Linear algebra determinant proof

[Mdhiggenz]

Homework Statement

Let A and B be nxn matrices. Prove that if AB=I, then BA=I

det(AB)=det(I)

1.det(A)*det(B)=1

det(BA)=det(I)

2.det(B)*det(A)=1

Equating 1 and two together I get det(B)*det(A)=det(A)*det(B)

Thus AB=I, then BA=I

Is this correct?

Homework Equations

The Attempt at a Solution

 

[Karnage1993]

What you showed is that the $det(AB) = det(BA)$ but that doesn't really imply that $AB = BA$. If you want to show that $AB = BA$, you need to use the properties of inverses and facts about them.

What you did is a start, though. Since $det(AB) = det(A)det(B) = 1$, that means neither $det(A)$ or $det(B)$ are 0. Ie, A and B are non-singular and have inverses. Now all you need to do is show that $A$ and $B$ are inverses of each other.

 

[Mdhiggenz]

Could I say that if B=A^-1 then

det(AB)=det(A*A^-1)=1

?

 

[micromass]

Could I say that if B=A^-1 then

det(AB)=det(A*A^-1)=1

?

Sure, that is correct if B=A^{-1}. The problem is that you don't know that B=A^{-1}. You need to prove that.

 

[Mdhiggenz]

Sure, that is correct if B=A^{-1}. The problem is that you don't know that B=A^{-1}. You need to prove that.

Would it be enough if I maybe do something like this

assume A^-1=B

then A*B=I == A*A^-1=I
I=I

Which would make it true?

Because if B=A^-1 then A*B=I would be true

 

[vela]

No, you can't start with an assumption, get a true result, and then claim that proves the assumption is true. It's possible to start with a false assumption and get a true result.

 

[Mdhiggenz]

What you're saying makes sense but I'm a bit unsure on how to use a false assumption in this case. For instance if I assume that B~=A-1 what can I do with that?

 

[vela]

The point is you can't assume B=A^-1. That's what you're trying to prove.

 

[Mdhiggenz]

That's where I'm stuck, How do I prove the assumption I made without using another assumption? Can you help me get started in the direction of proving B=A-1 so I don't run in circles.

 

[Karnage1993]

It's very simple. Start with your "If" statement and somehow isolate $B$. Remember that what you are given is assumed to be true.

 

[vela]

Let's back up for a second. You know what AB=I because that's a given. At this point, you don't know anything about A and B other than the fact that their product AB is the identity matrix. In particular, you don't know if A has an inverse or B has an inverse.

You then took the determinant of both sides and got to det(A)det(B)=1. Now det(A)det(B)=1 implies that det(A)≠0. See what Karnage1993 said up in post 2. det(A)≠0 tells you A does indeed have an inverse. So now you know that AB=I and A^-1 exists. Can you use this knowledge to solve for B?

 

[Mdhiggenz]

So the if statement is if AB=I

then A^-1*AB=A^-1I

Which is equal to B=A^-1

 

[Karnage1993]

Yes, that's correct, but you can only multiply both sides by $A^{-1}$ once you know for a fact that $A$ does indeed have an inverse, which we already went over above.

 

[Mdhiggenz]

So I would have to state it like this

Since det(A)det(b)=det(I)=1
det(AB) is nonsingular meaning it has an inverse, and then I can show the proof for B=A^-1

 

[vela]

Not exactly. Try again. Your logic is a bit muddied still.

 

[Mdhiggenz]

Looking at the book it says that a matrix A is termed invertable if there is a matrix B such that A*B=I.

So if I'm quoting the definition in the book technically I'm not assuming anything, just explaining the if statement. Right?