How to prove that the map $f: M_n(\Bbb R) \to M_n(\Bbb R)$ defined by $A \mapsto AA^t$ is continuous?

Question

This map has appeared several times on this site such as here, here and here. I thought of writing out a proof for the continuity of this map. Identify $M_n(\Bbb R)$ with $\Bbb R^{n^2}$.

I tried using sequential way to prove the continuity here.

So let $X_n$ be a converging sequence in $M_n(\Bbb R)$ and $||\cdot||$ be max-norm norm on $M_n(\Bbb R)$ i.e. $||A||=\max {(a_{ij})}.$ Then $\forall \; \delta \gt 0 \; \exists N \in \Bbb N$ such that $||X_n - X|| \lt \delta \; \forall \; n \ge N.$ This also means that $\forall \; \delta \gt 0 \; \exists N \in \Bbb N$ such that $||X_n^t - X^t|| \lt \delta \; \forall \; n \ge N.$

Consider $||f(X_n)-f(X)||=||X_n X_n^t-XX^t||=||X_n X_n^t - XX^t + X_nX^t - X_n X^t|| \le ||X_n X_n^t - X_n X^t||+ ||X_nX^t-XX^t||= ||X_n(X_n^t-X^t)||+||X^t(X_n-X)||.$

But I am stuck at the last step. What can I do from here?

Also I am intereseted in knowing other ways to show this.

EDIT : I have found that as per the norm that I have used, $||AB|| \not \le ||A||||B||.$ Take $$A = \left[ \begin{matrix} 1 & 2 \\ 3 & 4 \\ \end{matrix}\right]$$ and $$B = \left[ \begin{matrix} 4 & 5 \\ 6 & 7 \\ \end{matrix}\right].$$ So my way will not work.

This is because I was thinking whether $||X_n(X_n^t-X^t)|| \le ||X_n||||X_n^t-X^t||$ could be done. But my purpose is defeated.

Answer 1

Unfortunately your norm $||A||=\max {|a_{ij}|}$ is not submultiplikative. Since all norms on $M_n(\Bbb R)$ are equivalent, we can choose a norm which is submultiplikative. Let $ ||⋅|| $ such a norm.

Then we have

$||X_n(X_n^t-X^t)||+||X^t(X_n-X)|| \le ||X_n|| \cdot||X_n^t-X^t||+||X^t|| \cdot ||X_n-X||$.

Can you take it from here ? Observe that $(||X_n||)$ is bounded.

Answer 2

Here's a tricky solution. Consider a function

$$F:\mathbb{M}_n(\mathbb{R})\otimes \mathbb{M}_n(\mathbb{R})\to \mathbb{M}_n(\mathbb{R})$$ $$F(A\otimes B)=AB^{t}$$

where $\otimes$ stands for the tensor product of vector spaces. This function is linear, hence continuous.

Now consider

$$G:\mathbb{M}_n(\mathbb{R})\to\mathbb{M}_n(\mathbb{R})\otimes \mathbb{M}_n(\mathbb{R})$$ $$G(A)=A\otimes A$$

This function although not linear is also continuous:

Proof. Fix a basis $\{e_1,\ldots, e_m\}$ for $\mathbb{M}_n(\mathbb{R})$. It follows that

$$G(v)=G\big(\sum\lambda_i e_i\big)=\sum\lambda_i e_i\otimes \sum\lambda_i e_i=\sum_{i, j}\lambda_i\lambda_j (e_i\otimes e_j)$$

Define $\pi_{i,j}:\mathbb{M}_n(\mathbb{R})\otimes \mathbb{M}_n(\mathbb{R})\to\mathbb{R}$ to be $(i,j)$ projection, i.e. $\pi_{i,j}(\lambda e_i\otimes e_j)=\lambda$ and $0$ elsewehere. It follows that

$$\pi_{i,j}\circ G:\mathbb{M}_n(\mathbb{R})\to\mathbb{R}$$ $$\pi_{i,j}\circ G\big(\sum\lambda_k e_k\big)=\lambda_i\lambda_j$$ In other words $\pi_{i,j}\circ G$ is just a multiplication of two chosen coordinates. This is continuous because the multiplication is. Hence by the definition of product topology $G$ is continuous. $\Box$

Now it is easy to see that $f=F\circ G$ is continuous.