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Given a differential equation $\dot x = Ax$, $x \in \mathbb{R}^n$ we define its characteristic equation as $\chi(\lambda) = \det (\lambda I - A)$. Consider now the second order ODE $$ \ddot x + A x + B \dot x = 0, \;\;\; x \in \mathbb{R}^n $$ With substitution $u = x$, $v = \dot x$ we can rewrite this ODE as a system $$ \left\{\begin{array}{rcl} \dot u & = & v \\ \dot v & = & -Au - Bv. \end{array}\right. $$ This is an ODE with matrix $$ \begin{bmatrix} 0 & I \\ -A & -B \end{bmatrix} $$ and hence with characteristic equation $$ \chi(\lambda) = \det \begin{bmatrix} \lambda I & - I \\ A & \lambda I+B\end{bmatrix} . $$ I know that there is a representation of such determinant as a determinant of $n \times n$ matrix with $a_{ii} + \lambda^2$ on the diagonal. How to obtain such representation? How to find off-diagonal items explicitly?

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I thought your question was about ODEs but seems to be about matrices! – Mercy King Jun 23 '12 at 11:10
@Mercy you're right, I corrected tags. – Nimza Jun 23 '12 at 11:12
up vote 4 down vote accepted

We have for block matrices, with $A$ invertible, that $$\pmatrix{A&B\\ C&D}=\pmatrix{A&0\\ C&I}\cdot\pmatrix{I&A^{-1}B\\0&D-CA^{-1}B}.$$ In our case, when $\lambda\neq 0$, we get $$\chi(\lambda)=\lambda^n\det\left(\lambda I+B-A\frac 1{\lambda}I(-I)\right)=\lambda^n\det(\lambda I+B+\frac 1{\lambda} A)=\det(\lambda^2I+\lambda B+A).$$

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Thank you. I think that your formula works fine if $\det A = 0$. In this case we have to replace $A^{-1}$ by its pseudoinverse. Isn't it? – Nimza Jun 23 '12 at 11:28
In fact my notation is quite confusing, since the $A$ in the general formula is not the same as the $A$ in your problem. – Davide Giraudo Jun 23 '12 at 11:31
I know, it's just a general question. – Nimza Jun 23 '12 at 11:31
You meant in the general formula? I think yes, there should be more general equations. – Davide Giraudo Jun 23 '12 at 11:33

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