4 thoughts on “Q: Points in general position”

  1. $\begin{bmatrix}
    1 \\ 0 \\ 0
    \end{bmatrix}, \begin{bmatrix}
    0 \\ 0 \\ 1
    \end{bmatrix}, \begin{bmatrix}
    0 \\ 1 \\ 1
    \end{bmatrix}, \begin{bmatrix}
    -1 \\ 1 \\ 1
    \end{bmatrix}$ are not in general position in $P(\mathbb{R}^3) = \mathbb{R}^2 \cup P(\mathbb{R}^2)$. Furthermore $
    \begin{bmatrix}
    1 \\ 0
    \end{bmatrix}, \begin{bmatrix}
    -1 \\ 0
    \end{bmatrix} \left(= \begin{bmatrix}
    1 \\ 0
    \end{bmatrix}\right), \begin{bmatrix}
    0 \\ 1
    \end{bmatrix}$ are not in general position in $P(\mathbb{R}^2) = \mathbb{R} \cup P(\mathbb{R})$.

    Question. Let $U = \text{span}\left(\begin{pmatrix}
    1 \\ 0 \\ 0
    \end{pmatrix}, \begin{pmatrix}
    0 \\ 0 \\ 1
    \end{pmatrix}\right)$. Then $P(U)$ is a line. Is there a way to draw this line? Post your pictures 😀

      • Well, I think I have to rephrase my question (cos it doesnt express what it was intended to do)…
        When I imagine a line, I think of it as affine combinations of two points, which leads to my…
        Question. Take some arbitrary point and a point at infinity. Is there a line in the projective space s.t. both points lie on it? Can I visualize this line?

        • Talking about “points at infinity” implies using a particular model of a projective space.

          In the hemisphere model, you may see the points at the boundary as points at infinity. Clearly, there is a line through such a point on the boundary and a second point in the interior of the hemisphere.

          In the affine model, you may understand the points at infinity as directions. To require that a line passes through a certain point at infinity means prescribing a direction for the line. Again, a finite point and a direction determine a unique line.

          Question: What happens if you choose both points at infinity?

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