Another Linear Algebra Question

[mpm]

I am having to do a proof on a problem and am not really seeing it for some reason. Maybe it's because I have been doing the homework for so long.

Prove that if A is nonsingular then A^T is nonsingular and

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

Hint: (AB)^T = B^T*A^T

I understand the hint, but I can't seem to get an image of the actual problem.

Can anyone help me?

 

[quasar987]

Start from $(A^{-1}A)^T=I$.

 

[quicknote]

I can provide some guidance to help you with this proof. First, let's define what it means for a matrix to be nonsingular. A matrix is nonsingular if it has an inverse, meaning that there exists another matrix that, when multiplied with the original matrix, produces the identity matrix (a square matrix with 1s on the main diagonal and 0s everywhere else).

Now, let's consider the statement we are trying to prove: if A is nonsingular, then A^T is nonsingular and (A^T)^-1 = (A^-1)^T. To prove this, we will use the hint provided: (AB)^T = B^T*A^T.

First, let's take the transpose of both sides of the equation (A^T)^-1 = (A^-1)^T:

(A^T)^-1 = (A^-1)^T
Taking the transpose of both sides:
((A^T)^-1)^T = ((A^-1)^T)^T
Simplifying:
(A^-1)^T = A^T

Next, let's use the definition of a nonsingular matrix to prove that if A is nonsingular, then A^T is also nonsingular.

Assume A is nonsingular, meaning that it has an inverse A^-1. We want to show that A^T also has an inverse.

We know that A*A^-1 = I (where I is the identity matrix). Taking the transpose of both sides, we get:

(A*A^-1)^T = I^T
Using the hint, we can rewrite the left side as:

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

Now, we have shown that (A^-1)^T is the inverse of A^T, since (A^-1)^T*A^T = I. Therefore, if A is nonsingular, then A^T is also nonsingular and (A^T)^-1 = (A^-1)^T.

I hope this helps in understanding the problem and how to approach the proof. It is normal to struggle with proofs, but keep practicing and seeking help when needed. Good luck!