HW Section 1.4 p 81 (5a-e, 11, 12, 25, 27)
1.4 Column Space and Nullspace
Is the linear system solvable?
What is the size of the solution set?
First, we need to know what a subspace is.
Subspaces
Sometimes, a subset W of a vector space V is also a vector space. For example, 2 is a
subset of 3 and they are both vector spaces. A vector space inside another vector space
is called a subspace.
Subspace: A nonempty subset W of a vector space V is a subspace of V if it closed under
linear combinations. (I.e., if uW and vW then also s*u+t*vW for all scalars s, t.)
Note: This is all that is required because the subspace W of V inherits the other
properties of V!
Example: Let S be any subset of V. Then the span of S is a subspace.
By definition, the span of S includes all linear combinations of the elements of S.
Thus, it is closed under linear combinations.
Subspaces of 3
The subspaces of R3 are R3 itself, planes through the origin, lines through the origin, and
the origin.
Note: any subspace of 3 must contain the zero vector. WHY?
linear combination: a1 x1 + a2 x2 (you could choose a1 = a2 = 0)
Matrix Subspaces
Row Space: The span of the rows of a matrix. (section 2.3)
Column space: The span of the columns of a matrix.
Null space: Given an nxm matrix A, the set of all x in m such that Ax = 0 is a subspace.
Notes: (Exam I study hint.)
The row space does not change with elementary row operations. (Section 1.3)
We can use the column space to determine if a matrix is solvable. (Section 1.4)
We can use the null space to determine the size of the solution. (Section 1.4)
Column Space
Column Space and Solvability: A linear system is solvable if and only if the column
vector of constants belongs to the column space of the coefficient matrix.
difficulty: moderate
Proof: We want Ax=B -- what is required of B?
A general linear system looks like:
a11x+a12y+a13z=b1
a21x+a22y+a23z=b2
a31x+a32y+a33z=b3
The general matrix of coefficients looks like:
 a11 a12 a13 
a

 21 a22 a23 
a31 a32 a33 
The system can be rewritten as:
 b1   a 11x  a 12 y  a 13z 
 a11 
 a12   a13 
b   a x  a y  a z   x a   y a   z a 
22
23 
 2   21
 21 
 22   23 
b3  a 31x  a 32 y  a 33z 
a31 
a32  a33 
Notice that we have a linear
combination of the columns on
the right!
To prove “A if and only if B” you need to prove that A  B and B A.
A) If a linear system is solvable then the vector of constants belongs to the column space.
solvable  at least one solution for “x”, “y” and “z” satisfying the equation above
 vector of constants is a lin comb of columns
B) If the vector of constants belongs to the column space, then the linear system is
solvable.
If the vector of constants is in the span of the columns, there exist scalars such that a
linear combination of the columns will yield the vector of constants. These scalars would
yield a valid solution for “x”, “y” and “z”.
Example:
By inspection, the
following system is
inconsistent:
x+y=1
x=1
y=1
The coefficient
matrix is:
1 1
1 0


0 1
The
columns
are:
1  1 
1   0 
  
0  1 
Theorem: tells us the vector
of constants does not belong
in the column span.
1
1

1
x  0y  1
1 1 1 1 1 1 1 0 1 1 0 0
1 0 1  0  1 0  0  1 0  0 1 0  0 x  y  1


 
 
 
0 1 1 0 1 1 0 0 1 0 0 1
0x  0 y  1
Recall, the following system is inconsistent.
The coefficient
The columns are:
matrix is:
x-y=1
 1  1
 1    1
-x+y=1
 1  1 
 1 1 
  


Does the vector of constants
belong in the column span?
1
1

Nullspace
What is the size of the solution set? (Nullspace)
Homogeneous System: Ax=0
A linear system is homogeneous if all the constant terms are 0.
(Ax=B where B=0).
Significance: Closely related to the general system Ax=B.
a11x+a12y+a13z=0
a21x+a22y+a23z=0
a31x+a32y+a33z=0
Solutions to the homogeneous system
There is always at least one solution! (all zeros)
The solutions are closed under linear combinations. (proof depends on matrix
multiplication that we haven’t covered yet)
Nullspace: If A is a matrix, the nullspace of A is the set of vectors X such that AX=0.


Given a homogeneous system Ax=0, the nullspace is the set of solutions.
Given a general linear system Ax=B, the nullspace is the set of solutions of the
corresponding homogeneous system.
Work out HW 5a on the board
5a. Find the nullspace for each of the matrices in Exercise 3 on page 59. Express each
answer as a span.
2 7  5  3 13
A= 1 0 1
4 3  What is the nullspace of A?

1 3  2 2 6 
Just apply the definition: the nullspace is the solution to Ax=0.
2 7  5  3 13
1 0 1
4 3 

1 3  2 2 6 
 x1 
 x 2 0
   
 x3  = 0 
   
 x 4  0 
 x5 
Q: How do we know that there are 5 variables?
A: The matrix A would correspond to the
coefficient matrix, each column in the
coefficient for one variable.
2 7  5  3 13 0


4 3 0
Augmented matrix: 1 0 1
1 3  2 2 6 0
1 0 1 0 3 0 


Can use RREF from previous homework: 0 1  1 0 1 0
0 0 0 1 0 0
The solution reads:
x1  x3  3x5  0
x 2  x3  x5  0
x4  0
Pivots: x1, x2 and x4 are constrained
Free: x3 and x5
x1   x3  3 x5
x 2  x3  x5
x3  t
x4  0
x5  s
 x1   t  3s  0   1  3
 x 2   t  s   0   1    1
  
      
  x3  =  t  = 0  + t  1  + s  0 
  
      
 x 4   0  0   0   0 
 x5   s  0   0   1 
translation vector: origin (nullspace must contain the zero vector)
spanning vectors: [-1 1 1 0 0]t and [-3 -1 0 0 1]t
(Leave on board for later…)
Translation Theorem
(Relationship between a linear system and its corresponding homogeneous system.)
Translation Theorem: If p is any specific solution to the linear system Ax = b, then the
entire solution set can be described as
Geometrically, this says that the solution set for Ax = b is a translation of the solution set
for Ax = 0.
Most importantly, they have the same size and shape!
Translation Theorem Corollary: The system AX=B has exactly one solution if and
only if the nullspace of A has exactly one solution. (This solution would be the trivial
solution of course.)
Example:
1 0 1 0 3 
2 7  5  3 13


A= 1 0 1
RREF(A) = 0 1  1 0 1  .
4 3
0 0 0 1 0 
1 3  2 2 6 
How many solutions does the system Ax=0 have? The system Ax=B?
The system Ax=0:
no problem with consistency
two free variables x3, x5  infinite # slns
The system Ax=B:
will also have infinite # solutions
IF it is consistent
(if B is in the column space of A!!)
Check at home:
5
Suppose you have B=  1 .
 2 
Is it consistent?
(yes)
Expected # of solutions?
()
 x1   1  1  3
 x 2   1   1    1
       
After working out, actual solution is:  x3  =  0  + t  1  + s  0 
       
 x 4  0   0   0 
 x5   0   0   1 
Notice: infinite solutions, same size as nullspace
in fact, spanning vectors are the same!
Translation Theory: Example 1D
xR
ax=0 point x=0 only solution
ax=b point x=b/a only solution
‘v’
‘p’
general system solution = {p+v|v solves ax=0}
Translation Theory: Example 2D (with two constraining equations)
 x1 
x=    2
 x 2
ax1=0
bx2=0
 x1  0 
solutions: point   =  
 x 2  0 
only one solution
‘v’
 b1 
solutions: point x=  a 
b 2 
 b
only one solution
‘p’
ax1=b1
bx2=b2
general system solution = {p+v|v solves ax=0}
Translation Theory: Example 2D (with one constraining equation)
 x1 
x=    2
 x 2
ax1=0
bx1=0
 x1  0 
solutions: point   =  
 x 2  s 
infinite solution set
b1 
solutions: point x=  a 
 s 
infinite solution set only if b2 = b*b1
‘v’
ax1=b1
bx1=b2
general system solution = {p+v|v solves ax=0}