Matrices question please help?

Question

Let A ∈ Mn×n(R) be an invertible matrix. Prove that AT is also invertible,
and express the inverse of AT in terms of A−1.

Please can someone explain how to go about this?? Appreciate it

Answer 1

Lemma 1: an invertible matrix has non-zero determinant.



=> det(A) ≠ 0, since A is invertible.



Theorem: det(A) = det(AT)

Hence if det(A) ≠ 0 <=> det(AT) ≠ 0.

By Lemma 1, AT is invertible.



The general idea is that if A has non-zero determinant, then so does A^T.



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For the second proof, it's a bit more difficult, and requires you to know a second theorem:

Theorem: An invertible matrix has a unique inverse.



Begin by stating that:

(A*A^-1) = I

(A*A^-1)^T = I^T = I

(A^-1)^T * A^T = I



This is since (AB)^T = B^T * A^T.



However, we have shown that (A^-1)^T is the inverse matrix of A^T, since they multiply to give the identity matrix, and this matrix form is unique from the Theorem.



So we can say fully, that the last line has the implication that (A^T)^-1 = (A^-1)^T.

This means the inverse of the transpose is the transpose of the inverse, and is a property of the transpose matrix.

Answer 2

by AT i assume you mean transpose

i will use A^t for convenience



(A^-1A)=I

(A^-1A)^t=I^t=I

(A^t)*(A^-1)^t=I

this is a proof that the transpose is invertible and that its inverse is the transpose of the inverse or

(A^t)^-1=(A-1)^t



hope this helps

Answer 3

Yes.