## Show that diagonal entries of a skew symmetric matrix are zero.

I'm pretty inexperienced in proof writing. So not sure if this was valid.

If a matrix is skew symmetric then A^T = - A, that is the transpose of A is equal to negative A.

This implies that if A = a(i,j), then a(j,i) = -a(i,j). If we're referring to diagonal entries, we can say a(j,j) = -a(j,j). The only way for this to be true is if a(j,j) = 0. So therefore all the diagonal entries of a skew symmetric matrix are 0.

Is this good enough?

Thanks.
 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity
 Recognitions: Homework Help I think it would work as a valid proof.

 Quote by inknit I'm pretty inexperienced in proof writing. So not sure if this was valid. If a matrix is skew symmetric then A^T = - A, that is the transpose of A is equal to negative A. This implies that if A = a(i,j), then a(j,i) = -a(i,j). If we're referring to diagonal entries, we can say a(j,j) = -a(j,j). The only way for this to be true is if a(j,j) = 0. So therefore all the diagonal entries of a skew symmetric matrix are 0. Is this good enough? Thanks.
It's great. Thanks!

Recognitions:
Gold Member