Download MA 3280 Lecture 12 - Bases for a Homogeneous System Monday

MA 3280 Lecture 12 - Bases for a Homogeneous System
Monday, February 24, 2014.
Objectives: Material from 3.10: Basis for a solution space to a homogeneous system.
Homogeneous systems
A homogeneous system of linear equations is a system with 0’s on the right sides of the equation like
2x1
x1
x1
(1)
+
+
+
+
+
+
x2
2x2
2x2
3x3
3x3
5x3
= 0
= 0 .
= 0
As we’ll see the set of solutions to a system AX = 0 are closely related to the solutions to AX = B.
Example 1. Find the set of solutions to the system
(2)
+
x1
x2
+
3x3
x3
+
+
2x4
x4
=
=
0
0
We’ll do what did before when we had free variables. We will solve for the pivot variables and back substitute.
First solve for x3 to get
(3)
x3 = −x4 .
Then substitute into the first equation
(4)
x1 + x2 + 3(−x4 ) + 2x4 = 0,
and then solve for x1 to get
(5)
x1 = x4 − x2 .
Giving the free variables parameter names, x2 = a and x4 = b, we get
(6)
(x1 , x2, x3 , x4 ) = (b − a, a, −b, b).
This gives us the general form for all of the solutions. If we flip these vertically into column vectors, we can
rewrite the solutions as

 
 
 





x1
b−a
−a
b
−1
1
 x2  

 

 1 
 0 
a 

 
  a   0 




(7)
 x3  =  −b  =  0  +  −b  = a  0  + b  −1  .
x4
b
0
b
0
1
In other words, all of the solutions to this system of equations can be expressed uniquely as a linear combination of the two vectors




−1
1
 1 
 0 




(8)
 0  and  −1  .
0
1
Bases for the solution space
Given a homogeneous system of equations, the set of solutions will be called the solution space. We use the
term space, because the solution space can be thought of as a vector space on its own. For a homogeneous
system, the solution space will be the origin, or a line passing through the origin, or a plane passing through
the origin, or a 3-D hyperplane passing through the origin, etc. of the appropriate Rn .
If W is the solution space for some homogeneous system that is more than just the origin, then there will
be vectors u1 , u2, . . . , us ∈ W such that every w ∈ W can be expressed uniquely as a linear combination
(9)
w = a1 u 1 + a2 u 2 + · · · + as u s .
1
MA 3280 Lecture 12 - Bases for a Homogeneous System
2
The set { u1 , u2 , . . . , us } is called a basis for W , and the dimension of W is defined to be s (the number
of vectors in the basis). If every vector w ∈ W can be written as a linear combination of the ui ’s, but
not necessarily uniquely, then the set { u1 , u2, . . . , ut } is called a spanning set for W . Some subset of any
spanning set will be a basis, and every basis for a particular W will be the same size, the dimension of W .
Example 2. This is Example 3.18 in our book. Find a basis and the dimension of the solution space for
the system
(10)
x1
2x1
5x1
+
+
+
2x2
4x2
10x2
−
−
−
3x3
5x3
13x3
+ 2x4
+ x4
+ 4x4
−
−
−
4x5
6x5
16x5
=
=
=
0
0 .
0
In order to determine the free and pivot variables, we need to get to echelon form. We start by doing
−2E1 + E2 and −5E1 + E3 to get
x1
+
2x2
−
3x3
x3
2x3
+
−
−
2x4
3x4
6x4
−
+
+
4x5
2x5
4x5
=
=
=
0
0 .
0
+
2x2
−
3x3
x3
+
−
2x4
3x4
−
+
4x5
2x5
0
=
=
=
0
0 .
0
(11)
Next, we do −2E2 + E3 to get
x1
(12)
Note that in a homogeneous system, we are guaranteed at least one solution, since it is impossible to get
a degenerate equation. Each free variable will be associated with a basis vector, so we now know that the
dimension of the solution space is three, since we have three free variables x2 , x4 , and x5 . We can find the
basis as we did before by solving for the pivot variables and back substituting. We have
(13)
x3 = 3x4 − 2x5 ,
and then substituting to get
(14)
x1 + 2x2 − 3(3x4 − 2x5 ) + 2x4 − 4x5 = 0,
and
(15)
x1 = −2x2 + 7x4 − 2x5 .
All solutions, therefore, will take the form (−2a + 7b − 2c, a, 3b − 2c, b, c). Flipping this up into a column
vector, we get




 


−2a + 7b − 2c
−2
7
−2

 1 
 0 
 0 
a 




 









(16)
3b − 2c  = a  0  + b  3  + c 

 −2  .







b
0
1
0 
c
0
0
1
We can get the basis another way, if we wish. If we go back to our equations and let x2 = 1, x4 = 0, and
x5 = 0, we get
x1
(17)
+
2(1) −
3x3
x3
+ 2(0)
− 3(0)



We get x3 = 0 and x1 = −2. This gives us the vector 


−2
1
0
0
0
− 4(0)
+ 2(0)
0
=
=
=
0
0 .
0



.


We can get the other two basis vectors by setting each of the free variables equal to 1 with the other free
variables equal to 0.
MA 3280 Lecture 12 - Bases for a Homogeneous System
Quiz 12
Find the other two basis vectors using this second method.
Homework 12
Find the dimension and basis for the solution space.
1.
x1
+
3x2
−
x3
x1
+
2x2
−
3x3
x3
x1
+
x2
x2
2.
3.
−
+
2x3
3x3
+
x4
x4
+ 2x4
− 3x4
x4
+
−
5x4
x4
+
+
3x5
2x5
−
+
+
=
=
4x5
2x5
3x5
=
=
0
.
0
=
=
=
0
0 .
0
0
0
Answers: Quiz: h 7, 0, 3, 1, 0 i and h −2, 0, −2, 0, 1 i.
HW: 1) { h −3, 1, 0, 0, 0 i, h 1, 0, 1, 0, 0 i , h −1, 0, 0, −2, 1 i}.
2) { h −2, 1, 0, 0, 0 i, h −23, 0, −11, −3, 1 i}.
3) { h 5, −3, 1, 0 i , h −6, 1, 0, 1 i}.
3