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I recently made a question regarding the measure of a linearly independent set of vectors in $\mathbb{R}$ over $\mathbb{Q}$. As answered, such types of sets have measure zero. That question was made in the hope of knowing how frequently one would choose a linearly independent set of vectors. On the other hand, the answer I got does not seem to be reasonable, since finding a linearly independent set of vectors frequently occurs, when made randomly in finite-dimensional spaces, like that question answered here and the other one here. Now, in order to the question make more sense,

What is the probability of choosing $n$ linearly independent vectors in $\mathbb{R}$ over $\mathbb{Q}$? Or over other infinite dimensional space?

R. W. Prado
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  • You ask about "probability". To answer that, first tell us what probability distribution to use. (There is no assumed probability distribution on $\mathbb R$, so you need to specify one.) You may also specify whether the $n$ vectors are to be independent. – GEdgar Aug 14 '22 at 21:12
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    Your second sentence is not an accurate summary of the other question, which talks about actual bases of $\Bbb R$ over $\Bbb Q$, not just linearly independent sets (nor is the other question answering a question about probability). – Greg Martin Aug 14 '22 at 21:15
  • Uniform continuous distribution, I think. – R. W. Prado Aug 14 '22 at 21:17
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    Your initial link goes to a question asked by "Ben Passer", not R. W. Prado. Did you link to the wrong question? – Jacob Manaker Aug 14 '22 at 22:10
  • Wow! I didn't know that duplicate questions are redirected to the other one whose such question is similar/duplicate. – R. W. Prado Aug 14 '22 at 22:13
  • Logged in my account, the question I first linked is that one I made. – R. W. Prado Aug 14 '22 at 22:16

2 Answers2

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If you restrict your choices to, say, the unit interval, and use the uniform distribution to choose $n$ real numbers then they will be independent over $\mathbb{Q}$ with probability $1$.

[This assertion needs a proof that I can't see immediately. Showing that any two particular irrational numbers are rationally independent is subtle, but the probability argument ought to be straightforward. My original answer below is to a slightly different question.]

The same is true for a set of $n$ vectors in $\mathbb{R}^n$ If you restrict your choices to, say, the unit sphere or the unit ball, and use the uniform distribution to choose vectors in ℝ then they will be independent with probability 1.

One way to see that is to see that the set where the determinant is $0$ has $0$ volume.

The same argument will work for any nice enough set on which you can define a reasonable distribution. You can't use the (nonexistent) uniform distribution on the whole space.

Ethan Bolker
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  • I would like some references regarding your answer. For me, a novice in probability theory, It seems hard to grasp all such details. Where such results come from? – R. W. Prado Aug 14 '22 at 21:54
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    @R.W.Prado See my edit. Sorry not to be more useful right now. Perhaps I will edit my answer later, or someone else will offer one. – Ethan Bolker Aug 14 '22 at 22:00
  • The proof is by induction on $n.$ After selecting $n-1$ independent, only countably many choices for $X_n!$ so the countably many have measure $0.$ – Thomas Andrews Aug 14 '22 at 22:54
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Claim. Let $\mathbb{P}$ be any probability distribution that is absolutely continuous w.r.t. Lebesgue measure (less fancily: "has a pdf"). Let $\{X_j\}_{j=1}^J$ be any sequence of iid draws from $\mathbb{P}$. Then $\{X_j\}_j$ is linearly independent a.s.
Proof by induction:

If $J=1$, the claim is obvious. Otherwise, suppose $\{X_j\}_{j=1}^{J-1}$ is linearly independent a.s. and condition on that event.

Since $\mathbb{Q}$ is countable, so is $\mathbb{Q}^{J-1}$. The latter is isomorphic (as a $\mathbb{Q}$-vector space) to and thus in bijection with $\operatorname{span}_\mathbb{Q}{\{X_j\}_{j=1}^{J-1}}$. But any countable set has Lebesgue measure $0$, so that the latter span has $\mathbb{P}$-measure $0$ as well. In particular, $X_J\notin\operatorname{span}_\mathbb{Q}{\{X_j\}_{j=1}^{J-1}}$ a.s. too. QED.

I do not know the answer in the case of an infinite-dimensional (topological) vector space of equal cardinality to the subspace.

Jacob Manaker
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