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 $ ?
Best Answer
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.