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Given the set $R^{n}$, what is the probability that 2 vectors drawn are linearly independent? I know the answer is almost 1 but I am not able to reason. Could someone please help me out?

computationally_curious
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    This depends on how you define probabilities. If you are using Lebesgue measure, the probability is 1 – M Turgeon Aug 23 '13 at 21:39
  • can you please elaborate? Im not very comfortable with probability – computationally_curious Aug 23 '13 at 21:41
  • What is your background in probability? Do you know what measure theory is? – M Turgeon Aug 23 '13 at 21:45
  • No, I am reading the wiki page on it. Its just that I am using this result to understand a certain algorithm that I am studying. – computationally_curious Aug 23 '13 at 21:46
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    For this to be a well posed question, you also need to say something about the distribution on $\mathbb{R}^n$ from which you are drawing these vectors. It obviously cannot be uniform. Assuming that the distribution has full support on $\mathbb{R}^n$ and is atomless will suffice to ensure that the probability of linear independence is 1. – exk Aug 23 '13 at 22:15
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    @exk, Given that the poster (Bambi) has no background in measure theory, it is perhaps not "obvious" that you cannot choose uniformly from $\mathbb{R}^n$... though of course I agree with your point. @ Bambi, the important point here is that without specifying how you're choosing your vectors, your question is not well defined... – owen88 Aug 23 '13 at 22:48
  • How sure (close to almost) would you need to be? Can this question be solved by simulations? If you're looking for an a priori proof then it may be that not be that any mathematical answer will be able to assuage your anxiety. This may be more a question of worldview than math. – dardisco Aug 24 '13 at 06:14

2 Answers2

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Some intuitive thoughts:

  1. The probability of choosing the zero vector is zero.

  2. Aside from the zero vector, it doesn't matter what vector you choose as the first one; there's loads of symmetry. So you can safely assume that the first vector is $(1,0)$.

  3. Now the probability that the second vector is linearly dependent on the first is the probability that its second coordinate is $0$, which happens with probability zero.

dfeuer
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The probability is 1, for the following reason. Once you have a vector, the second vector will be linearly dependent if it lies on the line spanned by the first vector, i.e. if it is a scalar multiple of the first vector. If $n>1$, this line is negligible with respect to the whole of $\mathbb R^n$ (negligible in a way measure theory can make precise). This means that the probability of the two vectors to be linearly dependent is 0. Therefore, the probability that the two vectors are linearly independent is 1.

M Turgeon
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