Matrix Proof

Show that if A is invertible, then A^t is invertible
and (A^t )^-1
= (A^-1)^t .


If A is invertible, then det(A)≠0
det(A)=det(A^t)
thus det(A^t)≠0



I'm not really sure how to prove the second part. It's an identity that I remembered, but don't know how to prove.

I'll take a crack at it though:

Multiply A^t to both sides
gives I=A^t (A^-1)^t
For the RHS, can i multiply A^t to A^-1 or does the ^t prevent me from doing that?

 

I'm not sure what you are asking.

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

If AB=I, then obviously, A is the inverse of B or vise versa.

So In this case I let A=(A^t)^-1
Thus A^=((A^t)^-1)^-1 ? <<<<multiply this to both sides? If so, the left said reduces to the I matrix, but the right side still looks complicated.

 

OH!

So in this form

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

To get to the identity matrix A^T has to be multiplied by its inverse (A^T)^-1. The quantity [(A^-1)^t] must equal (A^T)^-1.


So simple, yet hard.