Rank of the matrix made by the diagonal components of the exponential matrix of the block diagonal matrix

Question

I define the $9 \times 9$ matrix $\bf{K}$ as

$${\bf K}=\exp\left(\begin{bmatrix} \bf{A} & p\bf{I} & \bf{O} \\ p\bf{I} & \bf{B} & q\bf{I} \\ \bf{O} & q\bf{I} & \bf{C} \\ \end{bmatrix}\right)$$

where $3 \times 3$ matrices ${\bf A},{\bf B}$ and ${\bf C}$ are defined as

$${\bf A}={\rm diag}(x_1,x_2,x_3), \\ {\bf B}=b{\bf I} + {\bf A}, \\ {\bf C}=c{\bf I} + {\bf A}, $$ for real numbers $x_1,x_2,x_3,b,c,p$ and $q$. I also define $3 \times 3$ matrix $\bf{U}$ as

$$\bf{U}=\begin{bmatrix} {\bf K}_{11} & {\bf K}_{44} & {\bf K}_{77} \\ {\bf K}_{22} & {\bf K}_{55} & {\bf K}_{88} \\ {\bf K}_{33} & {\bf K}_{66} & {\bf K}_{99} \\ \end{bmatrix}.$$

As I see in numerical simulation, the matrix rank of ${\bf U}$ seem to be always $1$. But I cannot prove it mathematically. How can I do that?

Answer 1

For a certain permutation matrix ${\bf P}$ you get $${\bf P}{\bf K}{\bf P}^T=\exp\left(\begin{bmatrix} {\bf Y} & \bf{O} & \bf{O} \\ \bf{O} & {\bf Y} + (x_2-x_1)\bf{I} & \bf{O} \\ \bf{O} & \bf{O} & {\bf Y} + (x_3-x_1)\bf{I} \\ \end{bmatrix}\right) = \begin{bmatrix} \exp({\bf Y}) & \bf{O} & \bf{O} \\ \bf{O} & \exp({\bf Y} + (x_2-x_1)\bf{I}) & \bf{O} \\ \bf{O} & \bf{O} & \exp({\bf Y} + (x_3-x_1)\bf{I}) \\ \end{bmatrix} = \begin{bmatrix} \exp({\bf Y}) & \bf{O} & \bf{O} \\ \bf{O} & \exp(x_2-x_1)\exp({\bf Y}) & \bf{O} \\ \bf{O} & \bf{O} & \exp(x_3-x_1)\exp({\bf Y}) \\ \end{bmatrix} $$ where $$ {\bf Y} = \begin{bmatrix} x_1 & p & 0 \\ p & x_1+b & q \\ 0& q& x_1+c \\ \end{bmatrix}, \qquad {\bf Z}:= \exp({\bf Y}). $$ As a consequence, $$ \bf{U}=\begin{bmatrix} {\bf Z}_{11} & {\bf Z}_{22} & {\bf Z}_{33} \\ \exp(x_2-x_1){\bf Z}_{11} & \exp(x_2-x_1){\bf Z}_{22} & \exp(x_2-x_1){\bf Z}_{33} \\ \exp(x_3-x_1){\bf Z}_{11} & \exp(x_3-x_1){\bf Z}_{22} & \exp(x_3-x_1){\bf Z}_{33} \\ \end{bmatrix} $$ that has always rank exactly 1 since the diagonal of ${\bf Z}$ cannot be zero.