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Prove that $A^T + A$ is symmetric for any $n \times n$ matrix $A$. So I understand how matrix Transpose works but I'm not sure if there is an example I can use to prove that this works for ANY $n \times n$ matrix.

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    $\begingroup$ hint $(A^T)^T=A$ also $A+B=B+A$ $\endgroup$
    – jack
    Feb 21, 2016 at 22:45
  • $\begingroup$ Let the $(i,j)$ entry in $A$ be denoted by $a_{ij}$. The $(i,j)$-element of $A^\top$ is $a_{j,i}$. Then let $C=A+A^\top$. Then if the $(i,j)$ entry in $C$ is $c_{ij}$, it is $c_{ij}=a_{ij}+a_{ji}=c_{ji}$. $\endgroup$ Feb 21, 2016 at 22:49

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Consider the following identity:

$(A+B)^T=A^T+B^T$

Now apply it: $$ (A^T+A)^T=(A^T)^T+A^T=A+A^T.$$ A matrix is symmetric if it equals its transpose. Matrix addition is commutative, and so: $$ A^T+A=A+A^T=(A^T+A)^T.$$ Thus, $A^T+A$ is a symmetric matrix for any $n\times n$ matrix $A$.

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A square matrix $M$ is symmetric if (by definition) it coincides with its transpose, i.e. if $M^T=M$.

Now, if $M=A^T+A$, we have that $$ M^T=(A^T+A)^T=(A^T)^T+A^T=A+A^T=M\;\;\;\;, $$ where the second equality above, follows by linearity of the operator of transposition, and the third one follows because such operator is involutory.

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