Are Skew-Symmetric Matrices a Subspace of Mat2x2(ℝ)?

[mang733]

I was able to show that the set of all 2x2 "symmetric" matrices is a subspace of Mat2x2(lR) using the 3 axioms. However, I wasn't able to do the same with skew-symmetric (A^T=-A). anyone can help? Thanks.

 

[HallsofIvy]

There's reason you can't! If A is the matrix with entries [0 1] and [-1 0] it is skew symmetric. What is A²?

 

[matt grime]

Subspace, Halls, squaring it doesn't matter.

0^t=0=-0

If A^t =-A, then (kA)^t = -kA, for k a scalar

(A+B)^t = A^t+B^t, so if they're skew symmetric then this is -A-B = -(A+B)

so where did you proof go wrong?

 

[mathwonk]

the kernel of any linear map is a subspace.

consider the map taking A to (A + A^t). for symetric ones consider A goes to (A-A^t).

 

[mang733]

thanks for the help guys.