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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.

  1. We first prove that the object exists.

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

$\implies -A = -[a_{ij}]_{n \times n}$ 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.

$= A$

$Q.E.D.$

  1. We must now prove that $A$ is unique.

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

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.

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You cannot assume neither $A$ nor $B$ is zero in the second part. For this part, if $A$ is both symmetric and skew symmetric, then $A=(A^{T})^{T}=(-A)^{T}=-A^{T}=-A$, so $2A=0$, then $A=0$.

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  • $\begingroup$ Thanks for the response. I don't see why I cannot assume this; the assumption that $A$ is the zero matrix is part of the hypothesis, and so it is assumed by definition? $\endgroup$ – The Pointer Dec 9 '17 at 5:44
  • $\begingroup$ The assumption does not say that, rather, it means partly that, if $A$ is both symmetric and skew symmetric, prove that $A=0$, this is the keyword "the only". $\endgroup$ – user284331 Dec 9 '17 at 5:45
  • $\begingroup$ Hmm. I'm interpreting it conversely; namely that, If we have a zero square matrix, then the matrix is both symmetric and skew-symmetric. Therefore, the assumption would be that we have a zero square matrix (in this case, $A$), and we must prove the conclusion that it is both symmetric and skew-symmetric. Can anyone else please clarify this? $\endgroup$ – The Pointer Dec 9 '17 at 5:53
  • $\begingroup$ Actually there are two parts: First, show that the zero matrix fulfils the condition. Second, show that if any matrix $A$ satisfies the condition, then $A$ is the zero matrix. $\endgroup$ – user284331 Dec 9 '17 at 5:55
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    $\begingroup$ Now I know why you were confused. If the question says that, prove that the matrix which satisfies the condition of being symmetry and skew-symmetry is unique (so how does the matrix look like? We simply have no idea by just this kind of uniqueness), then your reasoning works. But now the question indicates particularly that it is unique and it is the zero matrix, so at the final stage you need to realise either $A$ or $B$ to be the zero matrix. $\endgroup$ – user284331 Dec 9 '17 at 17:30

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