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It is said that only a linear system of differential equations with all real and complete eigenvalues can be decoupled by applying a spectral decomposition to its coefficient matrix.

Since the singular value decomposition is a generalization of the spectral decomposition, can the SVD be used to decouple linear systems of differential equations with some complex and/or defective eigenvalues?

For defective eigenvalues, what about a spectral decomposition where the leftmost matrix includes generalized eigenvectors?

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  • $\begingroup$ There is no clear way in which the $SVD$ can be used in this context. While it's clear how one could use the decomposition $A = SDS^{-1}$ to compute $e^{tA}$, it is not clear how one could do the same with the decomposition $A = U\Sigma V^T$. $\endgroup$ – Omnomnomnom Jan 19 at 19:20
  • $\begingroup$ As far as defective eigenvalues: yes, your idea works. In other words, it suffices to find a basis that puts the matrix in its Jordan normal form. $\endgroup$ – Omnomnomnom Jan 19 at 19:25
  • $\begingroup$ @Omnomnomnom Hmmm, I tried $A = \begin{bmatrix} -3 & 1 \\ -1 & -1 \end{bmatrix}$ and $A = \begin{bmatrix} -2 & 1 \\ -9 & -4 \end{bmatrix}$, arriving at $\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix}$ and $\begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ -3 & 1 \end{bmatrix}$, respectively, but neither decomposition multiplies back into the original matrix. $\endgroup$ – user10478 Jan 20 at 0:34
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    $\begingroup$ I misunderstood your point for the second. No, you will not be able to get a spectral decomposition; the matrix in the middle can't be diagonal. You can get the next best thing, which is a matrix in Jordan form. $\endgroup$ – Omnomnomnom Jan 20 at 0:37

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