Note that from a common theorem we have that
Note though that since
we must have that
and so by
we know that
. But, since
we know that
. Thus, putting these two together gives
Thus,
equals either
or
. And so plugging this into
gives
But note that
and
(the first is clear, and the second follows since
and a subspace which has the same dimension as the ambient space must be the full space). Thus, this may be rewritten as
Thus, if
then
is an
-dimensional subspace of
and thus equal to
and if
then
is an
-dimensional subspace and not equal to
. The conclusion follows.