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The general linear group $GL_n(k)$ is the group of automorphisms (in a suitable sense) of the $n$-space $k^n$. Similarly, the projective linear group $PGL_n(k)$ is the group of automorphisms (in a suitable sense*) of the projective $n$-space $\mathbb{P}^n(k)$. In projective geometry, two nonzero points are identified if they are on a same line going through ...

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A hyperbola is given by a quadratic equation in $x$ and $y$, so assuming that a line $ax+by=c$ goes through a point on the hyperbola and it is also a tangent line to the hyperbola, there is a quadratic equation in $x$ (we get that by eliminating $y$) with three roots accounted with multiplicity. That obviously cannot happen, since $\mathbb{R}[x]$ is an ...

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First, note that a given triangle and a given point of perspectivity determine three rays making some set of angles; we'll say that the "outer" rays form respective angles $\alpha$ and $\beta$ with the "middle" ray. (Having taken opposite rays if necessary, we may assume $\alpha + \beta \leq 180^\circ$.) Also, if a target triangle has a "similar copy" that's ...

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Consider the equation $x_0^2 - x_1^2 = rx_2^2$ (visualize this as a cone in any way you choose) and then let $r\to 0$ to see how the cone degenerates into two intersecting planes.

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Let $S=k[x,y,z]$ and let $I$ be a homogeneous ideal such that the projective variety defined by $I$ has dimension $0$. Then $S/I$ has Krull dimension $1$ and so its Hilbert polynomial is a constant, i.e., the Hilbert function is eventually a constant. Suppose that $H_{S/I}(n) = H_{S/I}(n_0), \, \forall n \ge n_0$. Now let $R = k[x,y]$ and let $J$ be the ...

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The question is equivalent to its dual: Let $g_1,g_2,g_3$ be three distinct points on a line $A$, and let $h_1,h_2,h_3$ be three distinct points on a line $B$ and suppose $A\cap B\notin\{g_1,g_2,g_3,h_1,h_2,h_3\}$. Let $C_{ij}$ be the line $(g_i,h_j)$. Are $C_{12}\cap C_{21}$, $C_{13}\cap C_{31}, C_{23}\cap C_{32}$ collinear? But this is just ...

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You can look at Liu, Algebraic Geometry and Arithmetic Curves, prop 7.4.1 p285. The idea is use the Riemann-Roch theorem in order to show that the anticanonical divisor of the curve $C$ induces a closed immersion $C \hookrightarrow \mathbf{P}^2_K$. Thus $C$ is isomorphic to a plane curve $C'$. This curve is defined by a homogeneous polynomial of some degree ...

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The projective plane over $\mathbb F$ arises from $\mathbb F^3$ by excluding the origin and then identifying vectors that are multiples of each other. Similarly, $PGL_n(\mathbb F)$ arises from the set of $n\times n$ matrices by excluding the singular matrices and then identifying matrices that are multiples of each other (because the center of the general ...

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Okay then. One way to achieve what you want is via the following: First use a combination of $x$ and $y$ rotations to align the $x$ axis with the minor axis of the ellipse. Then rotate about the $x$-axis by angle $\phi$. (Remark: the following paragraph is new) To find out the angle $\phi$, we start by thinking about what happens if we wrote a flat ...

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Start by computing the plane spanned by the line and the point of the camera. If you don't know how to compute such a join using Plücker coordinates, I can expand on this. The projection of the line is the intersection of that plane with the plane of your image. If you can choose the embedding of your drawing plane, embed it at $z=0$ in such a way that the \$...

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It turns out that it doesn’t matter what units you use for the extrinsic matrix as long as they’re consistent with the units that you use for world coordinates. For simplicity, let’s assume that there’s no rotation, as is the case for your matrices. That part of the extrinsic matrix isn’t affected by the units of measurement, anyway. We have some ...

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Is the camera plane the projective space of the real world? It depends. Usually a physical camera has a limited sensor, so the thing where the image appears is too bounded to be an affine or projective plane. You can say “well, simply extend that plane infinitely”. Then it depends on how exactly you define that extension, whether you end up with an ...

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