ a_2b_1^*&{}a_1^*b_2&{}|b_2|^2+|a_2|^2&{}0&{}0&{}0&{}0&{}0\\ a_1^*b_4&{}a_2b_3^*&{}0&{}|b_4|^2+|a_2|^2&{}0&{}0&{}0&{}0\\ 0&{}0&{}0&{}0&{}|b_2|^2+|a_1|^2&{}0&{}a_1b_1^*&{}a_2^*b_2\\ 0&{}0&{}0&{}0&{}0&{}|b_4|^2+|a_1|^2&{}a_2^*b_4&{}a_1b_3^*\\ 0&{}0&{}0&{}0&{}a_1^*b_1&{}a_2b_4^*&{}|b_1|^2+|a_2|^2&{}0\\ 0&{}0&{}0&{}0&{}a_2b_2^*&{}a_1^*b_3&{}0&{}|b_3|^2+|a_2|^2\end{array}\right) \nonumber \\&\quad =C_1\oplus C_2, \end{aligned}$$

(14)

and

$$\begin{aligned}&D=\left( \begin{array}{cc}BB^\dag +|a_3|^2&{}a_4^*B+a_3B^\dag \\ a_4B^\dag +a_3^*B&{}|a_4|^2I+BB^\dag \end{array}\right) \nonumber \\&\quad \sim \left( \begin{array}{cccccccc}|b_1|^2+|a_3|^2&{}0&{}a_4^*b_1&{}a_3b_4^*&{}0&{}0&{}0&{}0\\ 0&{}|b_3|^2+|a_3|^2&{}a_3b_2^*&{}a_4^*b_3&{}0&{}0&{}0&{}0\\ a_4b_1^*&{}a_3^*b_2&{}|b_2|^2+|a_4|^2&{}0&{}0&{}0&{}0&{}0\\ a_3^*b_4&{}a_4b_3^*&{}0&{}|b_4|^2+|a_4|^2&{}0&{}0&{}0&{}0\\ 0&{}0&{}0&{}0&{}|b_2|^2+|a_3|^2&{}0&{}a_3b_1^*&{}a_4^*b_2\\ 0&{}0&{}0&{}0&{}0&{}|b_4|^2+|a_3|^2&{}a_4^*b_4&{}a_3b_3^*\\ 0&{}0&{}0&{}0&{}a_3^*b_1&{}a_4b_4^*&{}|b_1|^2+|a_4|^2&{}0\\ 0&{}0&{}0&{}0&{}a_4b_2^*&{}a_3^*b_3&{}0&{}|b_3|^2+|a_4|^2\end{array}\right) \nonumber \\&\quad =D_1\oplus D_2. \end{aligned}$$

(15)

Then, we have

$$\begin{aligned} XX^\dag =C_1\oplus C_2\oplus D_1 \oplus D_2, \end{aligned}$$

(16)

and

$$\begin{aligned} \mathrm{Tr}\,C_1= & {} \mathrm{Tr}\,C_2=\sum _{j=1}^4|b_j|^2+2(|a_1|^2+|a_2|^2), \nonumber \\ \mathrm{Tr}\,D_1= & {} \mathrm{Tr}\,D_2=\sum _{j=1}^4|b_j|^2+2(|a_3|^2+|a_4|^2). \end{aligned}$$

(17)

Let $\lambda$ and $\mu$ be two arbitrary eigenvalues of $XX^\dag$. Using Lemma 3, proving Conjecture 1 is equivalent to proving $\lambda +\mu \le \frac{1}{2}$. There are three cases for $\lambda$ and $\mu$.

Case 1. $\lambda$ is the eigenvalue of some $C_i$ and $\mu$ is the eigenvalue of some $D_j$ for any $i,j =1,2$.

Without loss of generality, we assume that $\lambda$ is the eigenvalue of $C_1$ and $\mu$ is the eigenvalue of $D_1$. Equation (14) implies that $\lambda \le \mathrm{Tr}\,C_1= \sum _{j=1}^4|b_j|^2+2(|a_1|^2+|a_2|^2)$. Equation (15) implies that $\mu \le \mathrm{Tr}\,D_1= \sum _{j=1}^4|b_j|^2+2(|a_3|^2+|a_4|^2)$. Then, $\lambda +\mu \le 2\sum _{j=1}^4(|b_j|^2+|a_j|^2)$, Eq. (2) shows that $\lambda +\mu \le \frac{1}{2}$.

Case 2. $\lambda$ and $\mu$ are the eigenvalues of same $C_i$ or $D_j$ .

Assume $\lambda$ and $\mu$ are the eigenvalue of $C_1$, Eq. (14) implies that $\lambda +\mu \le \mathrm{Tr}\,C_1= \sum _{j=1}^4|b_j|^2+2(|a_1|^2+|a_2|^2)\le 2[\sum _{j=1}^4(|b_j|^2+|a_j|^2)]=\frac{1}{2}$ .

Case 3. $\lambda$ and $\mu$ are the eigenvalues of $C_1$ and $C_2$ or $D_1$ and $D_2$.

Without loss of generality, we can assume that $\lambda$ is the maximum eigenvalue of $C_1$ and $\mu$ is the maximum eigenvalue of $C_2$. Let $x=(x_1, x_2, x_3, x_4), y=(y_1, y_2, y_3, y_4)$ and $\Vert x\Vert =\Vert y\Vert =1$, then

$$\begin{aligned} \begin{array}{ll} xC_1x^\dag &{}=(x_1, x_2, x_3, x_4)\left( \begin{array}{cccc}|b_1|^2+|a_1|^2&{}0&{}a_2^*b_1&{}a_1b_4^*\\ 0&{}|b_3|^2+|a_1|^2&{}a_1b_2^*&{}a_2^*b_3\\ a_2b_1^*&{}a_1^*b_2&{}|b_2|^2+|a_2|^2&{}0\\ a_1^*b_4&{}a_2b_3^*&{}0&{}|b_4|^2+|a_2|^2\end{array}\right) \left( \begin{array}{c}x_1^*\\ x_2^*\\ x_3^*\\ x_4^*\end{array}\right) \\ &{}=(|b_1|^2+|a_1|^2)|x_1|^2+(|b_3|^2+|a_1|^2)|x_2|^2\\ &{}\qquad +(|b_2|^2+|a_2|^2)|x_3|^2+(|b_4|^2+|a_2|^2)|x_4|^2\\ &{}\quad \quad +a_2b_1^*x_1^*x_3+a_1^*b_4x_1^*x_4+a_1^*b_2x_2^*x_3+a_2^*b_3x_2^*x_4\\ &{}\quad \quad +a_2^*b_1x_1x_3^*+a_1b_4^*x_1x_4^*+a_1b_2^*x_2x_3^*+a_2b_3^*x_2^*x_4\\ &{}=|b_1x_1+a_2x_3|^2+|a_1x_1+b_4x_4|^2+|a_1x_2+b_2x_3|^2+|b_3x_4+a_2x_2|^2\\ &{}\le (|b_1|^2+|a_2|^2)(|x_1|^2+|x_3|^2)+(|a_1|^2+|b_2|^2)(|x_2|^2+|x_3|^2)\\ &{}\quad +(|a_1|^2+|b_4|^2)(|x_1|^2+|x_4|^2)+(|b_3|^2+|a_2|^2)(|x_2|^2+|x_4|^2)\\ &{}=(|b_1|^2+|a_2|^2+|a_1|^2+|b_4|^2)|x_1|^2+(|a_1|^2+|b_2|^2+|a_2|^2+|b_3|^2)|x_2|^2\\ &{}\quad +(|b_1|^2+|a_2|^2+|a_1|^2+|b_2|^2)|x_3|^2+(|a_1|^2+|b_4|^2+|a_2|^2+|b_3|^2)|x_4|^2\\ &{}=k_1|x_1|^2+k_2|x_2|^2+k_3|x_3|^2+k_4|x_4|^2, \end{array}\nonumber \\ \end{aligned}$$

(18)

where $k_1=|b_1|^2+|a_2|^2+|a_1|^2+|b_4|^2, k_2=|a_1|^2+|b_2|^2+|a_2|^2+|b_3|^2, k_3=|b_1|^2+|a_2|^2+|a_1|^2+|b_2|^2, k_4=|a_1|^2+|b_4|^2+|a_2|^2+|b_3|^2,$ we have known that $|x_1|^2+|x_2|^2+|x_3|^2+|x_4|^2=1$, then

$$\begin{aligned} xC_1x^\dag \le k_1x_1^2+k_2x_2^2+k_3x_3^2+k_4x_4^2\le \max \limits _{i=1,\cdots , 4}\{k_1, k_2, k_3, k_4\}. \end{aligned}$$

(19)

Equation (2) shows that $k_i\le \sum _{i=1}^4 (|a_i|^2+|b_i|^2) = \frac{1}{4}$ for $i=1, 2, 3, 4$, this implies that the maximal eigenvalue $\lambda$ of matrix $C_1$ satisfies

$$\begin{aligned} \lambda \le \frac{1}{4}. \end{aligned}$$

(20)

Similarly, we have

$$\begin{aligned} \begin{array}{ll} yC_2y^\dag &{}=(y_1, y_2, y_3, y_4)\left( \begin{array}{cccc}b_2^2+a_1^2&{}0&{}a_1b_1&{}a_2b_2\\ 0&{}b_4^2+a_1^2&{}a_2b_4&{}a_1b_3\\ a_1b_1&{}a_2b_4&{}b_1^2+a_2^2&{}0\\ a_2b_2&{}a_1b_3&{}0&{}b_3^2+a_2^2\end{array}\right) \left( \begin{array}{c}y_1^*\\ y_2^*\\ y_3^*\\ y_4^*\end{array}\right) \\ &{}\le (|b_2|^2+|a_2|^2+|a_1|^2+|b_1|^2)|y_1|^2+(|a_2|^2+|b_4|^2+|a_1|^2+|b_3|^2)|y_2|^2\\ &{}\quad +(|a_1|^2+|b_1|^2+|a_2|^2+|b_4|^2)|y_3|^2+(|b_2|^2+|a_2|^2+|a_1|^2+|b_3|^2)|y_4|^2\\ &{}=l_1|y_1|^2+l_2|y_2|^2+l_3|y_3|^2+l_4|y_4|^2.\end{array}\nonumber \\ \end{aligned}$$

(21)

Since $|y_1|^2+|y_2|^2+|y_3|^2+|y_4|^2=1,$ we have

$$\begin{aligned} yC_2y^\dag \le \max \limits _{i=1,\cdots , 4}\{l_1, l_2, l_3, l_4\}. \end{aligned}$$

(22)

From Eq. (2), we can get that $l_i\le \sum _{i=1}^4 (|a_i|^2+|b_i|^2) = \frac{1}{4}$ for $i=1, 2, 3, 4$, this implies that the maximal eigenvalue $\mu$ of matrix $C_2$ satisfying

$$\begin{aligned} \mu \le \frac{1}{4}. \end{aligned}$$

(23)

Equations (20) and (23) imply that $\lambda +\mu \le \frac{1}{2}.$

This completes the proof.

The proof of Proposition 5

$$\begin{aligned} X=A\otimes I+I\otimes B= & {} \left( \begin{array}{cccc}0&{}a_1I &{}0&{}0\\ 0&{}0&{}a_2I &{}0\\ 0&{}0&{}0&{}a_3I\\ a_4I&{}0 &{} 0&{}0\end{array}\right) +\left( \begin{array}{cccc}B&{}0&{}0&{}0\\ 0&{}B&{}0&{}0\\ 0&{}0&{}B&{}0\\ 0&{}0&{}0&{}B\end{array}\right) \\= & {} \left( \begin{array}{cccc}B&{}a_1I&{}0&{}0\\ 0&{}B &{} a_2I&{}0\\ 0&{}0&{}B&{}a_3I\\ a_4I &{}0&{}0&{}B\end{array}\right) , \\ XX^\dag= & {} \left( \begin{array}{cccc}B&{}a_1I&{}0&{}0\\ 0&{}B &{} a_2I&{}0\\ 0&{}0&{}B&{}a_3I\\ a_4I &{}0&{}0&{}B\end{array}\right) \left( \begin{array}{cccc}B^\dag &{}0&{}0&{}a_4^*I\\ a_1^*I&{}B^\dag &{}0&{}0\\ 0&{}a_2^*I &{}B^\dag &{}0\\ 0&{}0&{}a_3^*I&{}B^\dag \end{array}\right) \\= & {} \left( \begin{array}{cccc}BB^\dag +|a_1|^2I&{}a_1B^\dag &{}0&{}a_4^*B\\ a_1^*B&{}BB^\dag +|a_2|^2I&{}a_2B^\dag &{}0\\ 0&{}a_2^*B&{}BB^\dag +|a_3|^2I &{} a_3B^\dag \\ a_4B^\dag &{}0&{}a_3^*B&{}|a_4|^2I+BB^\dag \end{array}\right) \\= & {} \left( \begin{array}{ll}Y_1&{}Y_2\\ Y_2^\dag &{}Y_3\end{array}\right) , \end{aligned}$$

where

$$\begin{aligned}&Y_1=\left( \begin{array}{cc}BB^\dag +|a_1|^2I&{}a_1B^\dag \\ a_1^*B&{}|a_2|^2I+BB^\dag \end{array}\right) \\&\quad =\left( \begin{array}{cccccccc}|b_1|^2+|a_1|^2&{}0&{}0&{}0&{}0&{}0&{}0&{}a_1b_4^*\\ 0&{}|b_2|^2+|a_1|^2&{}0&{}0&{}a_1b_1^*&{}0&{}0&{}0\\ 0&{}0&{}|b_3|^2+|a_1|^2&{}0&{}0&{}a_1b_2^*&{}0&{}0\\ 0&{}0&{}0&{}|b_4|^2+|a_1|^2&{}0&{}0&{}a_1b_3^*&{}0\\ 0&{}a_1^*b_1&{}0&{}0&{}|b_1|^2 +|a_2|^2&{}0&{}0&{}0\\ 0&{}0&{}a_1^*b_2&{}0&{}0&{}|b_2|^2+|a_2|^2&{}0&{}0\\ 0&{}0&{}0&{}a_1^*b_3&{}0&{}0&{}|b_3|^2 +|a_2|^2&{}0\\ a_1^*b_4&{}0&{}0&{}0&{}0&{}0&{}0&{}|b_4|^2 + |a_2|^2\end{array}\right) \\&\quad =P \left( \begin{array}{cccccccc}|b_1|^2+|a_1|^2&{}a_1b_4^*&{}0&{}0&{}0&{}0&{}0&{}0\\ a_1^*b_4&{}|b_4|^2+|a_2|^2&{}0&{}0&{}0&{}0&{}0&{}0\\ 0&{}0&{}|b_2|^2+|a_1|^2&{}a_1b_1^*&{}0&{}0&{}0&{}0\\ 0&{}0&{}a_1^*b_1&{}|b_1|^2+|a_2|^2&{}0&{}0&{}0&{}0\\ 0&{}0&{}0&{}0&{}|b_3|^2+|a_1|^2&{}a_1b_2^*&{}0&{}0\\ 0&{}0&{}0&{}0&{}a_1^*b_2&{}|b_2|^2+|a_2|^2&{}0&{}0\\ 0&{}0&{}0&{}0&{}0&{}0&{}|b_4|^2+|a_1|^2&{}a_1b_3^*\\ 0&{}0&{}0&{}0&{}0&{}0&{}a_1^*b_3&{}|b_3|^2+|a_2|^2\end{array}\right) P^\dag , \end{aligned}$$

where P is a permutation. We have

$$\begin{aligned}&Y_2=\left( \begin{array}{cc}0&{}a_4^*B\\ a_2B^\dag &{}0\end{array}\right) =\left( \begin{array}{cccccccc}0&{}0&{}0&{}0&{}0&{}a_4^*b_1&{}0&{}0\\ 0&{}0&{}0&{}0&{}0&{}0&{}a_4^*b_2&{}0\\ 0&{}0&{}0&{}0&{}0&{}0&{}0&{}a_4^*b_3\\ 0&{}0&{}0&{}0&{}a_4^*b_4&{}0&{}0&{}0\\ 0&{}0&{}0&{}a_2b_4^*&{}0&{}0&{}0&{}0\\ a_2b_1^*&{}0&{}0&{}0&{}0&{}0&{}0&{}0\\ 0&{}a_2b_2^*&{}0&{}0&{}0&{}0&{}0&{}0\\ 0&{}0&{}a_2b_3^*&{}0&{}0&{}0&{}0&{}0\end{array}\right) \\&\quad =P\left( \begin{array}{cccccccc}0&{}0&{}0&{}0&{}0&{}a_4^*b_1&{}0&{}0\\ 0&{}0&{}0&{}0&{}a_2b_3^*&{}0&{}0&{}0\\ 0&{}0&{}0&{}0&{}0&{}0&{}0&{}a_4^*b_2\\ 0&{}0&{}0&{}0&{}0&{}0&{}a_2b_4^*&{}0\\ 0&{}a_4^*b_3&{}0&{}0&{}0&{}0&{}0&{}0\\ a_2b_1^*&{}0&{}0&{}0&{}0&{}0&{}0&{}0\\ 0&{}0&{}0&{}a_4^*b_4&{}0&{}0&{}0&{}0\\ 0&{}0&{}a_2b_2^*&{}0&{}0&{}0&{}0&{}0\end{array}\right) P^t, \\&Y_3=\left( \begin{array}{cc}BB^\dag +|a_3|^2I&{}a_3B^\dag \\ a_3^*B&{}|a_4|^2I+BB^\dag \end{array}\right) \\&\quad =P \left( \begin{array}{cccccccc}|b_1|^2+|a_3|^2&{}a_3b_4^*&{}0&{}0&{}0&{}0&{}0&{}0\\ a_3^*b_4&{}|b_4|^2+|a_4|^2&{}0&{}0&{}0&{}0&{}0&{}0\\ 0&{}0&{}|b_2|^2+|a_3|^2&{}a_3b_1^*&{}0&{}0&{}0&{}0\\ 0&{}0&{}a_3^*b_1&{}|b_1|^2+|a_4|^2&{}0&{}0&{}0&{}0\\ 0&{}0&{}0&{}0&{}|b_3|^2+|a_3|^2&{}a_3b_2^*&{}0&{}0\\ 0&{}0&{}0&{}0&{}a_3^*b_2&{}|b_2|^2+|a_4|^2&{}0&{}0\\ 0&{}0&{}0&{}0&{}0&{}0&{}|b_4|^2+|a_3|^2&{}a_3b_3^*\\ 0&{}0&{}0&{}0&{}0&{}0&{}a_3^*b_3&{}|b_3|^2+|a_4|^2\end{array}\right) P^t. \end{aligned}$$

Here we can see that

$$\begin{aligned}&XX^\dag \sim \left( \begin{array}{cccccccc}N_1&{}0&{}0&{}0&{}0&{}0&{}N_2&{}0\\ 0&{}N_3&{}0&{}0&{}0&{}0&{}0&{}N_4\\ 0&{}0&{}N_5&{}0&{}N_6&{}0&{}0&{}0\\ 0&{}0&{}0&{}N_7&{}0&{}N_8&{}0&{}0\\ 0&{}0&{}N_6^\dag &{}0&{}N_9&{}0&{}0&{}0\\ 0&{}0&{}0&{}N_8^\dag &{}0&{}N_{10}&{}0&{}0\\ N_2^\dag &{}0&{}0&{}0&{}0&{}0&{}N_{11}&{}0\\ 0&{}N_4^\dag &{}0&{}0&{}0&{}0&{}0&{}N_{12}\end{array}\right) \nonumber \\&\quad \sim \left( \begin{array}{cccccccc}N_1&{}N_2&{}0&{}0&{}0&{}0&{}0&{}0\\ N_2^\dag &{}N_{11}&{}0&{}0&{}0&{}0&{}0&{}0\\ 0&{}0&{}N_3&{}N_4&{}0&{}0&{}0&{}0\\ 0&{}0&{}N_4^\dag &{}N_{12}&{}0&{}0&{}0&{}0\\ 0&{}0&{}0&{}0&{}N_5&{}N_6&{}0&{}0\\ 0&{}0&{}0&{}0&{}N_6^\dag &{}N_9&{}0&{}0\\ 0&{}0&{}0&{}0&{}0&{}0&{}N_7&{}N_8\\ 0&{}0&{}0&{}0&{}0&{}0&{}N_8^\dag &{}N_{10} \end{array}\right) =W, \end{aligned}$$

(24)

where $N_i, i=1, 2,\dots , 12$ are $2\times 2$ matrices. Then

$$\begin{aligned} XX^\dag \sim W=W_1\oplus W_2\oplus W_3 \oplus W_4, \end{aligned}$$

(25)

where

$$\begin{aligned} W_1=\left( \begin{array}{cc}N_1&{}N_2\\ N_2^\dag &{}N_{11}\end{array}\right) =\left( \begin{array}{cccc}|b_1|^2+|a_1|^2&{}a_1b_4^*&{}0&{}a_4^*b_1\\ a_1^*b_4&{}|b_4|^2+|a_2|^2&{}a_2b_3^*&{}0\\ 0&{}a_2^*b_3&{}|b_3|^2+|a_3|^2&{}a_3b_2^*\\ a_4b_1^*&{}0&{}a_3^*b_2&{}|b_2|^2+|a_4|^2\end{array}\right) , \end{aligned}$$

(26)

$$\begin{aligned} W_2=\left( \begin{array}{cc}N_3&{}N_4\\ N_4^t&{}N_{12}\end{array}\right) =\left( \begin{array}{cccc}|b_2|^2+|a_1|^2&{}a_1b_1^*&{}0&{}a_4^*b_2\\ a_1^*b_1&{}|b_1|^2+|a_2|^2&{}a_2b_4^*&{}0\\ 0&{}a_2^*b_4&{}|b_4|^2+|a_3|^2&{}a_3^*b_3\\ a_4b_2^*&{}0&{}a_3b_3^*&{}|b_3|^2+|a_4|^2\end{array}\right) , \end{aligned}$$

(27)

$$\begin{aligned} W_3=\left( \begin{array}{cc}N_5&{}N_6\\ N_6^t&{}N_{9}\end{array}\right) =\left( \begin{array}{cccc}|b_3|^2+|a_1|^2&{}a_1b_2^*&{}0&{}a_4^*b_3\\ a_1^*b_2&{}|b_2|^2+|a_2|^2&{}a_2b_1^*&{}0\\ 0&{}a_2^*b_1&{}|b_1|^2+|a_3|^2&{}a_3b_4^*\\ a_4b_3^*&{}0&{}a_3^*b_4&{}|b_4|^2+|a_4|^2\end{array}\right) , \end{aligned}$$

(28)

$$\begin{aligned} W_4=\left( \begin{array}{cc}N_7&{}N_8\\ N_8^t&{}N_{10}\end{array}\right) =\left( \begin{array}{cccc}|b_4|^2+|a_1|^2&{}a_1b_3^*&{}0&{}a_4^*b_4\\ a_1^*b_3&{}|b_3|^2+|a_2|^2&{}a_2b_2^*&{}0\\ 0&{}a_2^*b_2&{}|b_2|^2+|a_3|^2&{}a_3b_1^*\\ a_4b_4^*&{}0&{}a_3^*b_1&{}|b_1|^2+|a_4|^2\end{array}\right) ,\end{aligned}$$

(29)

and

$$\begin{aligned} \mathrm{Tr}\,W_i=\sum _{j=1}^4(|b_j|^2+|a_j|^2)=\frac{1}{4}, i=1,2,3,4. \end{aligned}$$

(30)

Let $\lambda$ and $\mu$ be two arbitrary eigenvalues of $XX^\dag$. Then, proving Conjecture 1 is equivalent to proving $\lambda +\mu \le \frac{1}{2}$. There are two cases for $\lambda$ and $\mu$.

Case 1. $\lambda$ and $\mu$ are the eigenvalues of different $W_i$.

Without loss of generality, we can assume that $\lambda$ is the maximum eigenvalue of $W_i$, and $\mu$ is the maximum eigenvalue of $W_j$, $1\le i, j\le 4, i\ne j$. $\lambda +\mu \le \mathrm{Tr}\,W_i+\mathrm{Tr}\,W_j= 2[\sum _{j=1}^4(|b_j|^2+|a_j|^2)]=\frac{1}{2}$ .

Case 2. $\lambda$ and $\mu$ are the eigenvalues of same $W_i, i=1, 2, 3, 4$. Equation (30) implies that $\lambda +\mu \le \mathrm{Tr}\,W_i= \sum _{j=1}^4(|b_j|^2+|a_j|^2)=\frac{1}{4}$.

This completes the proof.

The proof of Proposition 6

$$\begin{aligned} X= & {} A\otimes I+I\otimes B=\left( \begin{array}{cccc}a_1I &{}0&{}0 &{}0\\ 0&{}-a_1I&{}0&{}0\\ 0&{}0&{}0&{}a_2I\\ 0&{}0 &{} a_3I&{}0\end{array}\right) +\left( \begin{array}{cccc}B&{}0&{}0&{}0\\ 0&{}B&{}0&{}0\\ 0&{}0&{}B&{}0\\ 0&{}0&{}0&{}B\end{array}\right) \\= & {} \left( \begin{array}{cccc}a_1I+B&{}0&{}0&{}0\\ 0&{}-a_1I+B &{}0&{}0\\ 0&{}0&{}B&{}a_2I\\ 0&{}0&{}a_3I&{}B\end{array}\right) . \end{aligned}$$

So we have

$$\begin{aligned} \begin{array}{lll} XX^\dag &{}= &{} \left( \begin{array}{cccc}a_1I+B&{}0&{}0&{}0\\ 0&{}-a_1I+B &{}0&{}0\\ 0&{}0&{}B&{}a_2I\\ 0&{}0&{}a_3I&{}B\end{array}\right) \left( \begin{array}{cccc}a_1^*I+B^\dag &{}0&{}0&{}0\\ 0&{}-a_1^*I+B^\dag &{}0&{}0\\ 0&{}0&{}B^\dag &{}a_3^*I\\ 0&{}0&{}a_2^*I&{}B^\dag \end{array}\right) \\ &{}=&{}\left( \begin{array}{cccc}BB^\dag +|a_1|^2I+a_1B^\dag +a_1^*B&{}0&{}0&{}0\\ 0&{}BB^\dag -a_1^*B-a_1B^\dag +|a_1|^2I&{}0&{}0\\ 0&{}0&{}BB^\dag +|a_2|^2I &{} a_2B^\dag +a_3^*B\\ 0&{}0&{}a_3B^\dag +a_2^*B&{}|a_3|^2I+BB^\dag \end{array}\right) \\ &{}=&{}\left( \begin{array}{cc}C&{}0\\ 0&{}D\end{array}\right) =C\oplus D,\end{array} \end{aligned}$$

where

$$\begin{aligned}&C=\left( \begin{array}{cc}BB^\dag +|a_1|^2I+a_1B^\dag +a_1^*B&{}0\\ 0&{}BB^\dag -a_1^*B-a_1B^\dag +|a_1|^2I \end{array}\right) \nonumber \\&=C_1\oplus C_2\oplus C_3\oplus C_4, \nonumber \\&C_1= \left( \begin{array}{cc} |b_1+a_1|^2&{}0\\ 0&{}|b_1-a_1|^2 \end{array}\right) , \nonumber \\&C_2= \left( \begin{array}{cc} |b_2|^2+|a_1|^2&{}a_1^*b_2+a_1b_3^*\\ a_1^*b_3+a_1b_2^*&{}|b_3|^2+|a_1|^2\end{array}\right) , \nonumber \\&C_3= \left( \begin{array}{cc} |b_1-a_1|^2&{}0\\ 0&{}|b_1+a_1|^2\end{array}\right) , \nonumber \\&C_4=\left( \begin{array}{cc} |b_2|^2 +|a_1|^2&{}-a_1^*b_2-a_1b_3^*\\ -a_1b_2^*-a_1^*b_3&{}|b_3|^2 + |a_1|^2\end{array}\right) , \end{aligned}$$

(31)

and

$$\begin{aligned}&D=\left( \begin{array}{cc}BB^\dag +|a_2|^2I&{}a_3^*B+a_2B^\dag \\ a_3B^\dag +a_2^*B&{}|a_3|^2I+BB^\dag \end{array}\right) \nonumber \\&\quad =\left( \begin{array}{cccccccc}|b_1|^2+|a_2|^2&{}0&{}0&{}0&{}a_3^*b_1+a_2b_1^*&{}0&{}0&{}0\\ 0&{}|b_1|^2+|a_2|^2&{}0&{}0&{}0&{}-b_1^*a_2-b_1a_3^*&{}0&{}0\\ 0&{}0&{}|b_2|^2+|a_2|^2&{}0&{}0&{}0&{}0&{}a_3^*b_2+a_2b_3^*\\ 0&{}0&{}0&{}|b_3|^2+|a_2|^2&{}0&{}0&{}a_3^*b_3+a_2b_2^*&{}0\\ a_3b_1^*+a_2^*b_1&{}0&{}0&{}0&{}|b_1|^2 +|a_3|^2&{}0&{}0&{}0\\ 0&{}-b_1a_2^*-b_1^*a_3&{}0&{}0&{}0&{}|a_3|^2+|b_1|^2&{}0&{}0\\ 0&{}0&{}0&{}a_3b_3^*+a_2^*b_2&{}0&{}0&{}|b_2|^2+|a_3|^2&{}0\\ 0&{}0&{}a_3b_2^*+a_2^*b_3&{}0&{}0&{}0&{}0&{}|b_3|^2 + |a_3|^2\end{array}\right) \nonumber \\&\quad \sim \left( \begin{array}{cccccccc}|b_1|^2+|a_2|^2&{}a_3^*b_1+a_2b_1^*&{}0&{}0&{}0&{}0&{}0&{}0\\ a_3b_1^*+a_2^*b_1&{}|b_1|^2+|a_3|^2&{}0&{}0&{}0&{}0&{}0&{}0\\ 0&{}0&{}|b_1|^2+|a_2|^2&{}-b_1^*a_2-b_1a_3^*&{}0&{}0&{}0&{}0\\ 0&{}0&{}-b_1a_2^*-b_1^*a_3&{}|b_1|^2+|a_3|^2&{}0&{}0&{}0&{}0\\ 0&{}0&{}0&{}0&{}|b_2|^2+|a_2|^2&{}a_3b_2^*+a_2^*b_3&{}0&{}0\\ 0&{}0&{}0&{}0&{}a_3^*b_2+a_2b_3^*&{}|b_3|^2+|a_3|^2&{}0&{}0\\ 0&{}0&{}0&{}0&{}0&{}0&{}|b_3|^2+|a_2|^2&{}a_3^*b_3+a_2b_2^*\\ 0&{}0&{}0&{}0&{}0&{}0&{}a_3b_3^*+a_2^*b_2&{}|b_2|^2+|a_3|^2\end{array}\right) \nonumber \\&\quad =D_1\oplus D_2\oplus D_3\oplus D_4. \end{aligned}$$

(32)

One can formulate the characteristic polynomial of C as follows.

$$\begin{aligned} \det (\lambda I-C)=f_1^2(\lambda )\cdot f_2^2(\lambda ), \end{aligned}$$

(33)

where

$$\begin{aligned} \begin{array}{lll} f_1(\lambda )&{}=&{}(|b_1|^2+|a_2|^2+2Re a_1^*b_1-\lambda )(|b_1|^2+|a_2|^2-2Re a_1^*b_1-\lambda ),\\ f_2(\lambda )&{}=&{}(|b_2|^2+|a_1|^2-\lambda )(|a_1|^2+|b_3|^2-\lambda )-|a_1^*b_2+a_1b_3^*|^2,\\ \end{array} \end{aligned}$$

(34)

and Rez denotes the real part of the complex number z.

We first claim the larger root of $f_i(\lambda )=0, \forall i=1,2,3,4$, is not greater than $\frac{1}{4}$.Take $f_1(\lambda )$ as an example. It is clear that the sum of the two roots of $f_1(\lambda )=0$ is $|a_1|^2+|a_2|^2+2|b_1|^2$. Since $XX^t=C\oplus D$ is a semipositive definite matrix. All the eigenvalues of C and D are non-negative. This implies that all the roots of $f_i(\lambda )=0, i=1, 2$ are nonnegative. So we have the larger root of $f_1(\lambda )=0$ is not greater than the sum of two roots, i.e., $|a_1|^2+|a_2|^2+2|b_1|^2$. Recall the $\sum _{i=1}^3(|a_i|^2+|b_i|^2)+|a_1|^2+|b_1|^2=\frac{1}{4}.$ Therefore, we conclude that the larger root of $f_1(\lambda )=0$ is not larger than $\frac{1}{4}.$ One can draw the same conclusion of $f_2(\lambda )$. Then, our claim holds.

This implies the largest eigenvalue of C is not greater than $\frac{1}{4}.$

The characteristic polynomial of D can be expressed as follows:

$$\begin{aligned} \det (\lambda I-D)=g_1 (\mu )\cdot g_2(\mu )\cdot g_3(\mu )\cdot g_3(\mu ), \end{aligned}$$

(35)

where

$$\begin{aligned} \begin{array}{lll} g_1(\mu )&{}=&{}(|b_1|^2+|a_2|^2-\mu )(|a_3|^2+|b_1|^2-\mu )-|a_3^*b_1+a_2b_1^*|^2,\\ g_2(\mu )&{}=&{}(|b_1|^2+|a_2|^2-\mu )(|a_3|^2+|b_1|^2-\mu )-|b_1^*a_2+b_1a_3^*|^2,\\ g_3(\mu )&{}=&{}(|b_2|^2+|a_2|^2-\mu )(|a_3|^2+|b_3|^2-\mu )-|a_3b_2^*+a_2^*b_3|^2,\\ g_4(\mu )&{}=&{}(|b_3|^2+|a_2|^2-\mu )(|a_3|^2+|b_2|^2-\mu )-|a_3^*b_3+a_2b_2^*|^2. \end{array} \end{aligned}$$

(36)

By the same way, we conclude that the largest eigenvalue of D is not greater than $\frac{1}{4}.$ Since $XX^t=C\oplus D$, the sum of the largest eigenvalues of $XX^t$ is at most $\frac{1}{2}$. This complete the proof.

The proof of Proposition 7

$$\begin{aligned} \begin{array}{lll}X&{}=&{}A\otimes I+I\otimes B=\left( \begin{array}{cccc}a_1I &{}0&{}0 &{}0\\ 0&{}-a_1I&{}0&{}0\\ 0&{}0&{}0&{}a_2I\\ 0&{}0 &{} a_3I&{}0\end{array}\right) +\left( \begin{array}{cccc}B&{}0&{}0&{}0\\ 0&{}B&{}0&{}0\\ 0&{}0&{}B&{}0\\ 0&{}0&{}0&{}B\end{array}\right) \\ &{}=&{}\left( \begin{array}{cccc}a_1I+B&{}0&{}0&{}0\\ 0&{}-a_1I+B &{}0&{}0\\ 0&{}0&{}B&{}a_2I\\ 0&{}0&{}a_3I&{}B\end{array}\right) . \end{array} \end{aligned}$$

Hence

$$\begin{aligned} \begin{array}{lll}XX^\dag &{}= &{} \left( \begin{array}{cccc}a_1I+B&{}0&{}0&{}0\\ 0&{}-a_1I+B &{}0&{}0\\ 0&{}0&{}B&{}a_2I\\ 0&{}0&{}a_3I&{}B\end{array}\right) \left( \begin{array}{cccc}a_1^*I+B^\dag &{}0&{}0&{}0\\ 0&{}-a_1^*I+B^\dag &{}0&{}0\\ 0&{}0&{}B^\dag &{}a_3^*I\\ 0&{}0&{}a_2^*I&{}B^\dag \end{array}\right) \\ &{}=&{}\left( \begin{array}{cccc}BB^\dag +|a_1|^2I+a_1B^\dag +a_1^*B&{}0&{}0&{}0\\ 0&{}BB^\dag -a_1^*B-a_1B^\dag +|a_1|^2I&{}0&{}0\\ 0&{}0&{}BB^\dag +|a_2|^2I &{} a_2B^\dag +a_3^*B\\ 0&{}0&{}a_3B^\dag +a_2^*B&{}|a_3|^2I+BB^\dag \end{array}\right) \\ &{}=&{}\left( \begin{array}{cc}C&{}0\\ 0&{}D\end{array}\right) =C\oplus D,\end{array} \end{aligned}$$

where

$$\begin{aligned}&C=\left( \begin{array}{cc}BB^\dag +|a_1|^2I+a_1^*B+a_1B^\dag )&{}0\\ 0&{}BB^\dag +|a_1|^2I-a_1^*B-a_1B^\dag \end{array}\right) \nonumber \\&\quad =\left( \begin{array}{cccccccc}|b_1|^2+|a_1|^2&{}a_1^*b_1&{}0&{}a_1b_4^*&{}0&{}0&{}0&{}0\\ a_1b_1^*&{}|b_2|^2+|a_1|^2&{}a_1^*b_2&{}0&{}0&{}0&{}0&{}0\\ 0&{}a_1b_2^*&{}|b_3|^2+|a_1|^2&{}a_1^*b_3&{}0&{}&{}0&{}0\\ a_1^*b_4&{}0&{}a_1b_3^*&{}|b_4|^2+|a_1|^2&{}0&{}0&{}0&{}0\\ 0&{}0&{}0&{}0&{}|b_1|^2 +|a_1|^2&{}-a_1^*b_1&{}0&{}-a_1b_4^*\\ 0&{} 0&{}0&{}0&{}-a_1b_1^*&{}|b_2|^2+|a_1|^2&{}-a_1^*b_2&{}0\\ 0&{}0&{}0&{}0&{}0&{}-a_1b_2^*&{}|b_3|^2 +|a_1|^2&{}-a_1^*b_3\\ 0&{}0&{}0&{}0&{}-a_1^*b_4&{}0&{}-a_1b_3^*&{}|b_4|^2 +|a_1|^2\end{array}\right) \nonumber \\&\quad =C_1\oplus C_2, \end{aligned}$$

(37)

and

$$\begin{aligned}&D=\left( \begin{array}{cc}BB^\dag +|a_2|^2I&{}a_3^*B+a_2B^\dag \\ a_3B^\dag +a_2^*B&{}|a_3|^2I+BB^\dag \end{array}\right) \nonumber \\&\quad =\left( \begin{array}{cccccccc}|b_1|^2+|a_2|^2&{}0&{}0&{}0&{}0&{}a_3^*b_1&{}0&{}a_2b_4^*\\ 0&{}|b_2|^2+|a_2|^2&{}0&{}0&{}a_2b_1^*&{}0&{}a_3^*b_2&{}0\\ 0&{}0&{}|b_3|^2+|a_2|^2&{}0&{}0&{}a_2b_2^*&{}0&{}a_3^*b_3\\ 0&{}0&{}0&{}|b_4|^2+|a_2|^2&{}a_3^*b_4&{}0&{}a_2b_3^*&{}0\\ 0&{}a_2^*b_1&{}0&{}a_3b_4^*&{}|b_1|^2 +|a_3|^2&{}0&{}0&{}0\\ a_3b_1^*&{}0&{}a_2^*b_2&{}0&{}0&{}|b_2|^2+|a_3|^2&{}0&{}0\\ 0&{}a_3b_2^*&{}0&{}a_2^*b_3&{}0&{}0&{}|b_3|^2 +|a_3|^2&{}\\ a_2^*b_4&{}0&{}a_3b_3^*&{}0&{}0&{}0&{}0&{}|b_4|^2 + |a_3|^2\end{array}\right) \nonumber \\&\quad \sim \left( \begin{array}{cccccccc}|b_1|^2+|a_2|^2&{}0&{}a_3^*b_1&{}a_2b_4^*&{}0&{}0&{}0&{}0\\ 0&{}|b_3|^2+|a_2|^2&{}a_2b_2^*&{}a_3^*b_3&{}0&{}0&{}0&{}0\\ a_3b_1^*&{}a_2^*b_2&{}|b_2|^2+|a_3|^2&{}0&{}0&{}0&{}0&{}0\\ a_2^*b_4&{}a_3b_3^*&{}0&{}|b_4|^2+|a_3|^2&{}0&{}0&{}0&{}0\\ 0&{}0&{}0&{}0&{}|b_2|^2+|a_2|^2&{}0&{}a_2b_1^*&{}a_3^*b_2\\ 0&{}0&{}0&{}0&{}0&{}|b_4|^2+|a_2|^2&{}a_3^*b_4&{}a_2b_3^*\\ 0&{}0&{}0&{}0&{}a_2^*b_1&{}a_3b_4^*&{}|b_1|^2+|a_3|^2&{}0\\ 0&{}0&{}0&{}0&{}a_3b_2^*&{}a_2^*b_3&{}0&{}|b_3|^2+|a_3|^2\end{array}\right) \nonumber \\&\quad =D_1\oplus D_2. \end{aligned}$$

(38)

Then, we have

$$\begin{aligned} XX^t\sim C_1\oplus C_2\oplus D_1 \oplus D_2, \end{aligned}$$

(39)

and

$$\begin{aligned} \mathrm{Tr}\,C_1=\mathrm{Tr}\,C_2=\sum _{j=1}^4|b_j|^2+4|a_1|^2, \mathrm{Tr}\,D_1=\mathrm{Tr}\,D_2=\sum _{j=1}^4|b_j|^2+2(|a_2|^2+|a_3|^2).\nonumber \\ \end{aligned}$$

(40)

Let $\lambda$ and $\mu$ be two arbitrary eigenvalues of $XX^t$. Then, proving Conjecture 1 is equivalent to proving $\lambda +\mu \le \frac{1}{2}$. There are three cases for $\lambda$ and $\mu$.

Case 1. $\lambda$ is the eigenvalue of some $C_i$ and $\mu$ is the eigenvalue of some $D_j$ for any $i,j =1,2$.

Without loss of generality, we assume $\lambda$ is the eigenvalue of $C_1$ and $\mu$ is the eigenvalue of $D_1$. Equations (37) and (40) imply that $\lambda \le \mathrm{Tr}\,C_1= \sum _{j=1}^4|b_j|^2+4|a_1|^2$. Equations (38) and (40) imply that $\mu \le \mathrm{Tr}\,D_1= \sum _{j=1}^4|b_j|^2+2(|a_2^2+|a_3|^2)$. Then $\lambda +\mu \le 2\sum _{j=1}^3(|b_j|^2+|a_j|^2)+2|a_1|^2+2|b_4|^2$, Eq. (7) shows that $\lambda +\mu \le \frac{1}{2}$.

Case 2. $\lambda$ and $\mu$ are the eigenvalues of same $C_i$ or $D_j$ .

Assume $\lambda$ and $\mu$ are the eigenvalue of $C_1$, Eq. (40) implies that $\lambda +\mu \le \mathrm{Tr}\,C_1= \sum _{j=1}^4|b_j|^2+4|a_1|^2\le 2[\sum _{j=1}^3(|b_j|^2+|a_j|^2)+|a_1|^2+|b_4|^2]=\frac{1}{2}$ .

Case 3. $\lambda$ and $\mu$ are the eigenvalues of $C_1$ and $C_2$ or $D_1$ and $D_2$.

Without loss of generality, we assume that $\lambda$ is the maximum eigenvalue of $C_1$ and $\mu$ is the maximum eigenvalue of $C_2$.

Let $x=(x_1, x_2, x_3, x_4), y=(y_1, y_2, y_3, y_4)$ and $\Vert x\Vert =\Vert y\Vert =1$, then

$$\begin{aligned} \begin{array}{ll} xC_1x^t&{}=(x_1, x_2, x_3, x_4)\left( \begin{array}{cccc}|b_1|^2+|a_1|^2&{}a_1^*b_1&{}0&{}a_1b_4^*\\ a_1b_1^*&{}|b_2|^2+|a_1|^2&{}a_1^*b_2&{}0\\ 0&{}a_1b_2^*&{}|b_3|^2+|a_2|^2&{}a_1^*b_3\\ a_1^*b_4&{}0&{}a_1b_3^*&{}|b_4|^2+|a_1|^2\end{array}\right) \left( \begin{array}{c}x_1^*\\ x_2^*\\ x_3^*\\ x_4^*\end{array}\right) \\ &{}=|b_1x_1+a_1x_2|^2+|a_1x_1+b_4x_4|^2+|a_1x_3+b_2x_2|^2+|b_3x_3+a_1x_4|^2\\ &{}\le (|b_1|^2+|a_1|^2)(|x_1|^2+|x_2|^2)+(|a_1|^2+|b_4|^2)(|x_1|^2+|x_4|^2)+(|a_1|^2+|b_2|^2)(|x_2|^2+|x_3|^2)\\ &{}\quad +(|b_3|^2+|a_1|^2)(|x_3|^2+|x_4|^2)\\ &{}=(|b_1|^2+2|a_1|^2+|b_4|^2)|x_1|^2+(2|a_1|^2+|b_1|^2+|b_2|^2)|x_2|^2+(|b_2|^2+|b_3|^2+2|a_1|^2)|x_3|^2\\ &{}\quad +(2|a_1|^2+|b_3|^2+|b_4|^2)|x_4|^2\end{array}.\nonumber \\ \end{aligned}$$

(41)

Set $k_1=|b_1|^2+2|a_1|^2+|b_4|^2, k_2=2|a_1|^2+|b_1|^2+|b_2|^2, k_3=|b_2|^2+|b_3|^2+2|a_1|^2, k_4=2|a_1|^2+|b_3|^2+|b_4|^2,$ we have known that $|x_1|^2+|x_2|^2+|x_3|^2+|x_4|^2=1$, then

$$\begin{aligned} xC_1x^t\le k_1|x_1|^2+k_2|x_2|^2+k_3|x_3|^2+k_4|x_4|^2\le \max \limits _{i=1,\cdots , 4}\{k_1, k_2, k_3, k_4\}. \end{aligned}$$

(42)

Equation (2) shows that $k_i\le \sum _{i=1}^3 (|a_i|^2+|b_i|^2) +|a_1|^2+|b_4|^2= \frac{1}{4}$ for $i=1, 2, 3, 4$, this implies that the maximal eigenvalue $\lambda$ of matrix $C_1$ satisfying

$$\begin{aligned} \lambda \le \frac{1}{4}. \end{aligned}$$

(43)

By the same way, we can show that the maximal eigenvalue $\mu$ of matrix $C_2$ satisfying

$$\begin{aligned} \mu \le \frac{1}{4}. \end{aligned}$$

(44)

Equations (43) and (44) imply that $\lambda +\mu \le \frac{1}{2}.$

This completes the proof.

The proof of Proposition 8

$$\begin{aligned} \begin{array}{lll}X&{}=&{}A\otimes I+I\otimes B=\left( \begin{array}{cccc}a_1I &{}0&{}0 &{}0\\ 0&{}-a_1I&{}0&{}0\\ 0&{}0&{}0&{}a_2I\\ 0&{}0 &{} a_3I&{}0\end{array}\right) +\left( \begin{array}{cccc}B&{}0&{}0&{}0\\ 0&{}B&{}0&{}0\\ 0&{}0&{}B&{}0\\ 0&{}0&{}0&{}B\end{array}\right) \\ &{}=&{}\left( \begin{array}{cccc}a_1I+B&{}0&{}0&{}0\\ 0&{}-a_1I+B &{}0&{}0\\ 0&{}0&{}B&{}a_2I\\ 0&{}0&{}a_3I&{}B\end{array}\right) . \end{array} \end{aligned}$$

We have

$$\begin{aligned} \begin{array}{lll}XX^\dag &{}= &{} \left( \begin{array}{cccc}a_1I+B&{}0&{}0&{}0\\ 0&{}-a_1I+B &{}0&{}0\\ 0&{}0&{}B&{}a_2I\\ 0&{}0&{}a_3I&{}B\end{array}\right) \left( \begin{array}{cccc}a_1^*I+B^\dag &{}0&{}0&{}0\\ 0&{}-a_1^*I+B^\dag &{}0&{}0\\ 0&{}0&{}B^\dag &{}a_3^*I\\ 0&{}0&{}a_2^*I&{}B^\dag \end{array}\right) \\ &{}=&{}\left( \begin{array}{cccc}BB^\dag +|a_1|^2I+a_1B^\dag +a_1^*B&{}0&{}0&{}0\\ 0&{}BB^\dag -a_1^*B-a_1B^\dag +|a_1|^2I&{}0&{}0\\ 0&{}0&{}BB^\dag +|a_2|^2I &{} a_2B^\dag +a_3^*B\\ 0&{}0&{}a_3B^\dag +a_2^*B&{}|a_3|^2I+BB^\dag \end{array}\right) ,\\ &{}=&{}C\oplus D.\end{array} \end{aligned}$$

where

$$\begin{aligned}&C=C_1\oplus C_2, \\&C_1=\left( \begin{array}{cccc} |a_1|^2+|b_1|^2&{}a_1b_2^*+a_1^*b_1&{}0&{}0\\ a_1b_1^*+a_1^*b_2&{}|a_1|^2+|b_2|^2&{}0&{}0\\ 0&{}0&{}|a_1|^2+|b_3|^2&{}a_1b_4^*+a_1^*b_3\\ 0&{}0&{}a_1b_3^*+a_1^*b_4&{}|a_1|^2+|b_4|^2\end{array}\right) , \\&C_2=\left( \begin{array}{cccc} |a_1|^2+|b_1|^2&{}a_1b_2^*-a_1^*b_1&{}0&{}0\\ a_1b_1^*-a_1^*b_2&{}|a_1|^2+|b_2|^2&{}0&{}0\\ 0&{}0&{}|a_1|^2+|b_3|^2&{}a_1b_4^*-a_1^*b_3\\ 0&{}0&{}a_1b_3^*-a_1^*b_4&{}|a_1|^2+|b_4|^2\end{array}\right) , \\ \end{aligned}$$

$$\begin{aligned}&D=\left( \begin{array}{cc}BB^\dag +|a_2|^2I&{}a_3^*B+a_2B^\dag \\ a_3B^\dag +a_2^*B&{}|a_3|^2I+BB^\dag \end{array}\right) \nonumber \\&\quad =\left( \begin{array}{cccccccc}|b_1|^2+|a_2|^2&{}0&{}0&{}0&{}0&{}a_3^*b_1+a_2b_2^*&{}0&{}0\\ 0&{}|b_2|^2+|a_2|^2&{}0&{}0&{}a_2b_1^*+a_3^*b_2&{}0&{}0&{}0\\ 0&{}0&{}|b_3|^2+|a_2|^2&{}0&{}0&{}0&{}0&{}a_2b_4^*+a_3^*b_3\\ 0&{}0&{}0&{}|b_4|^2+|a_2|^2&{}0&{}0&{}a_3^*b_4 +a_2b_3^*&{}0\\ 0&{}a_2^*b_1+a_3b_2^*&{}0&{}0&{}|b_1|^2 +|a_3|^2&{}0&{}0&{}0\\ a_3b_1^*+a_2^*b_2&{}0&{}0&{}0&{}0&{}|b_2|^2+|a_3|^2&{}0&{}0\\ 0&{}0&{}0&{}a_3b_4^*+a_2^*b_3&{}0&{}0&{}|b_3|^2 +|a_3|^2&{}\\ 0&{}0&{}a_2^*b_4 +a_3b_3^*&{}0&{}0&{}0&{}0&{}|b_4|^2 + |a_3|^2\end{array}\right) \nonumber \\&\quad \sim \left( \begin{array}{cccccccc}|b_1|^2+|a_2|^2&{}a_3^*b_1 +a_2b_2^*&{}0&{}0&{}0&{}0&{}0&{}0\\ a_3^*b_1+a_2^*b_2&{}|b_2|^2+|a_3|^2&{}0&{}0&{}0&{}0&{}0&{}0\\ 0&{}0&{}|b_2|^2+|a_2|^2&{}a_2b_1^* +a_3^*b_2&{}0&{}0&{}0&{}0\\ 0&{}0&{}a_2^*b_1 +a_3b_2^*&{}|b_1|^2+|a_3|^2&{}0&{}0&{}0&{}0\\ 0&{}0&{}0&{}0&{}|b_3|^2+|a_2|^2&{}a_2b_4^*+a_3^*b_3&{}0&{}0\\ 0&{}0&{}0&{}0&{} a_3b_3^*+a_2^*b_4&{}|b_4|^2+|a_3|^2&{}0&{}0\\ 0&{}0&{}0&{}0&{}0&{}0&{}|b_4|^2+|a_2|^2&{}a_3^*b_4+a_2b_3^*\\ 0&{}0&{}0&{}0&{}0&{}0&{} a_3b_4^*+a_2^*b_3&{}|b_3|^2+|a_3|^2\end{array}\right) \nonumber \\&\quad =D_1\oplus D_2\oplus D_3\oplus D_4. \end{aligned}$$

(45)

One can formulate the characteristic polynomial of C as follows.

$$\begin{aligned} \det (\lambda I-C)=f_1(\lambda )\cdot f_2(\lambda )\cdot f_3(\lambda )\cdot f_4(\lambda ), \end{aligned}$$

(46)

where

$$\begin{aligned} \begin{array}{lll} f_1(\lambda )&{}=&{}(|b_1|^2+|a_1|^2-\lambda )(|a_1|^2+|b_2|^2-\lambda )-|a_1b_2^*+a_1^*b_1|^2,\\ f_2(\lambda )&{}=&{}(|b_3|^2+|a_1|^2-\lambda )(|a_1|^2+|b_4|^2-\lambda )-|a_1b_4^*+a_1^*b_3|^2,\\ f_3(\lambda )&{}=&{}(|b_1|^2+|a_1|^2-\lambda )(|a_1|^2+|b_2|^2-\lambda )-|a_1b_2^*-a_1^*b_1|^2,\\ f_4(\lambda )&{}=&{}(|b_3|^3+|a_1|^2-\lambda )(|a_1|^2+|b_4|^2-\lambda )-|a_1b_4^*+a_1^*b_3|^2. \end{array} \end{aligned}$$

(47)

We first claim the larger root of $f_i(\lambda )=0, \forall i=1,2,3,4$, is not greater than $\frac{1}{4}$. Take $f_1(\lambda )$ as an example. It’s clear that the sum of the two roots of $f_1(\lambda )=0$ is $2|a_1|^2+|b_1|^2+|b_2|^2$. Since $XX^t=C\oplus D$ is a semipositive definite matrix. All the eigenvalues of C and D are nonnegative. This implies that all the roots of $f_i(\lambda )=0, i=1, 2, 3, 4$ are nonnegative. So we have the larger root of $f_1(\lambda )=0$ is not greater than the sum of two roots, i.e., $2|a_1|^2+|b_1|^2+|b_2|^2$. Recall the $\sum _{i=1}^3(|a_i|^2+|b_i|^2)+|a_1|^2+|b_4|^2=\frac{1}{4}.$ Therefore, we conclude that the larger root of $f_1(\lambda )=0$ is not larger than $\frac{1}{4}.$ One can draw the same conclusion of $f_i(\lambda ), i=2, 3, 4$. Then, our claim holds.

This implies the largest eigenvalue of C is not greater than $\frac{1}{4},$i.e.,

$$\begin{aligned} \lambda \le \frac{1}{4}. \end{aligned}$$

(48)

The characteristic polynomial of D can be expressed as follows:

$$\begin{aligned} \det (\lambda I-D)=g_1 (\mu )\cdot g_2(\mu )\cdot g_3(\mu )\cdot g_3(\mu ), \end{aligned}$$

(49)

where

$$\begin{aligned} \begin{array}{lll} g_1(\mu )&{}=&{}(|b_1|^2+|a_2|^2-\mu )(|a_3|^2+|b_2|^2-\mu )-|a_3^*b_1+a_2b_2^*|^2,\\ g_2(\mu )&{}=&{}(|b_2|^2+|a_2|^2-\mu )(|a_3|^2+|b_1|^2-\mu )-|b_1^*a_2+b_2a_3^*|^2,\\ g_3(\mu )&{}=&{}(|b_3|^2+|a_2|^2-\mu )(|a_3|^2+|b_4|^2-\mu )-|a_3^*b_3+a_2b_4^*|^2,\\ g_4(\mu )&{}=&{}(|b_4|^2+|a_2|^2-\mu )(|a_3|^2+|b_3|^2-\mu )-|a_3^*b_4+a_2b_3^*|^2. \end{array}\end{aligned}$$

(50)

By the same way, we can show that the maximal eigenvalue $\mu$ of matrix D satisfying

$$\begin{aligned} \mu \le \frac{1}{4}. \end{aligned}$$

(51)

Equations (48) and (51) imply that $\lambda +\mu \le \frac{1}{2}.$ This completes the proof.