Space of matrices with non-zero determinant

[brunob]

Hi there!
How can I prove that the space of matrices (2x2) nonzero determinant is dense in the space of matrices (2x2) ?

I've already proved that it's an open set.
Thanks.

PD: Sorry about the mistake in the title.

 

[economicsnerd]

The space you named isn't dense or open. In fact, it's closed with empty interior, i.e. its complement is open and dense.

[For any people reading, the named space has since changed. Now it is open and dense.]

 

[brunob]

Sorry, you're right it's not the matrices with determinant = 0, I should say nonzero determinant.

Thanks!

 

[economicsnerd]

There are lots of different ways you could show it. One would be to consider, for any matrix $M$ and small nonzero number $\epsilon$ the nearby matrix $M_\epsilon = M + \epsilon I$. What is the determinant of $M_\epsilon$? Can you show it's nonzero for small but nonzero $\epsilon$?

 

[brunob]

The determinant of M_\epsilon is always nonzero. Unless M = \begin{pmatrix} -\epsilon & 0\\ 0 & -\epsilon \end{pmatrix}.
Is \epsilon fixed or just a number decreasing to zero?

 

[Fredrik]

The determinant of M_\epsilon is always nonzero. Unless M = \begin{pmatrix} -\epsilon & 0\\ 0 & -\epsilon \end{pmatrix}.
Is \epsilon fixed or just a number decreasing to zero?

I'm pretty sure numbers never decrease. But $\epsilon$ is an arbitrary positive real number here, so it could certainly be an arbitrary term in a decreasing sequence that converges to 0.

Since $\epsilon$ is a number you choose, the possibility that M is that specific matrix is not a concern. (If M is that matrix, just change your choice of $\epsilon$).

What you need to show is that for all 2×2 matrices M and all $\varepsilon>0$, there's a 2×2 matrix M' with non-zero determinant such that $d(M',M)<\varepsilon$. I guess the idea here is that if you choose a small enough $t>0$ and define $M'=M+tI$, then we will have $d(M',M)<\varepsilon$. (I haven't tried it, and I'm going to bed now).

(economicsnerd's $\epsilon$ is my t).

 

[economicsnerd]

Fix your matrix $M = \begin{pmatrix} a & b\\ c & d \end{pmatrix}$, and notice that $M+\epsilon I \longrightarrow M$ as $\epsilon \longrightarrow 0$.

If we can show that the determinant of $M+\epsilon I$ is nonzero for small enough nonzero $\epsilon$, then we know that matrices with nonzero determinant are dense.* So let's show it.

The determinant of $M+\epsilon I$ is just $(a+\epsilon)(d+\epsilon) - bc$, which equals (ad-bc) + (a+d)\epsilon + \epsilon^2

-Exercise: If $ad-bc\neq 0$, then $(ad-bc) + (a+d)\epsilon + \epsilon^2\neq 0$ for small enough nonzero $\epsilon$. (Though you don't really need to do this one, since $ad-bc\neq 0$ means $M$ already has nonzero determinant.)
-Exercise: If $ad-bc=0$ but $a+d\neq 0$, then $(ad-bc) + (a+d)\epsilon + \epsilon^2\neq 0$ for small enough nonzero $\epsilon$.
-Exercise: If $ad-bc= a+d = 0$, then $(ad-bc) + (a+d)\epsilon + \epsilon^2\neq 0$ for small enough nonzero $\epsilon$. (This one is basically immediate.)

*[Indeed, for some $n\in\mathbb N$, we would know $M+\frac1k I$ has nonzero determinant for every $k\geq n$. Then $(M+\frac1k I)_{k=n}^\infty$ is a sequence of matrices with nonzero determinant, converging to $M$.]

 

[brunob]

Great, got it! I'll write it.

Thank you so much!