Linear Algebra

ian.f

Looking through a past exam paper here and found a question that I don't know the proof for... Anyone care to help?

Let A,B,C be invertible square matrices, prove that A(transpose) x B x C(transpose) is invertible.

Thanks

LeixlipRed

Is that all the question tells you? Are we assuming the matrices are all n x n? You need to show that product has a non-zero derminant is what I would attempt to do anyway.

ian.f

Thats all your told but I suppose if its a square matrix then it is (nxn)

LeixlipRed

Well they have to be square to be invertible. And you'd also have to assume A,B,C are the same size as you couldn't multiply them otherwise. What you're looking for here is to use the fact that det(AB)=detA x detB. You need to compute a general expression for each of the determinants and then multiply it and show it's non zero.

MathsManiac

I would suggest that how much work is involved here depends on what you can assume has already been proved.

I assume we're talking about matrices over R or C (or other field - or indeed commutative ring with no zero divisors).

If one can assume that it has already been proved that det(XY)=det(X)det(Y) and that det((transpose(X))=det(X), then the proof is straightforward, since the determinant of the given expression is equal to det(A)det(B)det(C), which can't be zero unless one of those three determinants is zero.

If those two results can't be assumed, you'd just prove them first. You could google for those proofs, if you don't have them in your notes. (e.g. here: http://algebra.math.ust.hk/determinant/05_proof/lecture3.shtml#product)

Fremen

The result follows easily if you can show (A is invertible) implies (A transpose is invertible). Try proving that and post back here with your efforts if you get stuck.

Edit: Bloody MM. Always one step ahead. And slightly better.

LeixlipRed

Ah lads, you just done all the work for him

MathsManiac

Thinking about this again, I think you can do it without making any reference to determinants.

One can prove easily enough that the transpose of a product is the product of the transposes in reverse order [i.e. (AB)t = BtAt]. Using only this, along with associativity of matrix multiplication, you can simply verify by multiplication that the inverse of the given matrix is: [C(inverse)transpose][B(inverse)][A(inverse)transpose].

This means that the result is true even if these are matrices over a commutative ring WITH zero divisors.

Michael Collins

Or if you can prove that (M^T)^-1 = (M^-1)^T in general, you should be able to construct the inverse by inspection too - and be sure it will exist. Actually, how did you do it without assuming that MM?

MathsManiac

Sorry for not being clearer. What I meant was that the result (M^T)^-1 = (M^-1)^T follows from the product result that I mentioned. Suppose M is invertible. Then consider the matrix (M^-1)^T. You can show that it is the inverse of M^T:

(M^-1)^T * M^T = (M * M^-1)^T (using the product result I referred to)
= I^T = I.
and
M^T * (M^-1)^T = (M^-1 * M)^T = I^T = I.

Michael Collins

Ah ha, that's it. I wasn't sure how you worked the transpose of a product into it. Thanks