Linear Algebra Theorem Proof

1. Jun 20, 2012

RWalden21

1. The problem statement, all variables and given/known data

Let A and B be square matrices, with B invertible. Show that det(BAB^-1) = det(A)

2. Relevant equations

I think its based off the theorem: If A and B are nxn matrices, then det(AB)= det(A)det(B)

3. The attempt at a solution

I started by simplifying BAB^-1det(A) to just try to get det(A) but I'm just not sure how to do the proof.

Last edited: Jun 20, 2012
2. Jun 20, 2012

rock.freak667

Right so how you would expand de(BAB-1) ? And how does det(B) relate to det(B-1)?

3. Jun 20, 2012

RWalden21

I suppose you can expand by det(B)det(A)det(B^-1) but I dont know how det(B) relates to det(B^-1). What I was thinking was by expanding and somehow getting B and B^-1 to = I somehow det(A) would remain

4. Jun 20, 2012

rock.freak667

Hint: B-1B= I so if you find then the determinant of I should be the same as the determinant of B-1B right?

5. Jun 20, 2012

RWalden21

is this correct?

det(BAB^-1)= det(B)det(A)det(B^-1)= det(BB^-1)det(A)= det(I)det(A)= det(A)

det(I) = 1

6. Jun 20, 2012

rock.freak667

Yep, that will work. So now you know that det(B-1) = 1/det(B), which may help you somewhere down the road.

7. Jun 20, 2012

dimension10

Didn't you realise you've like already done the proof? You know that if $\mathbf A,\;\mathbf B\in\mathbb R$, then $\det\mathbf{AB}=\det\mathbf A\det\mathbf B$ and $\det\left(\mathbf{A}^{-1}\right)=\det\left(\mathbf{A}\right)^{-1}$. So if you just simplify $\det\left(\mathbf{BAB^{-1}}\right)$, what do you get?

Edit: Why is the $\LaTeX$ equation not rendering?
Edit2: Ok, now it renders. I just realised there is an even simpler method though. You could just break up the determinant into three parts and the join two of them and you are done.

Last edited: Jun 20, 2012
8. Jun 20, 2012

Staff: Mentor

A and B aren't real numbers; they're matrices.

9. Jun 21, 2012

dimension10

Sorry, I meant $\mathbf{A},\;\mathbf{B}\in\mathbb R^{n\times n}$

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