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Express matrix as the sum of a symmetric and a skew-symmetric matrix: [ rr 3 & 5 1 & -1 ]

Vidyadip · Accepted Answer

As per Theorem “Any square matrix can be expressed as the sum of a symmetric and skew-symmetric matrix.” So in order to prove this, we will be using Theorem which states that “For any square matrix A with real number entries, A + A’ is a symmetric matrix and A – A’ is a skew-symmetric matrix.” Now, Let A = [ cc 3 & 5 1 & -1 ] Therefore, A' = [ cc 3 & 1 5 & -1 ] Now, on adding A and A’ we will get, A+A^ = [ cc 3 & 5 1 & -1 ]+ [ cc 3 & 1 5 & -1 ] A+A^ = [ cc 3+3 & 5+1 1+5 & -1+(-1) ] A+A^ = [ cc 6 & 6 6 & -2 ] Now, Let M = 1 2 (A+A^ ) Therefore, M=1 2 [ cc 6 & 6 6 & -2 ] = [ cc 3 & 3 3 & -1 ] Now, M^ = [ cc 3 & 3 3 & -1 ] M' = M Thus M = 1 2 (A+A^ ) is a symmetric matrix as M' = M. Now on subtracting A’ from A we will get, A-A^ = [ cc 3 & 5 1 & -1 ]- [ cc 3 & 1 5 & -1 ] A-A^ = [ cc 0 & 4 -4 & 0 ] Now, Let N = 1 2 (A-A^ ) Therefore, N=1 2 [ cc 0 & 4 -4 & 0 ] N= [ cc 0 & 2 -2 & 0 ] Now, N' = [ cc 0 & -2 2 & 0 ] N^ =(-1) [ cc 0 & 2 -2 & 0 ] N' = - N Thus M = 1 2 (A+A^ ) is a skew-symmetric matrix as N' = -N Now, Add M and N, we get, M+N= [ cc 3 & 3 3 & -1 ]+ [ cc 0 & 2 -2 & 0 ] M+N= [ cc 3+0 & 3+2 3+(-2) & -1+0 ] M+N= [ cc 3 & 5 1 & -1 ] So we see here, M + N = [ cc 3 & 5 1 & -1 ] = A Thus, A is represented as the sum of a symmetric matrix M and a skew-symmetric matrix N. Hence proved.

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