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In mathematical control theory a Hurwitz matrix or stability matrix $A$ for a asymtotically stable differential equation $ \dot{x} = A x $ has strictly negative real parts of eigenvalues $ \Re(\lambda_i) < 0 \;\forall \lambda_i(A)$. I am interested in leading principal submatrices $A_r$ of $A$.

Situation

In a control lecture it was proven that such submatrices are yet again Hurwitz. The proof was slightly convoluted and used the context of control. Hence I assume this or a similar statement can be made in a more general setting.

From reading other posts on here I found, that

  • for a Hermitian matrix A that is positive definite [$ \Re (x^T A x) > 0 $] there are results derived from the Fischer-Courant minimax principle: $ \lambda_k(A)\le\lambda_k(A_r)\le\lambda_{k+n-r}(A),\; 1\le k\le r.$ See there and there
    (I tried to find the critical point in the proof of the cited book by Horn and Johnson where symmetry is actually needed but did not find anything clearly stated.)
  • $\quad A$ positive definite $\implies$ $A$ has strictly positive eigenvalues
    but
    $\quad A$ has strictly positive eigenvalues $\not\implies$ $A$ positive definite
    See If eigenvalues are positive, is the matrix positive definite?
    (I suppose, the symmetry or Heritian property might be needed for that result)

Question

Can estimates be stated about the eigenvalues of the leading principal submatrices of Hurwitz matrices?

or at least

Can the mentioned proof be generalized to yield that leading principal submatrices of Hurwitz matrices are yet again Hurwitz?
$ \Re(\lambda_i) < 0 \;\forall \lambda_i(A) \quad \implies \quad \Re( \lambda_{max}(A_r) ) < 0 $ ?

BadAtLaTeX
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  • I think not. Suppose $a_0=1$, then $A$ is stable if and only if the determinant of every leading principle minor is positive. However, the determinant is also equal to the product of the eigenvalues. This means that if the size of the leading principle minor is odd, there should be at least one positive eigenvalue to ensure the product can be positive. As not every eigenvalue in this submatrix is negative, the leading principal submatrix is not hurwitz. – Petrus1904 May 19 '21 at 11:13
  • @Petrus1904 Your explanation sounds right, though I currently can not rephrase the question to make it more sensible. Moreover, user1551's example left me puzzled, as to why the theorem I referred to could be correct. Now I am also unsure whether to edit the question or accept the answer and state a new one. – BadAtLaTeX May 19 '21 at 15:25

1 Answers1

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No. Consider $$ A=\pmatrix{1&-1&-3\\ 1&1&2\\ 4&0&-3}. $$ Its three eigenvalues $-0.2561$ and $-0.3720\pm2.7697i$ have negative real parts, but both eigenvalues (namely, $1\pm i$) of its leading principal $2\times2$ submatrix have positive real parts.

user1551
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  • You are absolutely right. However, I wonder what general condition I missed that renders the theorem from the lecture correct. Your example should work as a stable system matrix. In the lecture, there was only additional information given by the system and positive Hankel singular values (also depend on the system; but they were positive for at least one test with this A). – BadAtLaTeX May 19 '21 at 15:17
  • Ok, I misunderstood a notation (The theorem holds for this $A$ as well, because $A_r$ was not simply a leading principal matrix). This question can stay unaltered though. – BadAtLaTeX May 19 '21 at 19:59