Linear Algebra, Theory And Applications, 2012a

118 ROW OPERATIONS
Corollary 4.5.4 Let A be an m × n matrix. Then A maps R n onto R m if and only if the
only solution to A T x = 0 is x = 0.
Proof: If the only solution to A T x = 0 is x = 0, then ker ( A T ) = {0} and so ker ( A T ) ⊥
=
R m because every b ∈ R m has the property that b · 0 =0. Therefore, Ax = b has a solution
for any b ∈ R m because the b forwhichthereisasolutionarethoseinker ( A T ) ⊥
by
Theorem 4.5.3. In other words, A maps R n onto R m .
Conversely if A is onto, then by Theorem 4.5.3 every b ∈ R m is in ker ( A T ) ⊥
and so if
A T x = 0, then b · x =0foreveryb. In particular, this holds for b = x. Hence if A T x = 0,
then x = 0.
Here is an amusing example.
Example 4.5.5 Let A be an m × n matrix in which m>n.Then A cannot map onto R m .
The reason for this is that A T is an n × m where m>nand so in the augmented matrix
(
A T |0 )
there must be some free variables. Thus there exists a nonzero vector x such that A T x = 0.
4.6 Exercises
1. Let {u 1 , ··· , u n } be vectors in R n . The parallelepiped determined by these vectors
P (u 1 , ··· , u n ) is defined as
{ n
}
∑
P (u 1 , ··· , u n ) ≡ t k u k : t k ∈ [0, 1] for all k .
k=1
Now let A be an n × n matrix. Show that
is also a parallelepiped.
{Ax : x ∈ P (u 1 , ··· , u n )}
2. In the context of Problem 1, draw P (e 1 , e 2 )wheree 1 , e 2 are the standard basis vectors
for R 2 . Thus e 1 =(1, 0) , e 2 =(0, 1) . Now suppose
( ) 1 1
E =
0 1
where E is the elementary matrix which takes the third row and adds to the first.
Draw
{Ex : x ∈ P (e 1 , e 2 )} .
In other words, draw the result of doing E to the vectors in P (e 1 , e 2 ). Next draw the
results of doing the other elementary matrices to P (e 1 , e 2 ).
3. In the context of Problem 1, either draw or describe the result of doing elementary
matrices to P (e 1 , e 2 , e 3 ). Describe geometrically the conclusion of Corollary 4.3.7.
4. Consider a permutation of {1, 2, ··· ,n}. This is an ordered list of numbers taken from
this list with no repeats, {i 1 ,i 2 , ··· ,i n }. Define the permutation matrix P (i 1 ,i 2 , ··· ,i n )
as the matrix which is obtained from the identity matrix by placing the j th column
of I as the i th
j column of P (i 1 ,i 2 , ··· ,i n ) . Thus the 1 in the i th
j column of this permutation
matrix occurs in the j th slot. What does this permutation matrix do to the
column vector (1, 2, ··· ,n) T ?