UTM Ideals Varieties and Algorithm - Chapter 2 Section 1 Exercise 3

Problem:

Solution:

For part (a), note that according to the errata, the last equation should be $ x_3 = -t + 6 $.

First note that all the equations are linear and there is one parameter, it is a line. We should expect two linear equations defining two planes.

From the first equation, it is obvious that $ t = x_1 + 5 $, so we get from the second equation $ x_2 = 2t + 1 = 2 (x_1 + 5) + 1 = 2x_1 + 11 $, and from the third equation $ x_3 = -t + 6 = -(x_1 + 5) + 6 = 1 - x_1 $, so the solution is:

$ x_2 = 2x_1 + 11 $
$ x_3 = 1 - x_1 $

Part (b) is more complicated, it is a two dimensional 'plane' in a 4 dimensional space. Adhoc methods like the previous problem does not work great, so we will try something different. We model the problem using this matrix.

$ \begin{array}{ccccccc}t & u & x_1 & x_2 & x_3 & x_4 & t\\2 & -5 & -1 & 0 & 0 & 0 & 0\\1 & 2 & 0 & -1 & 0 & 0 & 0\\-1 & 1 & 0 & 0 & -1 & 0 & 0\\1 & 3 & 0 & 0 & 0 & -1 & 0\end{array} $

Next, we find the reduced row echelon form using MATLAB (using the function rref), and we get this.

$ \begin{array}{ccccccc}t & u & x_1 & x_2 & x_3 & x_4 & t\\1 & 0 & 0 & 0 & 0.75 & -0.25 & 0\\0 & 1 & 0 & 0 & -0.25 & -0.25 & 0\\0 & 0 & 1 & 0 & 2.75 & 0.75 & 0\\0 & 0 & 0 & 1 & 0.25 & -0.75 & 0\end{array} $

We claim that the last two rows is our solution. To see that, any point that satisfied the parametrization must satisfy all four equations, and therefore it satisfy the last two. For any point satisfy the last two equations, we can always find $ t $ and $ u $ by equation 1 and 2, so we have proved the answer is

$ x_1 + 2.75x_3 + 0.75x_4 = 0 $

$ x_2 + 0.25x_3 - 0.75x_4 = 0 $

Part (c) is pretty easy, it is simple to inspect the answer is

$ x^4 = y $

$ x^7 = z $