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When I was solving linear differential equation systems, I encountered a big matrix as coefficient. The matrix is $$\begin{pmatrix} 1 & 2 & -3 & 0 & 1\\ -5 & 1 & -4 & 0 & 0\\ 0 & -2 & 4 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}$$ Finding the eigenvalues and eigenvectors step by step based on the definition just kills me. Is there any trick or strategy of solving problems like this?
Moverever, after I find the eigenvectors, and construct a matrix $\textbf{P}$ of eigenvectors, its inverse $\textbf{P}^{-1}$ is also difficult to compute. Any trick here?
After working for several minutes, I ended up with the diagonal eigenvalue matrix $$\textbf{D}=\begin{pmatrix} 0&0&0&0&0\\ 0&2&0&0&0\\ 0&0&3&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ \end{pmatrix}$$ and the corresponding eigenvector matrix is $$\textbf{P}=\begin{pmatrix} 2&-1&-2&-8&0\\ -10&1&1&15&0\\ -5&1&2&10&0\\ 0&0&0&0&1\\ 3&0&0&0&0 \end{pmatrix}$$ What if now I need to calculate $\textbf{x}=\textbf{P}e^{\textbf{D}t}\textbf{P}^{-1}\textbf{x}_0$, where $\textbf{x}_0$ is any constant vector.

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  • $\begingroup$ Trick: there's a row of $0$s. This tells us the matrix is not invertible and hence $0$ is an eigenvalue! $\endgroup$ – Theo Bendit Dec 2 '17 at 7:05
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This will just use determinant of a $(3 \times 3)$ matrix as,

$det\begin{pmatrix} 1-t & 2 & -3 & 0 & 1\\ -5 & 1-t & -4 & 0 & 0\\ 0 & -2 & 4-t & 0 & 0\\ 0 & 0 & 0 & 1-t & 0\\ 0 & 0 & 0 & 0 & 0-t \end{pmatrix}$ = $(-t)det\begin{pmatrix} 1-t & 2 & -3 & 0\\ -5 & 1-t & -4 & 0\\ 0 & -2 & 4-t & 0\\ 0 & 0 & 0 & 1-t \\ \end{pmatrix}$ = $(-t)(-(1-t))det\begin{pmatrix} 1-t & 2 & -3\\ -5 & 1-t & -4\\ 0 & -2 & 4-t\\ \end{pmatrix}$

Then just solve as you do for a $(3\times 3)$ matrix.

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  • $\begingroup$ I actually did what you said, and I edit my question. Can you check again? Thanks. $\endgroup$ – R.C Dec 2 '17 at 7:43

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