## Homework Assignment 4

### pg.95 prob. 1

Prove that the equivalence defined for homogeneous linear equations is an equivalence relation.

Two homogeneous linear equations xa + yb + zc = 0 and xe + yf + zg = 0 are equivalent if and only if there exists an m in D-{0} so that xe + yf + zg = xam +ybm + zcm.
This relation is:

• Reflexive: Since 1 in D, and xa + yb + zc = xa(1) + yb(1) + zc(1), every homogeneous linear equation is equivalent to itself.
• Symmetric: If xa + yb + zc = 0 is equivalent to xe + yf + zg = 0 then there exists an m in D-{0} so that xe + yf + zg = xam +ybm + zcm. Since m is not 0, m-1 exists and xe + yf + zg = 0 is equivalent to xem-1 + yfm-1 + zgm-1 = 0. But, xem-1 + yfm-1 + zgm-1 = xamm-1 + ybmm-1 + zcmm-1 = xa + yb + zc. Thus, xe + yf + zg = 0 is equivalent to xa + yb + zc = 0 and the relation is symmetric.
• Transitive: If xa + yb + zc = 0 is equivalent to xe + yf + zg = 0, and xe + yf + zg = 0 is equivalent to xh + yi + zj = 0, then there are m and k in D-{0} so that xe + yf + zg = xam + ybm + zcm and xh + yi + zj = xek + yfk + zgk = (xe +yf + zg)k. Thus, xh + yi + zj = (xe +yf + zg)k = (xam + ybm + zcm)k = xa(mk) + yb(mk) + zc(mk), so xa + yb + zc = 0 is equivalent to xh + yi + zj = 0.
Therefore, equivalence is an equivalence relation.

### pg.95 prob. 2

In the following D = Z7.
a) Show that there are 57 projective points <p,q,r>.

There are 73 = 343 vectors (p,q,r). One of these is (0,0,0) which does not represent a point. So, there are 342 nonzero vectors. As there are 6 nonzero elements of Z7, there will be 6 nonzero vectors in each 1 dimensional subspace. This gives, 342/6 = 57 1 dimensional subspaces, which are the projective points.

b) List the points satisfying the equation x + 5y + z = 0.
With z = 1 we get, (0,4,1), (1,1,1), (2,5,1), (3,2,1), (4,6,1), (5,3,1) and (6,0,1). And, if z = 0 we get (1,4,0).

c) Find the unique point that satisfies x +5y + z = 0 and 3x + 2y + z = 0.
Subtract the first from the second equation to get 2x -3y = 0, i.e., 2x = 3y. Now, x = 3 and y = 2 is clearly a solution. To find z, substitute these values into one of the original equations, say 3 + 5(2) + z = 0, z = -13 = 1. The point of intersection is thus, <3,2,1> which is in standard form.

d) Find a homogeneous linear equation satisfied by <1,5,1> and <3,2,1>. Compare with c).
A linear equation has the form ax + by + cz = 0. Substituting the two points gives two equations, a + 5b + c = 0 and 3a + 2b + c = 0, from which we must solve for the unknowns a,b and c. Note that these are the same two equations as in c), so a solution is a = 3, b = 2 and c = 1 and the line has the equation 3x + 2y + z = 0. This question is the dual of c).

### pg.98 prob.2

a) Let F be a field. Show that the equation for the line containing the distinct points <a,b,c> and <d,e,f> is
```                   a   b   c
det     d   e   f   = 0
x   y   z
```
The equation obtained from this determinant is (bf-ce)x -(af-cd)y + (ae-bd)z = 0. To show that this is the equation of the line determined by the two points we must show that each point satisfies the equation. Replacing x, y and z by a,b, and c respectively in the equation gives (bf-ce)a - (af-cd)b + (ae-bd)c = abf - ace =abf + bcd + ace - bcd = 0. Substituting d, e and f gives (bf-ce)d - (af-cd)e + (ae-bd)f = bdf - cde -aef + cde + aef - bdf = 0. Showing that this is the correct line. (Note, this is also easily seen from the determinant equation since either of these substitutions would give a matrix with two equal rows, and thus a determinant equal to 0).

b) Find a similar way of determining the point of intersection of two distinct lines in a projective plane over F.
If the equations are ax + by + cz = 0 and dx + ey + fz = 0, then the same determinant equation as in part a) can be used. The coefficients of x, y and z are the coordinates of the point of intersection, i.e., <bf-ce, cd -af, ae-bd>.

c) Do analogous results hold for a noncommutative division ring D?
Not in general, notice that in the computations of a) the commutativity of multiplication was used to verify that the points were on the line.