Question

please solve this ...
suppose A and B are n by n matrices,and AB=I.prove from rank(AB)<=rank(A) that the rank of A is n . So A is invertible and B must be its two-sided inverse. Therefore BA=I(Which is not obvious!).

Answer 1

So what we have to do .. ?

Answer 2

ok sorry wait

Answer 3

as given that A is n by n,

rank(A) \(\le\) n and conversely :
n = rank(\(I_n\))=rank(AB)\(\le\)rank(A)

Answer 4

got my point @mkumar441 ?

Answer 5

yes i got it
thank you for ur quick response

Answer 6

Your welcome dear..

also:
\[\huge{\mathbb{WE}\textbf{LC}\mathcal{OME}\space\textit{TO OPENSTUDY}}\]

Answer 7

i need small favour at wat timings do u present
inorder post questions.

Answer 8

i didn't get you ..
it seems that you are Indian. You can ask in Hindi?

Answer 9

yes i am indian but i dont Hindi.When do you be in online inorder to clarifiy my doubts?

Answer 10

ok i got you now:

actually i am present here from 3:00 PM to 9:00 PM (Indian Standard Time_ or there may be breaks but yes:

There are gr8 experts present you don't need to care for your clarification of doubts. In case, if your doubt is not answered then you can post the link of your question in the chat.
or tag some people having good smart score....

Answer 11

lemme introduce you to some experts for your future help:

experimentx
mukushla
Vaidehi09
amistre64
vishweshshrimali5
unklerhaukus

and many more

Answer 12

just put @ sign before there names and they will be looking on your question as soon as possible.

Answer 13

Another question ..Suppose A is an m by n matrix of rank r.Its reduced echelon form is R.
Describe exactly reduced row echelon form of R transpose (not A transpose).

Answer 14

Post that as a new question and tag me there.. I will help you there