Let, A be a given square matrix.

We can write A as-

$ A \; = \; \dfrac{1}{2}(A+A^{\theta}) + \dfrac{1}{2}(A-A^{\theta}) \; = \; say, \; P+Q \; where \; P=\dfrac{1}{2}(A+A^{\theta}) \; and \; Q=\dfrac{1}{2}(A-A^{\theta}) \\ \; \\ \; \\ Now, \;\; P^{\theta}=[\dfrac{1}{2}(A+A^{\theta})]^{\theta} \; = \; \dfrac{1}{2}(A^{\theta}+(A^{\theta})^{\theta}) \; =\; \dfrac{1}{2}(A^{\theta}+A) \; \ldots (\because (A^{\theta})^{\theta}) \\ \; \\ P^{\theta} \; = \; \dfrac{1}{2}(A^{\theta}+A) \; = \; \dfrac{1}{2}(A+A^{\theta}) \; = \; P \\ \; \\ \; \\ Hence, \; P \; is \; a \; Hermitian \; Matrix. \; \ldots(By \; Definition) \\ \; \\ \; \\ \;\\ Also , \; Q^{\theta}=[\dfrac{1}{2}(A-A^{\theta})]^{\theta} \; = \; \dfrac{1}{2}(A^{\theta}-(A^{\theta})^{\theta}) \; =\; \dfrac{1}{2}(A^{\theta}-A) \; \ldots (\because (A^{\theta})^{\theta}) \\ \; \\ Q^{\theta}=\dfrac{-1}{2}(A-A^{\theta}) \; = \; -Q \\ \; \\ \; \\ Hence, \; Q \; is \; a \;skew- Hermitian \; Matrix. \; \ldots(By \; Definition) \\ \; \\ \; $

To prove uniqueness: Let, A=R+S, where R is a Hermitian and S is a skew-Hermitian matrix, be another representation of A.

$ Now, A^{\theta} \; = \; (R+S)^{\theta} \; = \; R^{\theta}+S^{\theta} \; = \; R-S \ldots (\because R^{\theta} = R \; \& \; S^{\theta} \; = -S \ldots By \; Definition ) \\ \; \\ \therefore \dfrac{1}{2}(A+A^{\theta}) \; = \; \dfrac{1}{2}[(R+S)+(R-S)] \; = \; R. \; \; But \; \dfrac{1}{2}(A+A^{\theta}) \;=\;P. \; \; \; \; \therefore R=P \\ \; \\ \; \\ \; \\ Also, \; \dfrac{1}{2}(A-A^{\theta}) \; = \; \dfrac{1}{2}[(R+S)-(R-S)] \; = \; S. \; \; But \; \dfrac{1}{2}(A-A^{\theta}) \;=\;Q. \; \; \; \; \therefore S=Q \\ \; \\ \; $

Hence, every square matrix can be uniquely expressed as a sum of Hermitian and skew-Hermitian Matrix.