$P(Z)$ is matrix polynomials. Why is $s_n$ smooth in a neighbourhood of $Z$?

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 $Z$?

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.
• 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 $s_{n−1}(λ)$? – R.T MAN Mar 12 '16 at 2:24
• 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 $P(z)$. – Richard Zhang Mar 12 '16 at 3:58