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.