solving equation $X + \overline{X} Tr(X^2)-2XTr(X^2)-2X^2\overline{X}-2X^3+2\overline{X} X^2 =0$

Question

Hi I am wondering how to find a Hermitian matrix $X \in M_{4}(\mathbb{C})$ that satisfies the equation $$ X + \overline{X}\operatorname{tr}(X^2)-2X\operatorname{tr}(X^2)-2X^2\overline{X}-2X^3+2\overline{X} X^2 =0, $$ where $\overline{X}$ is complex conjugate of $X$ (since $X$ is Hermitian, $\overline{X}=X^\top$). Note that $$ Tr(X^2) = \sum_{i,j} X_{ij} X_{ji}. $$ Thanks!

Answer 1

As $X$ is Hermitian, $\overline{X}=X^T$. So, we may rewrite the equation as $\renewcommand{\tr}{\operatorname{tr}}$ $$X + \tr(X^2) (X^T-2X) - 2X^2X^T - 2X^3 + 2X^TX^2 = 0,\tag{1}$$ Taking the conjugate transpose of both sides, we get $$X + \tr(X^2) (X^T-2X) - 2X^TX^2 - 2X^3 + 2X^2X^T = 0.\tag{2}$$ So, the difference and sum of $(1)$ and $(2)$ give \begin{align} &X^2X^T = X^TX^2,\tag{3}\\ &X + \tr(X^2) (X^T-2X) - 2X^3 = 0.\tag{4} \end{align} Let $X=H+iK$, where $H,K\in M_4(\mathbb R)$ are respectively the real and imaginary parts of $X$. As $X$ is Hermitian, $H$ is real symmetric and $K$ is real skew-symmetric. Therefore $X^2 = (H^2-K^2) + i(HK+KH)$, $X^T = H-iK$ and $(3)$ reduces to \begin{align} &HK^2 = K^2H,\tag{5}\\ &KH^2 = H^2K.\tag{6} \end{align} In the special case that $H$ and $K$ commute, $(5)$ and $(6)$ are simultaneously satisfied. In this case, we may assume that $$ X=(a_1I+K_1)\oplus\cdots\oplus(a_mI+K_m),\tag{7} $$ where each $a_j$ is real and each $K_j$ is either zero or a nonzero $2\times2$ real skew symmetric matrix.

Let $t=\tr(X^2)$. Substitute $(7)$ into $(4)$, and compare the real and imaginary parts on both sides, we get, for each $j=1,\ldots,m$, \begin{align} a_j\left[(1-t - 2a_j^2)I + 6K_j^2\right] &= 0,\tag{8}\\ K_j\left[(1-3t - 6a_j^2)I + 2K_j^2\right] &= 0.\tag{9} \end{align} There are three possibilities:

1. $a_j=0$. Then $(9)$ gives $K_j=0$ or $0\ne K_j^2 = \frac{3t-1}2I$. Since $K_j^2$ is a non-positive multiple of the identity matrix, if the latter case occurs, we must have $t<\frac13$.
2. $a_j\ne0$ and $K_j\ne0$. Substitute $(8)$ into $(9)$, we get $(1-t - 2a_j^2)I - 3(1-3t - 6a_j^2)=0$. Hence $a_j = \pm\sqrt{\frac{1-4t}8}$ and in turn, $K_j^2=-\frac18I$. Note that since $a_j$ is real, if this case does occur, we must have $t<\frac14$.
3. $a_j\ne0$ and $K_j=0$. Then $(8)$ gives $a_j=\pm\sqrt{(1-t)/2}$. Since $a_j$ has to be real, if this case occurs, $t<1$.

It follows that, up to real orthogonal similarity, $X$ is equal to \begin{align*} X&=0\oplus iK_\ell\\ &\oplus\left(\sqrt{\frac{1-4t}8}\,I_p+iK_p\right) \oplus\left(-\sqrt{\frac{1-4t}8}\,I_q+iK_q\right)\\ &\oplus\left(\sqrt{\frac{1-t}2}\,I_r\right) \oplus\left(-\sqrt{\frac{1-t}2}I_s\right),\tag{10} \end{align*} where $K_\ell,K_p,K_q$ are two real skew symmetric matrices such that $K_\ell^2=\frac{3t-1}2I_\ell,\ K_p^2=-\frac18I_p$ and $K_q^2=-\frac18I_q$ ($\ell,p,q$ are even numbers). Hence \begin{align*} t=\tr X^2 &=\frac{1-3t}2\ell + \left[\left(\frac{1-4t}8\right)+\frac18\right](p+q) + \frac{1-t}2(r+s)\\ &=\frac{1-3t}2\ell + \left(\frac{1-2t}4\right)(p+q) + \frac{1-t}2(r+s). \end{align*} That is, $$ t = \frac{2\ell + (p+q) + 2(r+s)}{4 + 6\ell + 2(p+q) + 2(r+s)}.\tag{11} $$ In other words, if the real part and imaginary part of $X$ commute ($HK=KH$), then up to real orthogonal similarity, all Hermitian solutions $X$ to $(1)$ are given by $(10)$ and $(11)$, subject to the requirements that \begin{cases} \ell,p,q\in2\mathbb N\cup\{0\},\\ r,s\in\mathbb N\cup\{0\},\\ \ell+p+q+r+s\le4,\\ t<\frac13&\text{ if } \ell>0,\\ t<\frac14&\text{ if } p+q>0,\\ t<1&\text{ if } r+s>0. \end{cases} Using a brute-force search, we see that all feasible values of $(\ell, p+q, r+s, t)$ are given by $(0,0,0,0), (0,0,1,\frac13), (0,0,2,\frac12), (0,0,3,\frac35), (0,0,4,\frac23), (2,0,0,\frac14), (4,0,0,\frac27)$. The first five cases mean that for some real orthogonal matrix $Q\in O(4,\mathbb R)$, $$ Q^TXQ=0\oplus\left(\sqrt{\frac{1-t}2}\,I_r\right) \oplus\left(-\sqrt{\frac{1-t}2}I_s\right);\ 0\le r+s\le 4,\ t=\frac{r+s}{r+s+2}. $$ The case $(2,0,0,\frac14)$ means that $$ Q^TXQ=\frac{i}{\sqrt{8}}\pmatrix{0&-1\\ 1&0}\oplus0 $$ and the case $(4,0,0,\frac27)$ means that $$ Q^TXQ=\frac{i}{\sqrt{14}}\left[\pmatrix{0&-1\\ 1&0}\oplus\pmatrix{0&-1\\ 1&0}\right]. $$ These give all self-adjoint solutions $X$ to $(1)$ subject to the additional condition that the real part and imaginary part of $X$ commute.