Prove that invertible metrices set is an open set in a given space, and the determinant is continuous

Question

Given a matrix $M_{n\times m}$, we can think about it as a vector in $\mathbb{R}^{n\times m}$ (How come?).

How can I prove that the set of all the invertible metrices of size $n\times n$ is an open set in $\mathbb{R}^{n^2}$, and that the function $f:M\mapsto M^{-1}$ is continuous?

I understand that $f$ is the determinant of $M$, and $f = \det(M)\ne 0$, and becuase $f$ is a linear function, it's a continuous function. But I'm not 100% sure why $f$ is a linear.

The invertible metrices set is $\mathbb{R}^{n^2}\setminus\{0\}$ (am I wrong?), and that's why the set is an open set?

Thank you!

Answer 1

An $m\times n$ matrix (with real entries) is an array of $m\times n$ real numbers so we can associate this with $\mathbb{R}^{mn}$. The function $f$ (the determinant) is linear in each column of the matrix. So interpretting as $f:\mathbb{R}^{n^{2}}\to\mathbb{R}$ as a function on $\mathbb{R}^{n}\times\mathbb{R}^{n}\times...\times\mathbb{R}^{n}$ (where the product is taken $n$ times since the matrix is $n\times n$ and hence has $n$ columns and each column has $n$ entries) we get a multilinear function in each variable.

The invertible matrices are matrices with determinant in $\mathbb{R}-\{0\}$ (not $\mathbb{R}^{n{2}}-\{0\})$ and since $f$ is continuous the preimage of this open set is open. The set $\mathbb{R}^{n^{2}}-\{0\}$ would be equivalent to the set of $n\times n$ matrices excluding the $0$ matrix.

To show that the function that sends a matrix to its inverse is continuous note that $A^{-1}=\frac{1}{\det(A)}adj(A)$ which gives a matrix whose entries are rational functions on $\mathbb{R}$ (note that adj(A) denotes the adjugate matrix). If we restrict to $A$ such that $\det(A)\neq0$ then the denominator of the rational functions is not $0$.

To see that the determinant is linear in the first column I calculate the determinant by means of the cofactor expansion along column $1$ (a similar proof will work for other columns) suppose we have:

$\det\begin{pmatrix} (a_{1,1}+cb_{1})&a_{1,2}&...&a_{1,n}\\ (a_{2,1}+cb_{2})&a_{2,2}&...&a_{2,n}\\ ...&...&...&...\\ (a_{n-1,1}+cb_{n-1})&a_{n-1,2}&...&a_{n-1,n}\\ (a_{n,1}+cb)&a_{n,2}&...&a_{n,n}\\ \end{pmatrix}$

$=\sum_{i=1}^{n}(a_{i,1}+cb_{i})C_{i,1}=\sum_{i=1}^{n}a_{i,1}C_{i,1}+c\sum_{i=1}^{n}b_{i}C_{i,1}$

$=\det \begin{pmatrix} a_{1,1}&a_{1,2}&...&a_{1,n}\\ a_{2,1}&a_{2,2}&...&a_{2,n}\\ ...&...&...&...&...\\ a_{n-1,1}&a_{n-1,2}&...&a_{n-1,n}\\ a_{n,1}&a_{n,2}&...&a_{n,n}\\ \end{pmatrix}+c\det \begin{pmatrix} b_{1}&a_{1,2}&...&a_{1,n}\\ b_{2}&a_{2,2}&...&a_{2,n}\\ ...&...&...&...&...\\ b_{n-1}&a_{n-1,2}&...&a_{n,n-1}\\ n_{n}&a_{n,2}&...&a_{n,n}\\ \end{pmatrix}$

To see why the determinant maps $\mathbb{R}^{n^{2}}$ to $\mathbb{R}$ just remember that the determinant gives a number and its domain is the space of square matrices which we associate to $\mathbb{R}^{n^{2}}$.

Answer 2

Consider $M_{n\times m}$. Here each entry in a matrix is independently chosen, and we hav $mn$ such entries altogether. What I am saying in other words is that $M_{n\times m}$ is having dimension $mn$ as a vector space! So it is infact same as $\mathbb{R}^{n\times m}$. Or, you will agree to see any particular row in a matrix as a point in $\mathbb{R}^{m}$. Then why dont't you consider a whole matrix as $n$ such $m$ tuples kept one after the other in a row..! I think that solves the issue. Now the determinant map is simply a polynomial where each entry in the matrix is considered as a variable(Any matrix $A= (x_{ij}) $and $x_{ij}$ are our variables. Note that A is the $n^2$ tuple $(x_{11},x_{12},..,x_{1m},x_{21},x_{22},..,x_{nn})$) Now as this polynomial map is continuous the set of all non-singular matrices being the preimage of the open set $\mathbb{R}- \{0\}$ is open. This is a huge subset of $M_{n\times n}$, it's open & dense! $0$ in $\mathbb{R}^{n^2}$ represents the zero matrix having zero entries everywhere. So, $\mathbb{R}^{n^2}$- $0$ is obviously not our required set as u can see.