Is V a Vector Space with Transpose-Modified Scalar Multiplication?

[NullSpace0]

Homework Statement

Let V= set of 2x2 matrices with the normal addition, but where multiplication is defined as: β#A=β(A^T) where A^T is the transpose of A.

Homework Equations

The axiom about 1#A=A

The Attempt at a Solution

I think that because you can show that not ALL matrices satisfy A=A^T, you can't have a vector space since the multiplication by 1 doesn't hold up.

But then I'm wondering whether I'm assuming that the multiplicative identity should be the "normal" 1 (ie that 1 is just the scalar 1 in a normal R^n vector space).

How do you prove a multiplicative identity absolutely does NOT exist?

 

[STEMucator]

So are you trying to prove that V is a vector space if it obeys the normal matrix addition, but has a unique multiplication scalar multiplication defined as # which is sort of a mapping from A to A^t?

Your notation is a bit confusing to me.

 

[Dick]

Homework Statement

Let V= set of 2x2 matrices with the normal addition, but where multiplication is defined as: β#A=β(A^T) where A^T is the transpose of A.

Homework Equations

The axiom about 1#A=A

The Attempt at a Solution

I think that because you can show that not ALL matrices satisfy A=A^T, you can't have a vector space since the multiplication by 1 doesn't hold up.

But then I'm wondering whether I'm assuming that the multiplicative identity should be the "normal" 1 (ie that 1 is just the scalar 1 in a normal R^n vector space).

How do you prove a multiplicative identity absolutely does NOT exist?

If there were a scalar β that was a multiplicative identity it would have to satisfy β(A^T)=A for all matrices A. Show there isn't.