Let A be a given square matrix

Now, we can write,

$A=\dfrac {1}{2} (A+ A^\theta)+ \dfrac {1}{2} (A- A^\theta)= P + Q \ \text{ where,}$

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

Now $P^\theta = \dfrac {1}{2} (A+ A^\theta)^\theta = \dfrac {1}{2} \Big(A^\theta + (A^\theta)^\theta\Big) = \dfrac {1}{2} (A+A^{\theta})=P$

∴ P is Hermitian.

Also, $Q^\theta = \dfrac {1}{2} \Big((A-A^\theta)^\theta\Big)= \dfrac {1}{2} (-A+A^\theta) = -Q$

∴ Q is Skew-Hermitian.

Thus, we have expressed A as the sum of Hermitian and skew Hermitian matrix.

To prove Uniqueness

Let A=R+S,

Where R is hermitian and S is skew Hermitian be another representation of A,

Now, $A^\theta = (R+S)^\theta = R^\theta + S^\theta = R-S$

$ \therefore \dfrac {1}{2} (A+A^\theta) = \dfrac {1}{2} \Big [ (R+S) + (R-S) \Big ] = R \ \therefore R=P$

And, $\dfrac {1}{2}(A-A^\theta)=\dfrac {1}{2} \Big [ (R+S)-(R-S)\Big] = S \ \therefore S=Q$

Hence, Representation A=P+Q is unique.