For any square matrix C, prove that a.C+Ctranspose is symmetric b.C-Ctranspose is skew symmetric?

Question

c. C is a sum of a symmetric matrix and a skew-symmetric matrix

Answer 1

a) Let D = C + C(T) where C(T) means C transpose.
Then for every i, j we have Dij = Cij + Cji and Dji = Cji + Cij. So Dij = Dji and therefore D is symmetric.

b) Let E = C - C(T) where C(T) means C transpose.
Then for every i, j we have Eij = Cij - Cji and Eji = Cji - Cij. So Eij = -Eji and therefore E is skew-symmetric.

c) Define D and E as above and consider D + E.
D + E = (C + C(T)) + (C - C(T)) = 2C, so C = (1/2)D + (1/2)E.
D is symmetric, so (1/2) D is symmetric, and E is skew-symmetric, so (1/2) E is skew-symmetric. So C is a sum of a symmetric matrix and a skew-symmetric matrix.

Answer 2

Randy P gives one half of the proof (existence). For the other half, assume there are two representations as a sum of a symmetric matrix and a skew-symmetric matrix. That is, assume we have two symmetric matrices A and C and two skew-symmetric matrices B and D such that A + B = C + D Rewrite this as A - C = D - B Show (or simply note) that the left side is symmetric and the right side is skew-symmetric. Finally, show (if you haven't already) that the only matrix both symmetric and skew-symmetric is the zero matrix. This will show that A - C = D - B = 0, which leads to A = C and B = D, which proves uniqueness. As a sidenote for anyone reading between the lines, this does require that 2 isn't a zero divisor. However, if you accept the existence proof given by Randy P, then you're already assuming 2 is invertible, so obviously it's not a zero divisor. I don't personally know if any of this statement holds when 2 is not invertible, or even worse, if it's a zero divisor.

Answer 3

X=MC squared