# Linear Algebra/Topic: Projective Geometry/Solutions

## Solutions

Problem 1

What is the equation of this point?

${\displaystyle {\begin{pmatrix}1\\0\\0\end{pmatrix}}}$

From the dot product

${\displaystyle 0={\begin{pmatrix}1\\0\\0\end{pmatrix}}\cdot {\begin{pmatrix}L_{1}&L_{2}&L_{3}\end{pmatrix}}=L_{1}}$

we get that the equation is ${\displaystyle L_{1}=0}$.

Problem 2
1. Find the line incident on these points in the projective plane.
${\displaystyle {\begin{pmatrix}1\\2\\3\end{pmatrix}},\,{\begin{pmatrix}4\\5\\6\end{pmatrix}}}$
2. Find the point incident on both of these projective lines.
${\displaystyle {\begin{pmatrix}1&2&3\end{pmatrix}},\,{\begin{pmatrix}4&5&6\end{pmatrix}}}$
1. This determinant
${\displaystyle 0={\begin{vmatrix}1&4&x\\2&5&y\\3&6&z\end{vmatrix}}=-3x+6y-3z}$
shows that the line is ${\displaystyle L={\begin{pmatrix}-3&6&-3\end{pmatrix}}}$.
2. ${\displaystyle {\begin{pmatrix}-3\\6\\-3\end{pmatrix}}}$
Problem 3

Find the formula for the line incident on two projective points. Find the formula for the point incident on two projective lines.

The line incident on

${\displaystyle u={\begin{pmatrix}u_{1}\\u_{2}\\u_{3}\end{pmatrix}}\qquad v={\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}}$

can be found from this determinant equation.

${\displaystyle 0={\begin{vmatrix}u_{1}&v_{1}&x\\u_{2}&v_{2}&y\\u_{3}&v_{3}&z\end{vmatrix}}=(u_{2}v_{3}-u_{3}v_{2})\cdot x+(u_{3}v_{1}-u_{1}v_{3})\cdot y+(u_{1}v_{2}-u_{2}v_{1})\cdot z}$

The equation for the point incident on two lines is the same.

Problem 4

Prove that the definition of incidence is independent of the choice of the representatives of ${\displaystyle p}$ and ${\displaystyle L}$. That is, if ${\displaystyle p_{1}}$, ${\displaystyle p_{2}}$, ${\displaystyle p_{3}}$, and ${\displaystyle q_{1}}$, ${\displaystyle q_{2}}$, ${\displaystyle q_{3}}$ are two triples of homogeneous coordinates for ${\displaystyle p}$, and ${\displaystyle L_{1}}$, ${\displaystyle L_{2}}$, ${\displaystyle L_{3}}$, and ${\displaystyle M_{1}}$, ${\displaystyle M_{2}}$, ${\displaystyle M_{3}}$ are two triples of homogeneous coordinates for ${\displaystyle L}$, prove that ${\displaystyle p_{1}L_{1}+p_{2}L_{2}+p_{3}L_{3}=0}$ if and only if ${\displaystyle q_{1}M_{1}+q_{2}M_{2}+q_{3}M_{3}=0}$.

If ${\displaystyle p_{1}}$, ${\displaystyle p_{2}}$, ${\displaystyle p_{3}}$, and ${\displaystyle q_{1}}$, ${\displaystyle q_{2}}$, ${\displaystyle q_{3}}$ are two triples of homogeneous coordinates for ${\displaystyle p}$ then the two column vectors are in proportion, that is, lie on the same line through the origin. Similarly, the two row vectors are in proportion.

${\displaystyle k\cdot {\begin{pmatrix}p_{1}\\p_{2}\\p_{3}\end{pmatrix}}={\begin{pmatrix}q_{1}\\q_{2}\\q_{3}\end{pmatrix}}\qquad m\cdot {\begin{pmatrix}L_{1}&L_{2}&L_{3}\end{pmatrix}}={\begin{pmatrix}M_{1}&M_{2}&M_{3}\end{pmatrix}}}$

Then multiplying gives the answer ${\displaystyle (km)\cdot (p_{1}L_{1}+p_{2}L_{2}+p_{3}L_{3})=q_{1}M_{1}+q_{2}M_{2}+q_{3}M_{3}=0}$.

Problem 5

Give a drawing to show that central projection does not preserve circles, that a circle may project to an ellipse. Can a (non-circular) ellipse project to a circle?

The picture of the solar eclipse — unless the image plane is exactly perpendicular to the line from the sun through the pinhole — shows the circle of the sun projecting to an image that is an ellipse. (Another example is that in many pictures in this Topic, the circle that is the sphere's equator is drawn as an ellipse, that is, is seen by a viewer of the drawing as an ellipse.)

The solar eclipse picture also shows the converse. If we picture the projection as going from left to right through the pinhole then the ellipse ${\displaystyle I}$ projects through ${\displaystyle P}$ to a circle ${\displaystyle S}$.

Problem 6

Give the formula for the correspondence between the non-equatorial part of the antipodal modal of the projective plane, and the plane ${\displaystyle z=1}$.

A spot on the unit sphere

${\displaystyle {\begin{pmatrix}p_{1}\\p_{2}\\p_{3}\end{pmatrix}}}$

is non-equatorial if and only if ${\displaystyle p_{3}\neq 0}$. In that case it corresponds to this point on the ${\displaystyle z=1}$ plane

${\displaystyle {\begin{pmatrix}p_{1}/p_{3}\\p_{2}/p_{3}\\1\end{pmatrix}}}$

since that is intersection of the line containing the vector and the plane.

Problem 7

(Pappus's Theorem) Assume that ${\displaystyle T_{0}}$, ${\displaystyle U_{0}}$, and ${\displaystyle V_{0}}$ are collinear and that ${\displaystyle T_{1}}$, ${\displaystyle U_{1}}$, and ${\displaystyle V_{1}}$ are collinear. Consider these three points: (i) the intersection ${\displaystyle V_{2}}$ of the lines ${\displaystyle T_{0}U_{1}}$ and ${\displaystyle T_{1}U_{0}}$, (ii) the intersection ${\displaystyle U_{2}}$ of the lines ${\displaystyle T_{0}V_{1}}$ and ${\displaystyle T_{1}V_{0}}$, and (iii) the intersection ${\displaystyle T_{2}}$ of ${\displaystyle U_{0}V_{1}}$ and ${\displaystyle U_{1}V_{0}}$.

1. Draw a (Euclidean) picture.
2. Apply the lemma used in Desargue's Theorem to get simple homogeneous coordinate vectors for the ${\displaystyle T}$'s and ${\displaystyle V_{0}}$.
3. Find the resulting homogeneous coordinate vectors for ${\displaystyle U}$'s (these must each involve a parameter as, e.g., ${\displaystyle U_{0}}$ could be anywhere on the ${\displaystyle T_{0}V_{0}}$ line).
4. Find the resulting homogeneous coordinate vectors for ${\displaystyle V_{1}}$. (Hint: it involves two parameters.)
5. Find the resulting homogeneous coordinate vectors for ${\displaystyle V_{2}}$. (It also involves two parameters.)
6. Show that the product of the three parameters is ${\displaystyle 1}$.
7. Verify that ${\displaystyle V_{2}}$ is on the ${\displaystyle T_{2}U_{2}}$ line..
1. Other pictures are possible, but this is one.

The intersections ${\displaystyle T_{0}U_{1}\,\cap T_{1}U_{0}=V_{2}}$, ${\displaystyle T_{0}V_{1}\,\cap T_{1}V_{0}=U_{2}}$, and ${\displaystyle U_{0}V_{1}\,\cap U_{1}V_{0}=T_{2}}$ are labeled so that on each line is a ${\displaystyle T}$, a ${\displaystyle U}$, and a ${\displaystyle V}$.

2. The lemma used in Desargue's Theorem gives a basis ${\displaystyle B}$ with respect to which the points have these homogeneous coordinate vectors.
${\displaystyle {\rm {Rep}}_{B}({\vec {t}}_{0})={\begin{pmatrix}1\\0\\0\end{pmatrix}}\quad {\rm {Rep}}_{B}({\vec {t}}_{1})={\begin{pmatrix}0\\1\\0\end{pmatrix}}\quad {\rm {Rep}}_{B}({\vec {t}}_{2})={\begin{pmatrix}0\\0\\1\end{pmatrix}}\quad {\rm {Rep}}_{B}({\vec {v}}_{0})={\begin{pmatrix}1\\1\\1\end{pmatrix}}}$
3. First, any ${\displaystyle U_{0}}$ on ${\displaystyle T_{0}V_{0}}$
${\displaystyle {\rm {Rep}}_{B}({\vec {u}}_{0})=a{\begin{pmatrix}1\\0\\0\end{pmatrix}}+b{\begin{pmatrix}1\\1\\1\end{pmatrix}}={\begin{pmatrix}a+b\\b\\b\end{pmatrix}}}$
has homogeneous coordinate vectors of this form
${\displaystyle {\begin{pmatrix}u_{0}\\1\\1\end{pmatrix}}}$
(${\displaystyle u_{0}}$ is a parameter; it depends on where on the ${\displaystyle T_{0}V_{0}}$ line the point ${\displaystyle U_{0}}$ is, but any point on that line has a homogeneous coordinate vector of this form for some ${\displaystyle u_{0}\in \mathbb {R} }$). Similarly, ${\displaystyle U_{2}}$ is on ${\displaystyle T_{1}V_{0}}$
${\displaystyle {\rm {Rep}}_{B}({\vec {u}}_{2})=c{\begin{pmatrix}0\\1\\0\end{pmatrix}}+d{\begin{pmatrix}1\\1\\1\end{pmatrix}}={\begin{pmatrix}d\\c+d\\d\end{pmatrix}}}$
and so has this homogeneous coordinate vector.
${\displaystyle {\begin{pmatrix}1\\u_{2}\\1\end{pmatrix}}}$
Also similarly, ${\displaystyle U_{1}}$ is incident on ${\displaystyle T_{2}V_{0}}$
${\displaystyle {\rm {Rep}}_{B}({\vec {u}}_{1})=e{\begin{pmatrix}0\\0\\1\end{pmatrix}}+f{\begin{pmatrix}1\\1\\1\end{pmatrix}}={\begin{pmatrix}f\\f\\e+f\end{pmatrix}}}$
and has this homogeneous coordinate vector.
${\displaystyle {\begin{pmatrix}1\\1\\u_{1}\end{pmatrix}}}$
4. Because ${\displaystyle V_{1}}$ is ${\displaystyle T_{0}U_{2}\,\cap \,U_{0}T_{2}}$ we have this.
{\displaystyle g{\begin{pmatrix}1\\0\\0\end{pmatrix}}+h{\begin{pmatrix}1\\u_{2}\\1\end{pmatrix}}=i{\begin{pmatrix}u_{0}\\1\\1\end{pmatrix}}+j{\begin{pmatrix}0\\0\\1\end{pmatrix}}\qquad \Longrightarrow \qquad {\begin{aligned}g+h&=iu_{0}\\hu_{2}&=i\\h&=i+j\end{aligned}}}
Substituting ${\displaystyle hu_{2}}$ for ${\displaystyle i}$ in the first equation
${\displaystyle {\begin{pmatrix}hu_{0}u_{2}\\hu_{2}\\h\end{pmatrix}}}$
shows that ${\displaystyle V_{1}}$ has this two-parameter homogeneous coordinate vector.
${\displaystyle {\begin{pmatrix}u_{0}u_{2}\\u_{2}\\1\end{pmatrix}}}$
5. Since ${\displaystyle V_{2}}$ is the intersection ${\displaystyle T_{0}U_{1}\,\cap \,T_{1}U_{0}}$
{\displaystyle k{\begin{pmatrix}1\\0\\0\end{pmatrix}}+l{\begin{pmatrix}1\\1\\u_{1}\end{pmatrix}}=m{\begin{pmatrix}0\\1\\0\end{pmatrix}}+n{\begin{pmatrix}u_{0}\\1\\1\end{pmatrix}}\qquad \Longrightarrow \qquad {\begin{aligned}k+l&=nu_{0}\\l&=m+n\\lu_{1}&=n\end{aligned}}}
and substituting ${\displaystyle lu_{1}}$ for ${\displaystyle n}$ in the first equation
${\displaystyle {\begin{pmatrix}lu_{0}u_{1}\\l\\lu_{1}\end{pmatrix}}}$
gives that ${\displaystyle V_{2}}$ has this two-parameter homogeneous coordinate vector.
${\displaystyle {\begin{pmatrix}u_{0}u_{1}\\1\\u_{1}\end{pmatrix}}}$
6. Because ${\displaystyle V_{1}}$ is on the ${\displaystyle T_{1}U_{1}}$ line its homogeneous coordinate vector has the form
${\displaystyle p{\begin{pmatrix}0\\1\\0\end{pmatrix}}+q{\begin{pmatrix}1\\1\\u_{1}\end{pmatrix}}={\begin{pmatrix}q\\p+q\\qu_{1}\end{pmatrix}}\qquad (*)}$
but a previous part of this question established that ${\displaystyle V_{1}}$'s homogeneous coordinate vectors have the form
${\displaystyle {\begin{pmatrix}u_{0}u_{2}\\u_{2}\\1\end{pmatrix}}}$
and so this a homogeneous coordinate vector for ${\displaystyle V_{1}}$.
${\displaystyle {\begin{pmatrix}u_{0}u_{1}u_{2}\\u_{1}u_{2}\\u_{1}\end{pmatrix}}\qquad (**)}$
By (${\displaystyle *}$) and (${\displaystyle **}$), there is a relationship among the three parameters: ${\displaystyle u_{0}u_{1}u_{2}=1}$.
7. The homogeneous coordinate vector of ${\displaystyle V_{2}}$ can be written in this way.
${\displaystyle {\begin{pmatrix}u_{0}u_{1}u_{2}\\u_{2}\\u_{1}u_{2}\end{pmatrix}}={\begin{pmatrix}1\\u_{2}\\u_{1}u_{2}\end{pmatrix}}}$
Now, the ${\displaystyle T_{2}U_{2}}$ line consists of the points whose homogeneous coordinates have this form.
${\displaystyle r{\begin{pmatrix}0\\0\\1\end{pmatrix}}+s{\begin{pmatrix}1\\u_{2}\\1\end{pmatrix}}={\begin{pmatrix}s\\su_{2}\\r+s\end{pmatrix}}}$
Taking ${\displaystyle s=1}$ and ${\displaystyle r=u_{1}u_{2}-1}$ shows that the homogeneous coordinate vectors of ${\displaystyle V_{2}}$ have this form.