is $GL(n,\mathbb R)$ dense in $M(n,\mathbb R)$

Question

Is $GL(n,\mathbb R)$ dense in $M(n,\mathbb R)$? I have proved it to be open,not closed,not connected but not sure about this property .How to do this?

score 17 · Answer 1 · answered Nov 12 '14 at 07:52

17

Yes, $GL(n,\mathbb R)$ is dense in $M(n,\mathbb R)$.

Using the determinant function $\det:M(n,\mathbb R)\rightarrow \mathbb R$, check that for small values of $\varepsilon$, the matrix $A+\varepsilon I_n$ is invertible.

answered Nov 12 '14 at 07:52

Bebop

3,622

Is $A+nB$ invertible for any n if A and B invertible? – Learnmore Nov 12 '14 at 07:55
1

$n$ is quite a big number.... – Nov 12 '14 at 07:59
3

No, not in general. It is true for only finitely many values of $n$. Notice that $\varepsilon \mapsto \det(A+\varepsilon I_n)$ is actually a polynomial function. – Bebop Nov 12 '14 at 07:59
How to check that $A+\epsilon B$ is invertible .Any hints@Bebop – Learnmore Nov 12 '14 at 15:37
2

The function $\varepsilon \mapsto \det(A+\varepsilon B)$ vanishes at most $n$ times since it is a real polynomial function. Hence, there exists $N\in\mathbb N$ such that for any $p\geq N$, $\det(A+\frac{1}{p}B)\neq 0$ i.e. $A+\frac{1}{p}B$ is invertible. – Bebop Nov 12 '14 at 17:54

score 16 · Accepted Answer · answered Nov 12 '14 at 07:53

16

Hint: A matrix has finitely many eigenvalues. If none of them is $0$, ...

answered Nov 12 '14 at 07:53

Robert Israel

448,999

Sorry I did not get it – Learnmore Nov 12 '14 at 15:30
Add a small multiple of $I$ to your matrix, and it becomes invertible. – Robert Israel Nov 12 '14 at 15:50
How to prove that – Learnmore Nov 12 '14 at 15:59
Eigenvalues of $A + \epsilon I$ are (eigenvalues of $A$) $+ \epsilon$. – Robert Israel Nov 12 '14 at 17:41
@RobertIsrael where did you use "none of them are zero"(eigen values)? what if something is zero – Cloud JR K Sep 02 '18 at 15:57
@CloudJR If an eigenvalue is $0$, the matrix is not in $GL(n,\mathbb R)$. – Robert Israel Sep 02 '18 at 17:06
@RobertIsrael what is the metric used in this problem? I break my head for an hour please help! – Cloud JR K Sep 02 '18 at 17:09
@CloudJR Use the Euclidean metric on $M(n,\mathbb R)$, identified as $\mathbb R^{n^2}$. – Robert Israel Sep 02 '18 at 17:30
@RobertIsrael then why people are talking about determinant? To check it is invertible? – Cloud JR K Sep 02 '18 at 17:32

score 7 · Answer 3 · answered Nov 12 '14 at 08:08

Since this is posted under the tag metric spaces, I will show that the statement is not true for every metric:

Let $d$ be the discrete metric: $$ d(A,B) = \begin{cases} 0 & \text{ if } A=B\\ 1 & \text{ if } A\ne B.\end{cases} $$ Then $$ d(0,B) =1 $$ for all invertible $B$, hence $GL(n,\mathbb R)$ is not dense in $M(n,\mathbb R)$.

If the metric is induced by a norm, $d(A,B)=\|A-B\|$, then the statement is true. Let $A$ be not invertible. Then there are invertible matrices $S,T$ such that $$ SAT = \pmatrix{ I_r & 0 \\ 0 & 0 }, $$ with $r=rank(A)$. Now let $\epsilon\ne0$ be given. Set $$ A_\epsilon := S^{-1} \pmatrix{ (1+\epsilon)I_r & 0 \\ 0 & \epsilon I_{n-r} } T^{-1} = A + \epsilon S^{-1}T^{-1}. $$ This matrix is clearly invertible as a product of invertible matrices $S^{-1}$, $\pmatrix{ (1+\epsilon)I_r & 0 \\ 0 & \epsilon I_{n-r} }$, $T^{-1}$. Then $$ A-A_\epsilon = S^{-1}\pmatrix{ -\epsilon I_r & 0 \\ 0 & -\epsilon I_{n-r}} T^{-1} = - \epsilon S^{-1}T^{-1}, $$ which shows $$ \|A-A_\epsilon\| = |\epsilon|\cdot \|S^{-1}T^{-1}\|. $$ And $GL(n,\mathbb R)$ is dense in $M(n,\mathbb R)$.

score 2 · Answer 4 · edited Mar 02 '21 at 04:23

The way I like writing this is as follows: Consider $A\in M_{n}(\mathbb R)$ and let $\varepsilon>0$. Let $\lambda _1, \ldots, \lambda_n$ be the eigenvalues of $A$. Take $\delta$ such that $0<\delta<\frac{\varepsilon}{n^{1/2}}$ and $\delta\neq \lambda_j$ for every $j\in \{1, \ldots, n\}$. Define

$$A_\delta:=A-\delta I.$$ Then $A_\delta$ is invertible. In fact, on the contrary,

$$\det(A_\delta)=\det(A-\delta I)=0$$

and therefore $\delta$ would be an eigenvalue of $A$, so that $\delta =\lambda_j$ for some $j\in \{1, \ldots, n\}$. This contradicts the choice of $\delta$. Finally,

$$d_2(A, A_\delta)=\|A-A_\delta\|_2=\|A-(A-\delta I)\|_2=|\delta|\|I\|_2=\delta {n^{1/2}}<\frac{\varepsilon}{n^{1/2}} n^{1/2}=\varepsilon.$$

I assumed here that your metric is induced by norm:

$$\|A\|_2=\left(\sum_{i=1}^n \sum_{j=1}^n a_{ij}^2\right)^{1/2}.$$

score 1 · Answer 5 · edited Jun 01 '17 at 20:34

Here GLn(R) is subspace as a metric space in set of all metrices of order n. If I take any member in Mn(R): if it is in GLn(R) then clearly we get sequence of GLn(R) that converge to our chosen point. Now if it is not in GLn(R) then by changing exactly one element of the matrix I can make it a member of the set of all invertible matrices, so by this way I can get a sequence of GLn(R) which converges to set of our chosen point.

is $GL(n,\mathbb R)$ dense in $M(n,\mathbb R)$

5 Answers5