Solve Weird Matrix Equation to Get Values for P

[maistral]

So while I was trying to browse some notes with regard to Laplace transforms in solving systems of ODE, this matrix came up:

I could easily just use RK4 and call it a day to nuke the systems of ODE, but this actually made me curious. Apparently one can transform the matrix DE into another form, then this appeared.

How do I get the values for P?

 

[S.G. Janssens]

I don't think such ##P## exists. The numerical matrices on the left and right have different eigenvalues.

 

[DrClaude]

I don't think such ##P## exists. The numerical matrices on the left and right have different eigenvalues.

It exists and is not unique.

I was able to obtain a possible ##P## using Mathematica. However, I do not know how to do it in a rigorous fashion.

 

[S.G. Janssens]

What do you find for the eigenvalues of the left and right numerical matrices? For the one on the left I get one real eigenvalue and a complex-conjugate pair, while for the matrix on the right I get ##\{-3,-2,-1\}##. So these two matrices cannot be similar.

(If such ##P## would exist, it would not be unique indeed, because every nonzero multiple of ##P## would also qualify.)

 

[DrClaude]

What do you find for the eigenvalues of the left and right numerical matrices? For the one on the left I get one real eigenvalue and a complex-conjugate pair, while for the matrix on the right I get ##\{-3,-2,-1\}##. So these two matrices cannot be similar.

I made a mistake and used +11 instead of -11 in the right matrix. My guess is that there is an error in the OP and that the two matrices differ only by the arrangement of the numerical values.

 

[StoneTemplePython]

assuming the stated typo, consider using the reflection matrix

##\mathbf J = \left[\begin{matrix}0 & 0 & 1\\0 & 1 & 0\\1 & 0 & 0\end{matrix}\right]##

which is a permutation matrix (and involutary)

note:
For the nice case of diagonalizable matrices with the same spectrum (and in particular the extra nice case where all eigenvalues are distinct), the general approach is
##\mathbf S^{-1} \mathbf A \mathbf S= \mathbf D ##
and
##\mathbf U^{-1} \mathbf B \mathbf U= \mathbf D ##

so
##\mathbf U^{-1} \mathbf B \mathbf U= \mathbf S^{-1} \mathbf A \mathbf S ##
## \mathbf B = \mathbf U\mathbf S^{-1} \mathbf A \mathbf S\mathbf U^{-1} ##
setting ##\mathbf P:= \mathbf S\mathbf U^{-1}##
gives
## \mathbf B = \mathbf P^{-1} \mathbf A \mathbf P ##

the fact that the eigenvalues are distinct proves that ##\mathbf P## is unique (up to rescaling)

but in this case I could just eyeball the problem as a graph isomorphism and come up with ##\mathbf J##