# Why is a singular matrix rare?

I am exploring patterns of integers in $n\times n$ matrices. I have two matrices that have a determinant of $0$ and a circulant matrix that has positive determinants that differ depending on $n$.

I snipped this from Wikipedia and bolded the important part:

A square matrix that is not invertible is called singular or degenerate. A square matrix is singular if and only if its determinant is 0. Singular matrices are rare in the sense that if you pick a random square matrix over a continuous uniform distribution on its entries, it will almost surely not be singular.

The Good: $$\left( \begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 3 & 4 & 5 \\ \end{array} \right)$$ The Bad: $$\left( \begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 5 & 1 & 2 & 3 & 4 \\ 4 & 5 & 1 & 2 & 3 \\ 3 & 4 & 5 & 1 & 2 \\ 2 & 3 & 4 & 5 & 1 \\ \end{array} \right)$$ And the Ugly: $$\left( \begin{array}{ccccc} 11 & 12 & 13 & 14 & 15 \\ 16 & 17 & 18 & 19 & 20 \\ 21 & 22 & 23 & 24 & 25 \\ 26 & 27 & 28 & 29 & 30 \\ 31 & 32 & 33 & 34 & 35 \\ \end{array} \right)$$

The Good is same as Ugly mod n and both are singular. The Bad is the circulant Good and has determinant $>0$.

Two questions:

1)What makes a singular matrix rare?
2)Has anyone documented the differences? (preferably, using $n$ or $n^2$)

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Hint: construct equivalence classes of matrices determined by their determinant. You'll see that singular matrices are comparatively small (i.e., represented by a single class) out of uncountably many equivalence classes (assuming you are working over $\mathbb{R}$ or $\mathbb{C}$)... – Alex Nelson Dec 10 '12 at 3:32
I don't know what you mean by "the $n^{\rm th}$ column elements[...] are always multiples of $n.$" E.g., $(a,b;a,b)\in\operatorname{Mat}_{2\times 2}$ is singular for any choice of $a,b.$ – Andrew Dec 10 '12 at 3:33
What does "Each row is well-ordered up to a multiple of five" mean? – Gerry Myerson Dec 10 '12 at 3:59
@GerryMyerson, post has been updated. – Fred Kline Dec 10 '12 at 4:21
Nope, no closer to understanding what you mean by "well-ordered up to a multiple of $n$," but very close to concluding that you don't understand what you mean by that phrase, either. – Gerry Myerson Dec 10 '12 at 6:23

Thinking in terms of probability helps. If you have a continuous probability distribution defined on some space of matrices, then typically the singular matrices will have probability zero. Thinking in terms of the determinant: The determinant is a polynomial in the entries of the matrix. Setting it to zero gives a polynomial equation, which are defining (implicitely) some surface in the matrix space. This surface will have a reduced dimension , so its (Lebesgue) measure will be zero.

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The number of $2\times2$ matrices over a field of $q$ elements is $q^4$.

The number of non-singular $2\times2$ matrices over a field of $q$ elements is $$(q^2-1)(q^2-q)=q^4-q^3-q^2+q$$ which means only $q^3+q^2-q$ out of $q^4$ are singular.

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If you think about the interpretation of a matrix as a system of equations, and take the 2-variable case as an easy to visualize example, most pairs of equations are not parallel lines. If the 2nd column is a multiple of the first, then they are parallel. Inb this case, the 2nd column will be a multiple of the first.

Of course, for the 3 dimensional case, there are linear combinations, so this no longer holds in the same sense, but you still have the nth column as a "multiple" of one or both of the other columns.

Does that clarify, or is there something more complex you are noticing?

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Here is an extension of Gerry's argument. There are $q^{n^2}$ matrices in $M_{n\times n}(\mathbb{F}_q)$ and $\prod_{k=0}^{n-1}(q^n-q^k)$ elements in $GL_n(\mathbb{F}_q)$.

$$\lim_{q\to\infty}\frac{\prod_{k=0}^{n-1}(q^n-q^k)}{q^{n^2}}=\lim_{q\to\infty}(q^{-n};q)_n=\lim_{q\to\infty}\prod_{k=0}^{n-1}\left(1-q^{-n+k}\right)=1$$ where $(q^{-n};q)_n$ is the $q$-Pochammer symbol.

Note that the fact that $\mathbb{F}_q$ has nonzero characteristic doesn't affect this argument, since the formula $\prod_{k=0}^{n-1}(q^n-q^k)$ is based on picking $n$ linearly independent vectors combinatorially from $(\mathbb{F}_q)^n$, a process which extends to the infinite case independently of characteristic.

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Good has rank 1 and Ugly has rank 2. Bad has no rank. I'm new at this stuff, so I hope I've read it right. – Fred Kline Dec 10 '12 at 4:34