Download Number of Equations Different From Number of Unknowns

2.3 Number of Equations Different from Number of Unknowns.
In the previous two sections we discussed Gaussian elimination in the case where the number of equations was
equal to the number of unknowns. Now we want to look at the case where the number of equations differs from
the number of unknowns. The Gaussian elimination process that we discussed in the previous section works
with essentially no change. However, in this case the solution is guaranteed to either not be unique or not have a
solution for every right hand side or both. Also, this more general situation leads to some theory of linear
equations that also applies to the case where the number of equations and unknows is the same.
We illustrate this method by means of an example.
Example 1.
(1)
x + 3y + 3z + 2t =
b1
2x + 6y + 9z + 5t =
b2
- x - 3y + 3z
1
or Au = b with A = 2
- 1
3
6
-3
=
2

5 ,
0
3
9
3
x
 1 3 3 2   y 
 b1 
 2 6 9 5   z  =  b2 
- 1 - 3 3 0  t 
 b3 
or
b3
 xy 
 b1 
u =  z , and b =  b2 . Note that we are leaving the right hand side
 b3 
t
unspecified for the time being so b could be any vector of three numbers.
Recall that one interpretation of the system of equations (1) is we are writing the vector b on the right hand side
as a superposition of the columns of A, i.e.
 
 
 
 
x 2  + y 6  + z9 + t5
1
- 1 
3
- 3 
3
3
2
5
=
 bb12 
 
 b3 
So the vectors b for which there is a solution to (1) are the vectors b which are a superposition of the columns of
A.
Definition 1. If A is a matrix then the column space of A is the set of all superpositions of the columns of A. It
is denoted by CA or C(A). The nullspace of A is the set of vectors u such that Au = 0. It is denoted by NA or
N(A).
Remark. As indicated the column space of A is also the set of vectors b for which the equation Au = b has a
solution. Recall there is a linear mapping T associated with A, i.e. Tu = Au. The column space of A is the range
of this linear mapping and sometimes also denoted R(T). For the matrix in (1) the nullspace is the set of space is
 xy 
the set of vectors u =  z  which are solutions when the right hand sides are all zero, i.e. b1 = 0, b2 = 0 and
t
b3 = 0. While the column space is associated with existence of solutions to (1), the nullspace is associated with
uniqueness because of the following proposition.
2.3 - 1
Proposition 1. Suppose up is any solution of Au = b. Then the following are true.
1.
If un is any vector in the nullspace then u = up + un is another solution of Au = b.
2.
If u is another solution of Au = b then u = up + un where un is in the nullspace.
In particular, a solution of Au = b is unique if and only if the nullspace of A consists of just the zero vector.
Definition 2. The row space of A is the set of all superpositions of the rows of A. It is denoted by C(AT). The
left nullspace of A is the set of solutions p of pA = 0. It is denoted by N(AT).
Remark. If we ignore the distinction between row and column vectors, then the row space is the column space
of AT and left nullspace is the null space of AT. If we distinguish between row and column vectors, then the row
space and left nullspace of A consist of the transposes of the vectors in the column space and nullspace of AT.
The vectors in the row space are orthogonal to the vectors in nullspace and the vectors in the column space are
orthogonal to the vectors in the left nullspace assuming the following definition of orthogonal.
Definition 3. Let p and u be row and column vectors with the same number of components. Then p and u are
orthogonal if pu = 0.
Proposition 2. If p is in row space of A and u is in the nullspace of A then p and u are orthogonal. If b is in the
column space of A and p is in the left null space the p and b are orthogonal.
Proof. We prove the second; the proof of the first follows by taking transposes. If b is in the column space then
there is u such that Au = b. So pb = pAu = 0u = 0. //
Returning to the equations (1), we subtract 2 times row 1 from row 2 of A and b and subtract - 1 times row 1


from row 3 of A and b. Alternatively, we multiply the equation Au = b by E = - 2 1 0 . The new equations

1
0
0
1
0
1

are
x + 3y + 3z + 2t =
(2)
3z +
b1
t = - 2b1 + b2
6z + 2t
=
b1

or A(1)u = B(1)b where A(1) = EA =  0
1

(1) -1 (1)



0
3
0
0
1
0
0
3
0
0
3
3
6
+ b3
3
3
6
2

1
2
x
2  
y

1 
z
2
t
=
- 12

1
0 0

1 0
0 1
 bb12 
 
 b3 


and B(1) = E = - 2 1 0 . One has B(1)A = A(1) and
A = (B ) A . The row space and null space of A

(1)
1
0
0
1
0
1

is the same as the row space and null space of A. However,
this is not true for the column space and left nullspace.
Since y is missing from equations 2 and 3, there is nothing more we can do with y. We move on to z. We


subtract 2 times row 2 from row 3 of A and b. Alternatively, we multiply the equation by F =  0 1 0 . The
1

new equations are
2.3 - 2
0
0 -2
0
1
x + 3y + 3z + 2t =
(3)
3z +
b1



t = - 2b1 + b2
1
0
0
3
0
0
x
2  
 y
1 
z
0
t
3
3
0
= 5b1 – 2b2 + b3
0

or A(2)u = B(2)b where A(2) = FA(1) =  0
1

3
0
0
0
3
3
0
2

1
0
=
- 12 01

 5 -2
0

0
1
 bb12 
 
 b3 


and B(2) = FB(1) = - 2 1 0 . One has B(2)A = A(2) and
1

0
0
1
5 -2
(2) -1 (2)
A = (B ) A .
We have reached the point where the system (2) is triangular, and we can draw some conclusions. From the last
equation 0 = 5b1 – 2b2 + b3 we conclude that in order for there to be a solution the right hand side must satisfy
this equation. If the right hand side does satisfy this equation, then we can solve the first two equations for x and
w. y and t are free variables in that for any values of y and t we get a solution. So in this example, there is not a
solution for every right hand side and if there is a solution it is not unique.
In order to exhibit the structure of the solution, when there is one, let’s continue the Gaussian elimination
process a little further. We divide equation 2 by 3 to get the coefficient of z equal to 1. Alternatively, we


multiply the equation by G =  0 1/3 0 . The new equations are

1
0
0
0
0
1
x + 3y + 3z + 2t =
(4)
z +
1
t
3
b1
= -
2
b
3 1
+




1
b
3 2
2 x

1 y 
3 z 
0 t 
 12 01
= - 3 3

 5 -2

1 0 0
2 1


5 -2
1
3
3
0
0
1
0
0
0
  b1 
b2
  b3 
1
0
0
= 5b1 – 2b2 + b3
0



1
or A(3)u = B(3)b where A(3) = GA(2) =  0
0
3
3
0
1
0
0
2

1
3
0
and B(3) = GB(2) = - 3 3 0 . One has B(3)A = A(3) and


1
(3) -1 (3)
A = (B ) A .
Finally, we subtract 3 times row 2 from row 1 of A and b. Alternatively, we multiply the equation by


H =  0 1 0 . The new equations are

1 -3
0
0
1
0

x + 3y
(5)
+
z +
t =
1
t
3
0
3b1 - b2
= -
2
b
3 1
+




1
b
3 2
1
3
0
0
0
1
0
0
0
1 x

1 y 
3 z 
0 t 
 32 - 11
= - 3 3

 5 -2
  b1 
b2
  b3 
1
0
0
= 5b1 – 2b2 + b3



1
or A(4)u = B(4)b where A(4) = GA(3) =  0
0
3
0
0
1
0
0
1

1
3
0

3 -1
2 1

5 -2
0

and B(4) = HB(3) =- 3 3 0 . One has B(4)A = A(4) and


1
(4) -1 (4)
A = (B ) A .
If 5b1 – 2b2 + b3 = 0 then we can express x and z in terms of y and t as follows.
2.3 - 3
x =
3b1 - b2 - 3y - t
2
1
z = - 3b1 + 3b2
1
- 3t
2.3 - 4