Finding $\bf{P}$ such that $\bf{P^{-1}AP=B}$ for both fixed $\bf{A}，\bf{B}$.

Question

How can I find a matrix $\bf{P}\in \mathbb{{R}^{n\times n}}$, such that $\bf{P^{-1}AP=B}$，where

$$\bf{A}=\begin{bmatrix} \bf{A_2}& \bf{C_2}& \\ & \bf{A_2}& \bf{C_2}& \\ & & \ddots& \ddots& \\ & & & \bf{A_2}&\bf{C_2} \\ & & & &\bf{A_2} \end{bmatrix},$$

$$\bf{B}=\begin{bmatrix} \bf{B_2}& \bf{C_2}& \\ & \bf{B_2}& \bf{C_2}& \\ & & \ddots& \ddots& \\ & & & \bf{B_2}&\bf{C_2}\\ & & & &\bf{B_2} \end{bmatrix}.$$

$$\bf{C_2}=\begin{bmatrix} 0& 0\\ 1& 0 \end{bmatrix}，\bf{B_2}=\begin{bmatrix} a& -b\\ b& a \end{bmatrix}$$ and$$ \bf{A_2}=\begin{bmatrix} 0& 1\\ -(a^2+b^2)& 2a \end{bmatrix}.\quad \textit{a}\in\mathbb{R}, \textit{b}\left(\ne \textrm{0}\right)\in\mathbb{R}.$$

---

Let $\bf{P_2}=\begin{bmatrix} 0& 1\\ b& a \end{bmatrix}\left(\bf{P_2}^{-1}=\begin{bmatrix} -a/b& 1/b\\ 1& 0 \end{bmatrix}\right)$ ，we have $\bf{P_2}^{-1}\bf{A_2}\bf{P_2}=\bf{B_2}$.

But$$\bf{P^{-1}AP\ne B}, \textrm{we set }\bf{P}=\begin{bmatrix} \bf{P_2}& & \\ & \bf{P_2}& &\\ & & &\ddots& \\ & & & &\bf{P_2} \\ \end{bmatrix}.$$ In fact, $\bf{P_2}^{-1}\bf{C_2}\bf{P_2}\ne\bf{C_2}.$ So the above $\bf{P}$ is undesirable,we need some other ways to find $\bf{P}$ satisfying the requirement.

Answer 1

Let $X=\pmatrix{1&0\\ a&-b}$. Since $b\ne0$, $X$ is non-singular. Then \begin{aligned} A_2X &=\pmatrix{0&1\\ -(a^2+b^2)&2a}\pmatrix{1&0\\ a&-b}\\ &=\pmatrix{a&-b\\ a^2-b^2&-2ab}=\pmatrix{1&0\\ a&-b}\pmatrix{a&-b\\ b&a}=XB_2,\\ -bC_2X &=-\pmatrix{0&0\\ b&0}\pmatrix{1&0\\ a&-b}\\ &=\pmatrix{0&0\\ -b&0} =\pmatrix{1&0\\ a&-b}\pmatrix{0&0\\ 1&0}=XC_2. \end{aligned} It follows that if $n=2m$, \begin{aligned} &\phantom{=}\pmatrix{ A_2&C_2\\ &A_2&C_2\\ &&\ddots&\ddots\\ &&&A_2&C_2 \\ &&&&A_2} \pmatrix{X\\ &-bX\\ &&b^2X\\ &&&\ddots\\ &&&&(-b)^{m-1}X}\\ &=\pmatrix{ A_2X&-bC_2X\\ &-bA_2X&b^2C_2X\\ &&\ddots&\ddots\\ &&&(-b)^{m-2}A_2X&(-b)^{m-1}C_2X\\ &&&&(-b)^{m-1}A_2X}\\ &=\pmatrix{ XB_2&XC_2\\ &-bXB_2&-bXC_2\\ &&\ddots&\ddots\\ &&&(-b)^{m-2}XB_2&(-b)^{m-2}XC_2\\ &&&&(-b)^{m-1}XB_2}\\ &=\pmatrix{X\\ &-bX\\ &&b^2X\\ &&&\ddots\\ &&&&(-b)^{m-1}X} \pmatrix{ B_2&C_2\\ &B_2&C_2\\ &&\ddots&\ddots\\ &&&B_2&C_2 \\ &&&&B_2}. \end{aligned} Hence you may take $P=\operatorname{diag}\left(X,\,-bX,\,b^2X,\,\ldots,\,(-b)^{m-1}X\right)$.

---

How $P$ was found. One natural way to solve the problem is to find $P_1$ and $P_2$ that make both $P_1^{-1}AP_1$ and $P_2^{-1}BP_2$ equal to their common Jordan form. Then one may set $P=P_1P_2^{-1}$. But I was too lazy to do that. So, I tried to play around with $A$ and $B$ to see if there were any shortcut.

The first thing I noticed was that if $AP=PB$, we must also have $f(A)P=Pf(B)$. If we take $f$ to be the common characteristic polynomial of $A_2$ and $B_2$, i.e. if we take $f(x)=x^2-2ax+(a^2+b^2)$, then $$ f(A)=J=\pmatrix{0&I_2\\ &\ddots&\ddots\\ &&\ddots&I_2\\ &&&0} $$ and $f(B)=-bJ$. The equation $f(A)P=Pf(B)$ thus becomes $JP=-bPJ$, which forces $P$ to take the folloiwng form: $$ \pmatrix{X&Y&Z&\cdots&\cdots&\vdots\\ &-bX&-bY&-bZ&\cdots&\vdots\\ &&b^2X&b^2Y&\ddots&\vdots\\ &&&\ddots&\ddots\\ &&&&(-b)^{m-2}X&(-b)^{m-2}Y\\ &&&&&(-b)^{m-1}X}. $$ The equation $AP=PB$ now reduces to a system of equations \begin{aligned} A_2X-XB_2&=0,\\ A_2Y-YB_2&=XC_2+bC_2X,\\ A_2Z-ZB_2&=YC_2+bC_2Y,\\ &\vdots \end{aligned} and the obvious thing to try is to set $Y=Z=\cdots=0$ and solve $A_2X-XB_2=0=XC_2+bC_2X$. Now the rest is easy.

Answer 2

You find $Q$ such that $Q^{-1}AQ=\Lambda$ is in Jordan form, and you find $R$ such that $R^{-1}BR=\Lambda$ (if there is no $R$, that is, if $A,B$ don't have the same Jordan form, then $P$ doesn't exist), and then you have $RQ^{-1}AQR^{-1}=B$, so $P=QR^{-1}$.