Proof Let B = A + A′, then
B′ = (A + A′)′
= A′ + (A′)′ (as (A + B)′ = A′ + B′)
= A′ + A (as (A′)′ = A)
= A + A′ (as A + B = B + A)
= B
Therefore
B = A + A′ is a symmetric matrix
Now let
C = A – A′
C′ = (A – A′)′ = A′ – (A′)′     (Why )
= A′ – A    (Why )
= – (A – A′) = – C
Therefore
C = A – A′ is a skew symmetric matrix Theorem 2 Any square matrix can be expressed as the sum of a symmetric and a
skew symmetric matrix