# Proof $B-B^T$ is Skew Symmetric [closed]

How can one prove that if $B$ is an $n\times n$ matrix, then the matrix $B-B^T$ ($B$ minus its transpose) is skew-symmetric?

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## closed as off-topic by Najib Idrissi, amWhy, Behaviour, Jean-Claude Arbaut, hardmathSep 16 at 12:53

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Have you tried taking the transpose? Do you know what $(B^T)^T$ is? –  Jason DeVito Nov 9 '11 at 3:55
Did you try following the hint you got at MathOverflow? mathoverflow.net/questions/80447/… –  Gerry Myerson Nov 9 '11 at 4:45

The transpose has the following properties: $$(A + B)^T = A^T + B^T$$ $$(A^T)^T = A$$

Hence

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

So $B - B^T$ is skew-symmetric by definition.

To prove the above properties of the transpose simply consider the entries of the matrices. We have $$(A+B)^T_{ij} = (A+B)_{ji} = A_{ji} + B_{ji} = A^T_{ij} + B^T_{ij}$$ and $$(A^T)^T_{ij} = (A^T)_{ji} = A_{ij}$$

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The wiki tells me that $(A+B)^\intercal = A^\intercal + B^\intercal$. Use this property of the transpose and use the definition of skew symmetric and see whether you can show the required property.

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