Theorem 2.7.1Suppose A is an n × n matrix. Then A is oneto one (injective) if andonly if A is onto (surjective).Also, if B is an n × n matrix and AB = I, thenit followsBA = I.

Proof:First suppose A is one to one. Consider the vectors,

{Ae1,⋅⋅⋅,Aen}

where e_{k} is the
column vector which is all zeros except for a 1 in the k^{th} position. This set of vectors is linearly
independent because if

∑n
ckAek = 0,
k=1

then since A is linear,

(∑n )
A ckek = 0
k=1

and since A is one to one, it follows

∑n
ckek = 0
k=1

which implies each c_{k} = 0 because the e_{k} are clearly linearly independent.

Therefore,

{Ae1,⋅⋅⋅,Aen }

must be a basis for F^{n} because if not there would exist a vector,
y

would be an independent
set of vectors having n + 1 vectors in it, contrary to the exchange theorem. It follows that for
y ∈ F^{n} there exist constants, c_{i} such that

n ( n )
∑ ∑
y = ckAek = A ckek
k=1 k=1

showing that, since y was arbitrary, A is onto.

Next suppose A is onto. This means the span of the columns of A equals F^{n}. If these columns
are not linearly independent, then by Lemma 2.6.3 on Page 167, one of the columns is a linear
combination of the others and so the span of the columns of A equals the span of the n− 1 other
columns. This violates the exchange theorem because

{e1,⋅⋅⋅,en}

would be a linearly
independent set of vectors contained in the span of only n− 1 vectors. Therefore, the columns of
A must be independent and this is equivalent to saying that Ax = 0 if and only if
x = 0. This implies A is one to one because if Ax = Ay, then A

(x − y)

= 0 and so
x − y = 0.

Now suppose AB = I. Why is BA = I? Since AB = I it follows B is one to one since
otherwise, there would exist, x≠0 such that Bx = 0 and then ABx = A0 = 0≠Ix. Therefore, from
what was just shown, B is also onto. In addition to this, A must be one to one because if Ay = 0,
then y = Bx for some x and then x = ABx = Ay = 0 showing y = 0. Now from what
is given to be so, it follows

(AB )

A = A and so using the associative law for matrix
multiplication,

A(BA )− A = A (BA − I) = 0.

But this means

(BA − I)

x = 0 for all x since otherwise, A would not be one to one. Hence
BA = I as claimed. ■

This theorem shows that if an n × n matrix B acts like an inverse when multiplied on
one side of A, it follows that B = A^{−1}and it will act like an inverse on both sides of
A.

The conclusion of this theorem pertains to square matrices only. For example, let