Page 105 - 35Linear Algebra

5.1 Examples of Vector Spaces 105

Example 62 (Solution set to a homogeneous linear equation.)
Let
 
1 1 1
M =  2 2 2  .
3 3 3
The solution set to the homogeneous equation Mx = 0 is

     
−1 −1
 

c 1  1  + c 2  0  c 1 , c 2 ∈ R .

0 1
 
 
1
3
0 . The sum of
This set is not equal to R since it does not contain, for example,  
0
any two solutions is a solution, for example
             
−1 −1 −1 −1 −1 −1
2 1 + 3 0 + 7 1 + 5 0 = 9 1 + 8 0
             
0 1 0 1 0 1
and any scalar multiple of a solution is a solution

        
−1 −1 −1 −1
4 5 1  − 3  0  = 20  1  − 12  0  .
 
0 1 0 1

This example is called a subspace because it gives a vector space inside another vector
space. See chapter 9 for details. Indeed, because it is determined by the linear map
given by the matrix M, it is called ker M, or in words, the kernel of M, for this see
chapter 16.

Similarly, the solution set to any homogeneous linear equation is a vector
space: Additive and multiplicative closure follow from the following state-
ment, made using linearity of matrix multiplication:

If Mx 1 = 0 and Mx 2 = 0 then M(c 1 x 1 +c 2 x 2 ) = c 1 Mx 1 +c 2 Mx 2 = 0+0 = 0.

A powerful result, called the subspace theorem (see chapter 9) guarantees,
based on the closure properties alone, that homogeneous solution sets are
vector spaces.
More generally, if V is any vector space, then any hyperplane through
the origin of V is a vector space.

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