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 $Z$?



$P(Z)$ is matrix polynomials. 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.


