prove that the diagonal elements of a scew symmetric matrix are all zero
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.
a_ij = [the row i, column j entry of A]
= [the row i, column j entry of -A^t]
= -[the row i, column j entry of A^t]
= -[the row j, column i entry of A]
In particular, taking j = i we see that for all 1 <= i <= n one has
a_ii = -a_ii.
Adding a_ii to both sides we deduce 2 a_ii = 0 for all 1 <= i <= n and hence a_ii = 0 for all 1 <= i <= n. So every diagonal entry of A is 0.