Let, A be any square matrix.

Now, $ A = \dfrac{1}{2} (A+A') + \dfrac{1}{2} (A-A') \; = \; say, \; P+Q \; where \\ \; \\ P=\dfrac{1}{2} (A+A') \; \; and \; \; Q=\dfrac{1}{2} (A-A') \\ \; \\ To \; prove: \; P \; is \; symmetric, \; i.e. \; P' \; = P \; and \; Q \; is \; skew - symmetric, \; i.e. \; Q' \; = \; -Q \\ \; \\ \; \\ \therefore P'=\dfrac{1}{2} \Big(A+A'\Big)' \; = \; \dfrac{1}{2} \bigg(A'+(A')'\bigg) \; = \; \dfrac{1}{2} (A'+A) \; = \; P \\ \; \\ \therefore P \; is \; symmetric \\ \; \\ \; \\ \therefore Q'=\dfrac{1}{2} \Big(A-A'\Big)' \; = \; \dfrac{1}{2} \bigg(A'-(A')'\bigg) \; = \; \dfrac{1}{2} (A'-A) \; = \; -Q \\ \; \\ \therefore Q \; is \; skew-symmetric \\ \; \\ \; \\ \; \\ To \; prove \; uniqueness,\; let \; A=R+S \; where \; R \; is \;symmetric\; and\; S\; is\; skew- symmetric \\ \; \\ \; \\ \therefore A' \;=\; (R+S)' \; = \; R'+S' \; = \; R-S \\ \; \\ \{ \because R'=R \; and \; S'=-S \; by \; Definition \; of \; symmetric \; and \; skew-symmetric \; matrices \} \\ \; \\ \; \\ \therefore \dfrac{1}{2} (A+A') \; = \; \dfrac{1}{2} (R+S+R-S) \; = \; R\; = \; P \\ \; \\ \dfrac{1}{2} (A-A') \; = \; \dfrac{1}{2} (R+S-R+S) \; = \; S\; = \; Q $

Hence, the representation $ A=P+Q $ is unique. Hence, it is proved that every square matrix can be uniquely expressed as a sum of symmetric and skew-symmetric matrix.