Let A be a square matrix of order n.

Part I:

$A=\dfrac{1}{2}(A+A)$

$\therefore A=\dfrac{1}{2}(A+A^{\theta}+A-A^{\theta})$

Let $P=\dfrac{1}{2}(A+A^{\theta})and\ Q=\dfrac{1}{2}(A-A^{\theta})\rightarrow (1)$

Now, $P^{\theta}=\left [ \dfrac{1}{2}\left ( A+A^{\theta} \right ) \right ]^{\theta}=\dfrac{1}{2}\left ( A+A^{\theta} \right )^{\theta}=\dfrac{1}{2}\left [ A^{\theta}+\left ( A^{\theta} \right )^{\theta} \right ]=\dfrac{1}{2}\left [ \dfrac{A^{\theta}}+A \right ]=P$

$\therefore$ P is Hermition.

Similarly,

$Q^{\theta}=\left [ \dfrac{1}{2} \left ( A-A^{\theta} \right )\right ]^{\theta}=\dfrac{1}{2}\left ( A-A^{\theta} \right )^{\theta}=\dfrac{1}{2}\left [ A^{\theta}-(A^{\theta})^{\theta} \right ]=\dfrac{1}{2}[A^{\theta}-A]=\dfrac{-1}{2}\left [ A-A^{\theta} \right ]=-Q$

$\therefore$ Q is skew-Hermitian.

Hence, A=P+Q, where P is Hermitian and Q is skew-Hermitian.

$\therefore$ Every square matrix can be expressed as the sum off Hermitian and skew-Hermitian matrices.

Part II: Uniqueness

Let, if possible, $A=R+S \rightarrow (2)$ where R is Hermitian and S is skew -Hermitian.

$\therefore R=R^{\theta}\ and\ S=-S^{\theta}\rightarrow (3)$

Consider, $A^{\theta}=(R+S)^{\theta}=R^{\theta}+S^{\theta}=R-S (From\ 2)\rightarrow (4)$

$\therefore $ From (1),

$P=\dfrac{1}{2}(A+A^{\theta})=\dfrac{1}{2}\left [ (R+S)+(R-S) \right ]=\dfrac{1}{2}[2R]=R(From\ 2\ and\ 4)$

Similarly, From (1),

$Q=\dfrac{1}{2}(A-A^{\theta})=\dfrac{1}{2}\left [ (R+S)-(R-S) \right ]\dfrac{1}{2}=S(From\ 2\ and\ 4)$

$\therefore A=R+S=P+Q$

Hence, the representation is unique.