Convergence of determinant of a sequence of matrices

Question

Let $A_k \in M(n, \mathbb{R})$ converge to $A \in M(n, \mathbb{R})$. Where $\|A\|:= \left(\text{Tr}(A^{\mathrm t}A)\right)^{\frac{1}{2}}$. Show that $\det(A_k) \rightarrow \det(A)$ and $A_k^2\rightarrow A^2$.

Let $A_k:=(a^{(k)}_{i,j})$ and $A:=(a_{i,j})$. A sequence $(A_k)$ in $M(n, \mathbb{R})$ converges to $A \in M(n, \mathbb{R})$ iff the matrix entries at $(a^{(k)}_{i,j}) \rightarrow a_{i,j}$ for all $i,j$ as $k\to\infty$. We know that the det is a function from $\mathbb{R}^{n^2} \rightarrow \mathbb{R}$ . Since $\det(A^k)$ is a function of all $a^{(k)}_{i,j}$, and as $a^{(k)}_{i,j} \rightarrow a_{i,j}$ for all $i,j$, it can be seen that $\det(A_k) \rightarrow \det(A)$. But I don't know how to write a formal $\epsilon$-$\delta$ proof of this problem. For the second part of the problem, I don't know how to start. Any hints?

Answer 1

A little work shows that the (Frobenius) norm above is $\|A\| = \sqrt{\sum_{i,j} [A]_{ij}^2}$, so $A_k \to A$ iff the $[A_k]_{ij} \to [A]_{ij}$.

The easy proof is to invoke continuity.

The functions $\det$ and $A \mapsto A^2$ are continuous functions, so if $A_k \to A$ then $\det A_k \to \det A$ and $A_k^2 \to A^2$.

Addendum:

We have $\det A = \sum_\sigma \operatorname{sgn} \sigma [A]_{1 \sigma_1} \cdots [A]_{n \sigma_n}$. Each map $A \mapsto [A]_{ij}$ is continuous hence so is the product and hence so is the above sum.

Since $[A^2]_{ij} = \sum_k [A]_{ik}[A]_{kj}$, we similarly see that the map $A \mapsto [A^2]_{ij}$ is continuous and hence so is $A \mapsto A^2$.

Answer 2

The other approach, which I wouldn't necessarily recommend, is:

do the second part of your problem first and prove $A_k^2 \to A^2$ and extend it to $A_k^r \to A^r$ for positive integer $r$. The trace (summing diagonal elements) is somewhat trivially a continuous function of matrix entries. This then implies $\text{characteristic polynomial}(A_k) \rightarrow \text{characteristic polynomial}(A)$ by Newton's Identities.