Proof
Let
and
denote the bases of
and
respectively, and let denote the column vectors of . (1). The mapping has the property
-
where is the -th entry of the -th column vector. Therefore,
-
This is if and only if
for all , and this is equivalent with
-
For this vector equation, there exists a nontrivial tuple , if and only if the columns are linearly dependent, and this holds if and only if is not injective.
(2). See
exercise.
(3). Let
.
The first equivalence follows from (1) and (2). If is bijective, then there exists a
(linear)
inverse mapping
with
-
Let denote the matrix for , and the matrix for . The matrix for the identity is the
identity matrix.
Because of
fact,
we have
-
and therefore is invertible. The reverse implication is proved similarly.