Theorem A.4.1Suppose A is an n×n matrix. Then A is oneto one (injective) if and only if Ais onto (surjective).Also, if B is an n × n matrix and AB = I, thenit follows BA = 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, since multiplication by A is linear, if

( )
∑n ∑n
ckAek = 0, then A ckek = 0
k=1 k=1

and so, since A is one to one,

∑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. Say Ax = 0 where x≠0. But this would say that the columns of A are linearly
dependent. If A were onto, then the columns of A would be a spanning set of F^{n}. However, one column is a
linear combination of the others and so there would exist a spanning set of fewer than n vectors
contradicting the above exchange theorem.

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.