Let $C^{n \times n}$ be a square matrix.

Prove that $$C=\frac{1}{2}(C+C^T)+\frac{1}{2}(C-C^T)$$

What I have manage so far is:

a. Let $S$ be a Symmetric Matrix so $S=C+C^T$

b. Let $N$ be a Skew-Symmetric Matrix so $N=C-C^T$




  • $\begingroup$ I think you mean that you want to prove that the two terms you've divided $C$ into are correctly symmetric and skew-symmetric. Also, your last equality is wrong! $\endgroup$ – Semiclassical Jul 23 '14 at 13:33
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    $\begingroup$ BTW $N^t=\color{red}{-}N$ $\endgroup$ – Algebraic Pavel Jul 23 '14 at 16:07
  • $\begingroup$ @AlgebraicPavel now its ok? $\endgroup$ – gbox Jul 23 '14 at 16:11

If $A$ is a square matrix in $\mathcal M_n(\Bbb C)$ then $$A=\underbrace{\frac12(A+A^t)}_{\text{symmetric matrix}}+\underbrace{\frac12(A-A^t)}_{\text{skew-symmetric matrix}}$$

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    $\begingroup$ Why should it be multiple by $1/2$? $\endgroup$ – gbox Jul 23 '14 at 15:16
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    $\begingroup$ Without the coefficient $\frac12$ the sum gives $2A$. $\endgroup$ – user63181 Jul 23 '14 at 15:21

$$(A+A^t)_{ij} =A_{ij}+A^t_{ij} = A^t_{ji}+A_{ji} = (A^t+A)_{ji} = (A+A^t)_{ji}$$ Therefore $(A+A^t)$ is symmetric. $$(A-A^t)_{ij} =A_{ij}-A^t_{ij} = A^t_{ji}-A_{ji} = (A^t-A)_{ji} = -(A-A^t)_{ji}$$ Therefore $(A-A^t)$ is scew-symmetric. Since $$A = \frac12A+\frac12A+\frac12A^t-\frac12A^t=\frac12(A+A^t)+\frac12(A-A^t)$$ $A$ can be Written as a sum of a scew-symmetric and a symmetric Matrix.


Using indicial notation, show that the square matrix can be expressed as the sum of a symmetric and antisymmetric matrix by the decomposition:

$$A=\frac{A+A^T}{2} + \frac{A-A^T}{2}$$


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