-1
$\begingroup$

How to prove set $S:GL_n(\mathbb{R}),$ in ${\mathbb{M}}_n(\mathbb{R}$) is dense. Any link or help would be greatly appreciated.

  • 0
    What does the notation $\;S:GL_n(\Bbb R)\;$ mean ? Is it $\;S\;$, is $\;GL_n(\Bbb R)\;$ .... ? What is $\;S\;$ What topology do you have on $\;\Bbb M_n(\Bbb R)\;$ here? – 2017-01-29
  • 0
    @DonAntonio the standard one from the operator norm on the set of matrices, or equivalently the topology we get when we identify $M(n, \mathbb{R})$ with $\mathbb{R}^{n^2}$ in the obvious way.. – 2017-01-29
  • 0
    @Henno :(1) How do you know that's what the OP meant? (2) Even assuming this, what is that $\;S:\;$ thing there...? The OP is irresponsive... – 2017-01-29

2 Answers 2

1

Let $A$ be any matrix not in $GL(n;\mathbb{R})$; then $\mathrm{det}(tI - A)$ is a nonzero polynomial and therefore has an isolated zero at $0$, so matrices of the form $tI - A$ in $GL(n;\mathbb{R})$ approximate $A$ arbitrarily well.

  • 0
    what is $t$ here? – 2017-01-29
  • 0
    $t$ being a scalar – 2017-01-29
  • 0
    how can u say that it has an isolated 0 at 0 – 2017-01-29
  • 0
    Polynomials have finitely many zeros – 2017-01-29
  • 0
    can u explain little more simply as I am not able to catch your arguments – 2017-01-29
  • 0
    @SubhashChandBhoria see my expanded answer – 2017-01-29
0

To expand a lot more on user399601's fine answer, which contains the idea:

We have to show that any open ball (in the norm metric used for matrices) around every matrix $A \in M(n;\mathbb{R})$ contains a point of $\operatorname{GL}(n;\mathbb{R})$. This is what being dense means.

Suppose we have a matrix in $M(n;\mathbb{R})$. We can assume that it is non-invertible, as otherwise $A \in \operatorname{GL}(n;\mathbb{R})$ already, and the condition above is trivially satisfied. So $\det(A) = 0$. I assume you know the basic fact that $A \in \operatorname{GL}(n;\mathbb{R}) \leftrightarrow \det(A) \neq 0$. Fix $A$ from now on.

define $p: \mathbb{R} \rightarrow \mathbb{R}$ by $p(t) = \det(A + tI)$, where $I$ is the idenity matrix (but any invertible matrix would do).

Note that in the norm metric $d(A, A + tI) = \left||A+tI - A\right|| = \left||tI\right|| = |t|$, so any open ball around $A$ with radius $r$ contains all matrices $A + tI$ for which $|t| < r$.

Now look at $p(t)$. This is a polynomial in $t$. This follows from the standard formula for determinants in terms of the coeficients of a matrix (see wikipedia, or any good textbook) :

$$\det(B) =\sum_{\sigma \in S_n} \operatorname{sgn}(\sigma)\prod_{i=1}^n b_{i,\sigma(i)}$$

where in our case the $b_{i,j} = a_{i,j}$ for $i \neq j$ and $b_{i,i} = a_{i,i} + t$, where all $a_{i,j}$ are constant for now (as we fixed $A$). Expanding the above formula we then get a sum of products of constant and linear terms, which are all polynomials in $t$, so $p(t)$ is a polynomial in $t$ of degree maximally $n$ (as we take $n$-fold products in this formula at most).

Now $p(0) = 0$ as $\det(0I + A) = \det(A) = 0$. Now the polynomial $p(t)$, like any polynomial of finite degree, only has finitely many zeroes $0 = t_0,\ldots, t_m \in \mathbb{R}$ say. So taking $r > 0$ small enough, the neighbourhood $(-r,r)$ of $0$ in the reals contains no other zeroes $t_i$ of $p(t)$.

This means that for any $t$ with $|t| < r$, $\det(A + tI) \neq 0$, so for those $t$, $A +tI \in \operatorname{GL}(n;\mathbb{R})$, and we already know that $B(A, r)$ as well. So any ball around $A$ contains matrices from $\operatorname{GL}(n;\mathbb{R})$, so the latter set is dense as claimed.