0
$\begingroup$

Let $A_j \in \mathbb{C}^{n \times n}$, ($j = 0,1,2,\ldots,m$) and

$P(Z) = A_m Z^m + \cdots + A_1 Z + A_0$ is a matrix polynomial, and $Z $ is a complex variable.

$Z$ is eigenvalue of $P(Z )$ if $\det P(Z ) = 0$ and

$s_1 \ge s_2 \ge \cdots \ge s_n$ are singular values of $P(Z)$.

Let $s_n$ have multiplicity one.

Why is $s_n$ smooth in a neighbourhood of $Z$?

$\endgroup$
4
$\begingroup$

The smallest singular value $s_n$, which has been stated to have multiplicity one, is a simple eigenvalue of the matrix $M(z) = P(z)P(z)^*$, or equivalently, a simple root to the characteristic polynomial $\chi(\lambda;z) = \det (\lambda I - M(z))$.

As defined, each coefficient in the matrix $M(z)$ is polynomial with respect to $z$, so $\chi(\lambda;z)$ is a polynomial in $\lambda$ whose coefficients are polynomials of $z$. $$\chi(\lambda;z) = c_0(z) + c_1(z)\lambda + \cdots + c_n(z)\lambda^n$$

It is a basic result from analysis that the simple roots of a polynomial are smooth functions with respect to the coefficients of the polynomial. And in this case, the coefficients of $\chi(\lambda;z)$ are themselves polynomials of $z$, and hence also smooth. Therefore, the simple roots of $\chi(\lambda;z)$ are smooth with respect to $z$. QED.

$\endgroup$
  • $\begingroup$ Can we say that, the study of points where differentiability is lost, is confined to the points of the plane where, $s_n(λ)$ meets the one corresponding to $s_{n−1}(λ)$? $\endgroup$ – R.T MAN Mar 12 '16 at 2:24
  • 1
    $\begingroup$ Yes, assuming you mean $s_n(z)$ and $s_{n-1}(z)$. And that statement even generalizes to the eigenvalues of the possibly nonsymmetric matrix $P(z)$. $\endgroup$ – Richard Zhang Mar 12 '16 at 3:58
  • $\begingroup$ - Which reference say that, " the simple roots of a polynomial are smooth functions with respect to the coefficients of the polynomial."? Please, send me a god reference for this text. $\endgroup$ – R.T MAN Mar 20 '16 at 22:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.