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User:FUNG Cheok Yin/proof(i) A127793

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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.)