A function from a matrix to its characteristic polynomial. Maybe it's a known fact, but I would like to know whether the function sending a square matrix of order $n$ to its characteristic polynomial is continuous. If this is true, there are some resources where I can find this?
Thanks in advance.
 A: The characteristic polynomial is monic of degree $n$, so it has $n$ nontrivial coefficients. Therefore we can identify the image of $\phi$ with (a subspace of) $\mathbb{R}^n$; consider the map
$$\pi:\mathbb{R}[X]\longrightarrow\mathbb{R}^n:\sum_{i=0}^{n-1}c_iX^i\longmapsto(c_0,\ldots,c_{n-1}).$$
Now if we define $\phi(A)\in\Bbb{R}[X]$ to be the characteristic polynomial of $A$, for any $A\in M_n(\Bbb{R})$, then we get a map
$$\pi\circ\phi:\mathbb{R}^{n\times n}\longrightarrow\mathbb{R}^n:A\longmapsto(c_0(A),\ldots,c_{n-1}(A)),$$
where $c_i(A)$ is the $i$-th coefficient of the chararacteristic polynomial of $A$.
Each coefficent $c_i(A)$ is a polynomial in the coefficients of $A$, so as functions the $c_i$ are polynomials (in $n\times n$ variables!). In particular they are continuous maps from $\mathbb{R}^{n\times n}$ to $\mathbb{R}$. It then follows that $\phi$ is continuous, so the answer to your question is yes.
A: The function that takes you from an $n\times n$ matrix to its characteristic polynomial is simply
$$
\begin{align}
\phi:\mathbb R^{n\times n} &\to \mathbb R[x]\\
A &\mapsto \det(A-x I_n)
\end{align}
$$
Where $I_n$ is the $n\times n$ identity matrix.  As for whether this function is "continuous", well, how do you define continuity on such a map?  What topology do we impose on $\mathbb R[x]$?
