- #1
Frank Castle
- 580
- 23
Given a system of linear differential equations $$x_{1}'=a_{11}x_{1}+a_{12}x_{2}+\cdots a_{1n}x_{n}\\ x_{2}'=a_{21}x_{1}+a_{22}x_{2}+\cdots a_{2n}x_{n}\\ \ldots\\ x_{n}'= a_{n1}x_{1}+a_{n2}x_{2}+\cdots a_{nn}x_{n}$$ this can be rewritten in the form of a matrix equation $$\mathbf{X}'=A\mathbf{X}$$ Assuming the matrix ##A## is diagonalisable, such that ##A=SDS^{-1}##, where ##S## is a (constant) matrix formed from the eigenvectors of ##A## and ##D## is a diagonal matrix whose diagonal elements are the eigenvalues of ##A##, we can recast the differential matrix equation as $$\mathbf{X}'=( SDS^{-1})\mathbf{X}$$ and defining ##\mathbf{Y}=S^{-1}\mathbf{X}##, then $$(S^{-1}\mathbf{X})=D(S^{-1}\mathbf{X})\Rightarrow\mathbf{Y}'=D\mathbf{Y}$$and so we have mapped the complicated system of differential equations into a set of (equivalent) diagonalised differential equations.
My question is, what is the motivation for doing this? Is it simply that in doing so we are able to solve for a much simpler system of differential equations (we reduce the system from one containing ##n\times n## parameters to one containing ##n##) which we can then map back to the original set via a similarity transformation (as defined above), or are there other reasons for diagonalising the system?
My question is, what is the motivation for doing this? Is it simply that in doing so we are able to solve for a much simpler system of differential equations (we reduce the system from one containing ##n\times n## parameters to one containing ##n##) which we can then map back to the original set via a similarity transformation (as defined above), or are there other reasons for diagonalising the system?