How to solve for X this matrix equation $X=XP,\text{ where } X \text{ is } (x_1, x_2, x_3), P \text{ is matrix 3x3?}$?

Question

$$\text{ How to solve for X this matrix equation: }X=XP,\text{ where } X \text{ is } (x_1, x_2, x_3), P \text{ is matrix 3x3?I need not zero solution } x \neq(0,0,0)\text{.I got X(P-I)=0, but what to do next?}$$$$\text{For example for example }P = \left[\begin{array}{l}0.25&0.5&0.25\\0&0.5&0.5\\0.33&0.33&0.34\end{array}\right] \text{or }P=\left[\begin{array}{l}0.3&0.1&0.6\\0.5&0.2&0.3\\0.7&0.1&0.2\end{array}\right]$$

Answer 1

I see you have modified the text of your question. This is very unfortunate, because now it is a different question !

What I can say is that, for a fixed $X$, the initial solution below remains unchanged.

If vector $X$ becomes variable as well, you add $2$ more degrees of freedom, making $6+2=8$ degrees of freedom (not $3$, because a degree of freedom is lost by the fact that a multiplicative factor can be passed from the vector to the matrix, more exactly because $(k X)M$ is undisguishable from $X(kM)$ for any constant $k \neq 0$.

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Initial solution : If we consider $X$ as fixed, and the 9 entries of matrix $P$ as unknowns, constraint $XP=X$ induces three (a priori) independent equations, meaning that this issue has $9-3=6$ dof ("degrees of freedom") i.e., $6$ independent variables.

Let $P_a$ be the 3D family of affine planes defined by equation $x x_1 + y x_2 + zx_3 = a.$

Matrices with the following structure (defined by column vectors $U_1,U_2,U_3$) :

$$P=[U_1,U_2,U_3] \ \ \text{where} \ U_1 \in P_{x_1}, U_2 \in P_{x_2}, U_3 \in P_{x_3}\tag{1}$$

is such that $XP=[XU_1,XU_2,XU_3]=[x_1,x_2,x_3]$ (dot products) i.e., line vector $X$, as desired.

Therefore $P$ (given by (1)) appears as the general solution ; indeed, we check that each column $U_1,U_2,U_3$ has 2 degrees of freedom, giving the forecasted $2+2+2=6$ dof.

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Taking parametric equation of planes, one can get a parametric representation of matrix $P$.

Let us show it on the particular case where $(x_1,x_2,x_3)=(1,2,4)$.

It is not difficult to find parametric equations of planes :

$$(P_1) \ \begin{cases}x&=&1-2a+0b\\ y&=&0-a-2b\\ z&=&0+a+b\end{cases}, \ \ (P_2) \ \begin{cases}x&=&0+2c+0d\\ y&=&1-c-2d\\ z&=&0+0c+d\end{cases}, \ \ (P_4) \ \begin{cases}x&=&0+0e+2f\\ y&=&0+2e+f\\ z&=&1-e-f\end{cases}$$

giving the parametric representation :

$$P=\begin{pmatrix}1-2a&2c&2f\\ -a-2b&1-c-2d&2e+f\\ a+b&d&1-e-f\end{pmatrix}$$

Please note that is $a=b=c=d=e=f=0$ we find identity matrix as a natural particular solution.

Answer 2

With the new text of your question :

Taking the transpose of your relationship, and writing $Q=P^T, Y=X^T$, you have

$$QY=1Y$$

Therefore, $Y$ is an eigenvector of $Q$ associated with eigenvalue $1$.

As a consequence, the characteristic equation of $Q$ is

$$(1-\lambda)(\lambda^2+a\lambda+b)=-\lambda^3+(1-a)\lambda^2+(a-b)\lambda-b=0$$

Therefore, using Cayley-Hamilton theorem, a necessary condition for matrix $Q$ is that, for some $a,b$ :

$$Q^3=(1-a)Q^2+(a-b)Q-bI$$