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I'd appreciate if someone could find a better title for this question, for I'm short of ideas right now.

Given a matrix $A \in R^{n,n}$, show that $$ \frac{1}{2}(A + A^t) $$ is symmetric.

I see that it's symmetric and it seems obvious, but I don't really know how to show that in particular.

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3 Answers 3

up vote 2 down vote accepted

$(B+B^t)^t=B^t+(B^t)^t=B^t+B$

So $B+B^t$ is symmetric $\forall B\in R^{n\times n}$.

Take $B=A/2$ and you get the desired result.

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Are you familiar with 'entry notation' to represent arbitrary elements in a matrix? I am omitting the scalar, but the idea is to show $ent_{ij}(A+A^T) = ent_{ij}(A+A^T)^T$, where

$$ent_{ij}(A^T) = ent_{ji}(A).$$

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You should apply the known properties of transposition:

  1. $(cX)^t=cX^t$ (where $c$ is a scalar and $X$ is any matrix);

  2. $(X+Y)^t=X^t + Y^t$ (where $X$ and $Y$ are matrices and the sum is defined).

Moreover, recall that $X+Y=Y+X$. That's all you need: set $$ C=\frac{1}{2}(A+A^t) $$ and apply the properties to show that $C^t=C$.

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