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There exists an infinite subset \(S\subset \mathbb{R}^3 \) such that any three vectors in \(S\) are linearly independent.

*Hint:*

Can you find a sequence of vectors in \(\mathbb{R}^3\) which satisfy the following property? What does it mean to be linearly independent in terms of 3x3 matrices and their determinants?

**Discussion:**

We focus on finding a sequence as mentioned above. But first, we recall that:

If three vectors are linearly independent in \(\mathbb{R}^3\) then the columns of the matrix formed by these three vectors as columns are linearly independent. Which means the column rank (or rank) of this matrix is 3 (i.e, full rank), hence the matrix must be invertible and so determinant of this matrix should be zero.

So we have now a slightly different goal: To find a sequence in \(\mathbb{R}^3\) such that if any three of those are taken as columns of a 3x3 matrix then the determinant would be zero.

Now we begin our search. We start with simple integer valued 'nice-looking' sequences and very soon arrive at the following: \( (1,n,n^2) \) where \(n \in \mathbb{N}\). (Note: This is really a trial and error, at least that's how I arrived at it.)

We just need to verify: \( \begin{vmatrix} 1 & 1 & 1 \\ n & m & l \\ n^2 & m^2 & l^2 \end{vmatrix} \neq 0 \) where \(n,m,l\) are three different natural numbers. This is easy to check. (Hint: determinant will be product of differences taken two at a time).

**What is this topic:**Linear Algebra**What are some of the associated concept:**Linearly independent, Invertible Matrix, Rank**Book Suggestions:**Linear Algebra done Right by Sheldon Axler

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