[Math] Prove that the zero square matrices are the only matrices that are both symmetric and skew-symmetric.

Prove that the zero square matrices are the only matrices that are both symmetric and skew-symmetric.

My Proof

I will restate the proposition in a way that makes the proof easier to formulate:

There exists a unique matrix $A = 0_{n \times n}$, such that it is both symmetric and skew-symmetric.

$A = 0_{n \times n} = [a_{ij}]_{n \times n} = 0 \ \forall i,j$

$\implies -A = -[a_{ij}]_{n \times n}$ By the properties of scalar multiplication of matrices.

$\implies (-A)^T = -(A)^T = -[a_{ji}]_{n \times n}$ By the properties and definition of matrix transposition.

$= -0 \ \forall j,i$ By the definition of the zero matrix.

$= 0 = A$

and

$A^T = [a_{ji}]_{n \times n} = 0 \ \forall j,i$ By the definition of the zero matrix.

$= A$

$Q.E.D.$

$A = A^T = (-A)^T$

Let $B = B^T = (-B)^T$

We want to prove that $A = B$.

We now have $A = (-A)^T, A = A^T, B = (-B)^T, B = B^T$

Adding $A = (-A)^T, A = A^T$ and $B = (-B)^T, B = B^T$, we get

$2A = A^T + (-A)^T$ and $2B = B^T + (-B)^T$

$\implies A = \dfrac{1}{2}(A^T) + \dfrac{1}{2}(-A)^T$

and

$B = \dfrac{1}{2}(B^T) + \dfrac{1}{2}(-A)^T$

$A – B = \dfrac{1}{2}(A^T) + \dfrac{1}{2}(-A^T) – \dfrac{1}{2}(B^T) – \dfrac{1}{2}(-B^T)$

$= \dfrac{1}{2}(A^T) – \dfrac{1}{2}(A^T) – \dfrac{1}{2}(B^T) + \dfrac{1}{2}(B^T)$ Using the hypothesis.

$= 0$

$\implies A = B$

$Q.E.D.$

I would greatly appreciate it if people could please take the time to review my proof for correctness and provide feedback.