Let
where is the matrix unit.
Let the matrix involved written as , where if , 0 if or .
To prove that is the right inverse of , we are going to show
Since for all :
for , , we have ;
for , , we have .
For , ,
-
We list out the terms which will be OR MAY be involved in the above summation:
...
It is obvious that the first lines cancelled out.
It remains to determine when A, B, C and D will be involved in the summation.
Consider the inequality (*) :
If :
(that is, A is in the summation)
for even
, ;
for odd
, .
We can see that C will be in the summation, but neither D nor the even larger B.
If :
(that is, A is not in the summation)
we separate into two cases again. (I) and (II) .
(I):
Solve and .
If is odd,
Impossible for an integer .
If is even,
but is even, so , thus .
Then none of A, C, D, or B will be in the summation.
(II):
If ,
(that is, C is in the summation)
then
, since .
Therefore we can conclude B is not in the summation.
D will be in the summation because (II) and (from (*) ).
Then C and D cancel out.
If ,
(that is, C is not in the summation)
(i) Consider ,
we will show .
(That is, both D and B are in the summation.)
The middle greater than sign is obvious; the rightmost sign comes from (*); it remains to show the leftmost sign is valid.
If is even,
If is odd,
(ii) Consider ,
D = .
Since ,
,
D is in the summation.
B = .
Since ,
:
if is even, ,
B will be in the summation (and cancel D out) except the edge case ;
if is odd,
B will be in the summation (and cancel D out) except the edge case .
First Draft Finished on Sep 12 2022 (The first draft is neither well written nor concise, comments are welcome.)