MATH1038
Lecture 2: Geometric interpretation of linear equations and vector spaces
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Row Pictures: Intersections between lines and planes
1.1
2-dimensional
Example 1.1. Solve the equations:
(
2x − 3y = −5,
x + y = 5.
There are two variables and we can draw these equations in a (2-dimensional) plane.
Each (linear) equation represented by a straight line. For the line that corresponds to
2x − 3y = −5, it passes through the point (0, 5/3) (take x = 0 and one gets y = 5/3)
and another the point (−5/2, 0). Hence it is determined. Similarly, the second
straight line is the one passes through the points (0, 5) and (5.0). The solution of the
y
2x − 3y = −5
•
(2, 3)
x
x+y =5
Figure 1: Intersection between lines
two equations corresponds to the unique intersection (2, 3) between the two straight
lines.
Example 1.2. The equations (obviously?) has no solution.
(
2x − 3y = −5,
2x − 3y = 0.
This is because the corresponding lines are parallel to each other and there is no
intersections.
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y
2x − 3y = −5
2x − 3y = 0
x
Figure 2: Parallel lines
Example 1.3. How about the equations:
(
2x − 3y = −5,
2x − 3y = −5.
There are infinite solutions: for any real number t, (x, y) = (t, 2t+5
) is a solution.
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The two lines coincide and their intersection is just the whole line.
1.2
3-dimensional
z
2x + 3y + 3z = 3
y
2x − y = 0
x
Figure 3: intersection of planes
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Example 1.4. Solve the equations:
(
2x + 3y + 3z = 3,
2x − y + 0z = 0.
There are three variables and we work in 3-dimensional space. Each (linear)
equation corresponds to a plane. For instance, the ones above correspond to the blue
and red planes in Figure 3. Their intersection is the thick straight line, where the
points on it can be expressed as
{(
3 − 3t 3 − 3t
,
, t) | t ∈ R}.
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Example 1.5. The equations have no solution since the corresponding planes are
parallel.
(
2x − y + 0z = 2,
2x − y + 0z = 0.
z
2x − y = 2
y
2x − y = 0
x
Example 1.6 (Singular Case). Suppose that there are three (linear) equations with
three variables that corresponds to three distinguish planes. If they do not share a
unique intersection, then either they share no intersections or they intersection is a
line. More precisely, there are four cases.
I. All three planes are parallel to each other. We shall illustrate this by the fol-
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lowing picture.
II. Only two planes are parallel to each other.
III. The intersection of each pair of planes is a line and three such lines are parallel
to each other.
IV. Their intersection is a line.
V. Their intersection is a point, e.g. the xy-plane, yz-plane and zx-plane intersect
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at (0, 0, 0).
z
(0, 0, 0)
y
x
Remark 1.7 (Higher dimensions). How does this row picture extend into n dimensions? The n equations will contain n unknowns. The first equation still determines
a plane. It is no longer a two dimensional plane in 3-space; somehow it has dimension
n − 1. It must be flat and extremely thin within n-dimensional space, although it
would look solid to us.
If time is the fourth dimension, then the plane t = 0 cuts through four-dimensional
space and produces the three-dimensional universe we live in (or rather, the universe
as it was at t = 0). Another plane is z = 0, which is also three-dimensional; it is
the ordinary x − y plane taken over all time. Those three-dimensional planes will
intersect! They share the ordinary x − y plane at t = 0. We are down to two
dimensions, and the next plane leaves a line. Finally a fourth plane leaves a single
point. It is the intersection point of 4 planes in 4 dimensions, and it solves the 4
underlying equations.
I will be in trouble if that example from relativity goes any further. The point
is that linear algebra can operate with any number of equations. The first equation
produces an (n − 1)-dimensional plane in n dimensions, The second plane intersects
it (we hope) in a smaller set of dimension n − 2. Assuming all goes well, every new
plane (every new equation) reduces the dimension by one. At the end, when all n
planes are accounted for, the intersection has dimension zero. It is a point, it lies on
all the planes, and its coordinates satisfy all n equations. It is the solution!
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Column Pictures: Vector point of view
Skipped. Can be discussed after introducing column vectors
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Example 2.1. Back to the equations in Example 1.1, which can be written in a
single vector form equation:
2
−3
−5
x
+y
=
= b.
1
1
5
The problem is to find the combination of the column vectors on the left side that
produces the vector b on the right side. Note that the vector b is in fact identified with
the
pint whose coordinates
are −5 and 5. In this case,
we look for the combination of
2
−3
−5
(red vector) and
(blue vector) to produce
(black vector), cf. Figure 4
1
1
5
y
(−5, 5)
(−9, 3)
(4, 2)
(−3, 1)
(2, 1)
x
Figure 4: Combinations of vectors
Example 2.2. In Example 1.2 and Example 1.3, the corresponding equations are
2
−3
−5
x
+y
=
.
2
−3
0
and
2
−3
−5
x
+y
=
.
2
−3
−5
respectively. The two vectors colinear. When the target vector is not colinear with
them, there is no solution (Example 1.2) and when the target vector is colinear with
them, there are infinite solutions (Example 1.3).
Example 2.3. Let us check a ‘trivial’ 3-dimensional case.
 
 
   
1
0
0
3
x 0 + y 1 + z 0 = 5 .
0
0
1
4
6
y
(2, 2)
(−5, 0)
x
(−3, −3)
(−5, −5)
One sees that the solution is x = 3, y = 4, z = 5, which corresponds to the scaling
of the red vectors so that the sum of the rescaled vectors is the blue vector in Figure 6.
More precisely, we have multiplication by scalars
   
   
   
1
3
0
0
0
0
3 0 = 0 , 4 1 = 4 , 5 0 = 0 ,
0
0
0
0
1
5
and the vector addition
       
3
0
0
3
0 + 4 + 0 = 5 .
4
5
0
0
Example 2.4. Let us see a non-trivial case.
 
 
   
2
1
1
5
x  4  + y −6 + z 0 = −2 .
−2
7
2
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Example 2.5 (Singular Case). Suppose that there are three (linear) equations with
three variables. In the column picture, what it happens to the singular case. So we
have three vectors and try to combine them to produce vector b.
When the three column vectors are in the same plane P any combination of them
falls into P . For instance, the vectors in the following equation are in the plane
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z
(3, 4, 5)
(0, 0, 1)
(0, 1, 0)
•
(1, 0, 0)
y
x
Figure 5:
(5, −2, 9) z
(2, 0, 4)
•
y
(2, 4, −2)
x
(1, −6, −7)
Figure 6:
2x + 3y + 3z = 0.
 3 

0
2
2
x −1 + y  1  + z  0  = b.
−1
0
−1
 3 

Then there are two cases:
No solution: When
 
0

b = b0 = 1
1
is not in P , then there is no solution.
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Infinite solution: When


3
b = b∞ = −1
−1
is in P , it may have infinite solution.
Remark 2.6. In general, if the n planes have no point in common, or infinitely many
points, then the corresponding n columns lie in the same plane.
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