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I have a real $ 4\times4$ matrix of the form $$ C = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ c_{31} & c_{32} & 0 & c_{34} \\ c_{41} & c_{42} & c_{43} & c_{44} \end{pmatrix} $$ The coefficients satisfy \begin{align*} c_{31}, c_{32}, c_{41} &\ge 0,\\ c_{42} &> 0,\\ c_{34}, c_{43}, c_{44} &\le 0, \end{align*} and $\det(C) = c_{31} c_{42} - c_{32} c_{41} > 0.$

I want to show that the matrix has four distinct, real eigenvalues. My approaches so far:

  • Calculate the characteristic polynomial and use Mathematica to find the roots. The resulting formulas are too large to work with.
  • Show that the discriminant of the characteristic polynomial is positive. This works a little better, but again, the formulas are huge and ugly.

I would like to exploit the block structure of $A$, but I do not know how.

Edit: The coefficients I am actually working with are \begin{align*} &c_{31} = \frac{\rho(\rho p_0 + 2 \lambda_0)}{p_0}, ~~~~~ c_{32} = \frac{\rho \lambda}{p_0}, ~~~~~ c_{34} = - \frac{\lambda}{p_0},\\ &c_{41} = \frac{\rho \lambda_0}{p}, ~~~~~ c_{42} = \frac{\rho (\rho p + \lambda \frac{n+1}{n})}{p}, ~~~~~ c_{43} = - \frac{\lambda_0}{p}, ~~~~~ c_{44} = - \frac{\lambda \frac{n-1}{n}}{p} \end{align*} with $n \in \mathbb{N} \setminus\{0\}, ~~~\rho, p, p_0 > 0, ~~~\lambda, \lambda_0 \ge 0$ and $\max\{\lambda, \lambda_0\} > 0.$

Actually, one of the eigenvalues is $-\rho,$ so there are only three remaining (and none of these three is equal to $-\rho$). The three remaining eigenvalues are then the roots of a cubic equation. (This implies that at least one other root is real). The discriminant of the cubic equation, however, is still very complicated.

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    $\begingroup$ Use en.wikipedia.org/wiki/Determinant#Block_matrices for $\det(C-z)$, perhaps? $\endgroup$ – Pierpaolo Vivo Feb 9 '16 at 13:23
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    $\begingroup$ The eigenvalues are not necessarily distinct. For example, take $c_{31}=c_{42}=1$ and set all other entries to $0$. $\endgroup$ – Omnomnomnom Feb 9 '16 at 13:23
  • $\begingroup$ Thanks, you are right! I added the actual coefficients that I am working with. $\endgroup$ – Elias Strehle Feb 9 '16 at 14:09
  • $\begingroup$ It may be useful that the blocks in your matrix are all symmetric, so the eigenvalues of any block are real. Then again, I can't immediately see where to go with that information. $\endgroup$ – Omnomnomnom Feb 9 '16 at 14:26
  • $\begingroup$ Oh, sorry, are the $\lambda$ and $\lambda_0$ distinct quantities, or was that a typo? $\endgroup$ – Omnomnomnom Feb 9 '16 at 14:27
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First attempt:

$C$ has the block structure $$ C = \pmatrix{0&I\\A&B} $$ In particular, following the discussion here, we note that since the top two block commute, we have $$ \det(C - \lambda I) = \det \pmatrix{-\lambda I&I\\A&(B - \lambda I)} = \\ \det[(-\lambda I)(B - \lambda I) - (I)(A)]=\\ \det[\lambda^2 I - \lambda B - A] $$


Second attempt: Perhaps we can do a bit better with Newton's identities. In particular, $$ \operatorname{trace}(C) = c_{44}\\ \operatorname{trace}(C^2) = \operatorname{trace}\pmatrix{A&B\\AB&A+B^2} = 2(c_{31}+c_{42} + c_{34}c_{43}) + c_{44}^2 $$ Note that $\operatorname{trace}(C^n) = \sum_{i=1}^4 \lambda_i^n$.

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  • $\begingroup$ Isn't Sylvester's theorem about determinant of a blocks matrix true whenever all the blocks matrices commute with each other? I checked this possibility but the lower left 2x2 matrix doesn't necessarily commute with the right lower one. $\endgroup$ – DonAntonio Feb 9 '16 at 13:51
  • $\begingroup$ @Joanpemo you only need two of them to commute $\endgroup$ – Omnomnomnom Feb 9 '16 at 13:52
  • $\begingroup$ Thank you @Omnomnomnom . Do you have some reference to this result, please? $\endgroup$ – DonAntonio Feb 9 '16 at 13:53
  • $\begingroup$ @Joanpemo see the link in my answer $\endgroup$ – Omnomnomnom Feb 9 '16 at 13:54

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