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.(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# I Diagonalising a system of differential equations

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