Let A and B be two matrices such that AB is defined. If AB=A,

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No.
Counterexample: 0X = 0.

>>8452950
Plz give example. OX is always O when OX is defined, r-right? How is this related in any way? AB=AC does not imply that B and C are equal. So, going by that line of thought, if C is I, B need not be I. But I can't really think of an example to convince myself.

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>>8452971
Do I really have to spell it out?
Let B be any n*n square matrix other than the identity matrix.
Let A be the m*n zero matrix.
Then AB = A but B is not the identity.

>>8452937
AB=A implies B is identity only if A is invertible.

>>8452977
Gommenasai.

Everyone is kind of off.
Since AB=A and BA=B, A and B must both be nxn in size. So A=B=I or A=B=0

>>8452937
AB=A is not sufficient to imply B=I.

>>8452937

The rank : [math] X : \mapsto XB - X [/math] is not zero. So there are more than one solution.

>>8452937
The proof makes perfect sense. What part do you not follow?

AB=A. Right multiply both sides by A,
(AB)A=(A)A. Apply Associative property to left side of equation,
A(BA)=(A)A. Since BA=B,
A(B)=(A)A.
AB=A^2. Since AB=A,
A=A^2.

If you can't follow any proof, analyze it step-by-step.

>>8452937
what a shitty proof anyway

who cares? seems like an exercise in basic matrix multiplication properties

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>>8453325
>Since AB=A and BA=B, A and B must both be nxn in size.
>So A=B=I or A=B=0
No.
Counterexample: A = B = any idempotent matrix other than 0 or I.