# Linear Algebra/Topic: Projective Geometry/Solutions
## Solutions
### Problem 1

What is the equation of this point?
    ${\begin{pmatrix}1\\0\\0\end{pmatrix}}$

### Answer

From the dot product
    $0={\begin{pmatrix}1\\0\\0\end{pmatrix}}\cdot {\begin{pmatrix}L_{1}&L_{2}&L_{3}\end{pmatrix}}=L_{1}$
we get that the equation is $L_{1}=0$.

### Problem 2

* Find the line incident on these points in the projective plane.
    ${\begin{pmatrix}1\\2\\3\end{pmatrix}},\,{\begin{pmatrix}4\\5\\6\end{pmatrix}}$
* Find the point incident on both of these projective lines.
    ${\begin{pmatrix}1&2&3\end{pmatrix}},\,{\begin{pmatrix}4&5&6\end{pmatrix}}$

### Answer

* This determinant shows that the line is $L={\begin{pmatrix}-3&6&-3\end{pmatrix}}$.
    $0={\begin{vmatrix}1&4&x\\2&5&y\\3&6&z\end{vmatrix}}=-3x+6y-3z$
* ${\begin{pmatrix}-3\\6\\-3\end{pmatrix}}$

### Problem 3

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

### Answer

The line incident on
    $u={\begin{pmatrix}u_{1}\\u_{2}\\u_{3}\end{pmatrix}}\qquad v={\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}$
can be found from this determinant equation.
    $0={\begin{vmatrix}u_{1}&v_{1}&x\\u_{2}&v_{2}&y\\u_{3}&v_{3}&z\end{vmatrix}}=(u_{2}v_{3}-u_{3}v_{2})\cdot x+(u_{3}v_{1}-u_{1}v_{3})\cdot y+(u_{1}v_{2}-u_{2}v_{1})\cdot z$
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 $p$ and $L$. That is, if $p_{1}$, $p_{2}$, $p_{3}$, and $q_{1}$, $q_{2}$, $q_{3}$ are two triples of homogeneous coordinates for $p$, and $L_{1}$, $L_{2}$, $L_{3}$, and $M_{1}$, $M_{2}$, $M_{3}$ are two triples of homogeneous coordinates for $L$, prove that $p_{1}L_{1}+p_{2}L_{2}+p_{3}L_{3}=0$ if and only if $q_{1}M_{1}+q_{2}M_{2}+q_{3}M_{3}=0$.

### Answer

If $p_{1}$, $p_{2}$, $p_{3}$, and $q_{1}$, $q_{2}$, $q_{3}$ are two triples of homogeneous coordinates for $p$ 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.
    $k\cdot {\begin{pmatrix}p_{1}\\p_{2}\\p_{3}\end{pmatrix}}={\begin{pmatrix}q_{1}\\q_{2}\\q_{3}\end{pmatrix}}\qquad m\cdot {\begin{pmatrix}L_{1}&L_{2}&L_{3}\end{pmatrix}}={\begin{pmatrix}M_{1}&M_{2}&M_{3}\end{pmatrix}}$
Then multiplying gives the answer $(km)\cdot (p_{1}L_{1}+p_{2}L_{2}+p_{3}L_{3})=q_{1}M_{1}+q_{2}M_{2}+q_{3}M_{3}=0$.

### 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?

### Answer

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 $I$ projects through $P$ to a circle $S$.

### Problem 6

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

### Answer

A spot on the unit sphere
    ${\begin{pmatrix}p_{1}\\p_{2}\\p_{3}\end{pmatrix}}$
is non-equatorial if and only if $p_{3}\neq 0$. In that case it corresponds to this point on the $z=1$ plane
    ${\begin{pmatrix}p_{1}/p_{3}\\p_{2}/p_{3}\\1\end{pmatrix}}$
since that is intersection of the line containing the vector and the plane.

### Problem 7

(Pappus's Theorem) Assume that $T_{0}$, $U_{0}$, and $V_{0}$ are collinear and that $T_{1}$, $U_{1}$, and $V_{1}$ are collinear. Consider these three points: (i) the intersection $V_{2}$ of the lines $T_{0}U_{1}$ and $T_{1}U_{0}$, (ii) the intersection $U_{2}$ of the lines $T_{0}V_{1}$ and $T_{1}V_{0}$, and (iii) the intersection $T_{2}$ of $U_{0}V_{1}$ and $U_{1}V_{0}$.
* Draw a (Euclidean) picture.
* Apply the lemma used in Desargue's Theorem to get simple homogeneous coordinate vectors for the $T$'s and $V_{0}$.
* Find the resulting homogeneous coordinate vectors for $U$'s (these must each involve a parameter as, e.g., $U_{0}$ could be anywhere on the $T_{0}V_{0}$ line).
* Find the resulting homogeneous coordinate vectors for $V_{1}$. (Hint: it involves two parameters.)
* Find the resulting homogeneous coordinate vectors for $V_{2}$. (It also involves two parameters.)
* Show that the product of the three parameters is $1$.
* Verify that $V_{2}$ is on the $T_{2}U_{2}$ line..

### Answer

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