prove that if B is a skew-symmetric matrix, then its diagonal entries are zero?

Answer 1

B is skew-symmettric iff for all i, j, Bij = - Bji.
The diagonal entries are Bii for some i. Since B is skew-symmetric, Bii = -Bii (set j = i in the first equation). But if Bii = -Bii then 2Bii = 0 so Bii = 0. So all the diagonal entries are 0.

Answer 2

if B is skew-symmetric
then B= -B^T (the transpose)
so the diagonal entries: bii have to be equal to =- bii:
bii= -bii
2bii =0
=> bii=0 .

Answer 3

Skew-symmetric means A^T = -A. Try using the transpose operation on ABA^T and see what you get. Hopefully it comes out to -ABA^T. If you show that steps that's good enough for a proof.