Proof that $(A^t)^t=A$, $(A+B)^t=A^t+B^t$, $(AB)^t=B^tA^t$, and deduce that $BB^t$ is symmetric and $B-B^t$ is skew-symmetric In a linear algebra textbook, I was given the following problem:

If $B$ is a $n \times n$ square matrix, show that $BB^t$ is symmetric and $B-B^t$ is skew-symmetric.


I know that there are relatively quick solutions using elementary matrix arithmetic. But I came up with a different way and wanted to ask if this is correct. Thereby I hope it is accepted to post two problems/solutions into one post, since these are quite connected. If not, please correct me.
To show that $BB^t$ is symmetric, let $A=BB^t$. I denote entries by $a$, $b$ and $b^t$ and show that $a_{ij}=a_{ji}$ for every entry of $A$. By the definition of matrix multiplication $a_{ij}=\sum_{k=0}^n b_{ik}b^t_{kj}$. But $b_{ik}=b^t_{ki}$ and $b^t_{kj}=b_{jk}$ by matrix transposition. So we have $$a_{ij}=\sum_{k=0}^n b_{ik}b^t_{kj}=\sum_{k=0}^n b_{jk}b^t_{ki}=a_{ji}$$ as required.
To show that $B-B^t$ is skew-symmetric, let $A=B-B^t$. We show that $a_{ij}=-a_{ji}$ for every entry of $A$. By definition of a difference of matrices $a_{ij}=b_{ij}-b^t_{ij}$. But $b_{ij}=b^t_{ji}$ and $b^t_{ij}=b_{ji}$ by matrix transposition. So we have
$$a_{ij}=b_{ij}-b^t_{ij}=b^t_{ji}-b_{ji}=-b_{ji}+b^t_{ji}=-(b_{ji}-b^t_{ji})=-a_{ji}$$
which completes our proof.
 A: In the following, I try to use the comments of @Branimir Ćaćić and @Paul Sinclair to expand my original solution by proofing some general rules of matrix arithmetic and deducing the results on symmetry and skew-symmetry from them.
Proposition: Let $A$ and $B$ be two matrices, whose size allows the following operations. Then
$$ \begin{align} &A) & (A^t)^t &= A \\
&B) & (A+B)^t &= A^t+B^t \\
&C) & (AB)^t &= B^tA^t \end{align}$$
Proof: For the first proposition notice that $a_{ij}=a_{ji}^t$ for every entry in $A$ and $A^t$ respectively. Transposing $A^T$ again yields $a_{ji}^t=a_{ij}$ as required.
To proof $B)$ let $A + B = C$ and $A^t + B^t = D$. We show that $c_{ji}^t = d_{ij}$ for every $c_{ji}^t$ and $d_{ij}$ in $C^t$ or $D$, respectively. Now, $c_{ji}^t = c_{ij} = a_{ij} + b_{ij}$ and $d_{ij} = a_{ji}^t + b_{ji}^t$ by definition of a sum of matrices. Since we also have $a_{ij} = a_{ji}^t$ as well as $b_{ij} = b_{ji}^t$ due to transposition of a matrix, we get
$$c_{ji}^t = c_{ij} = a_{ij}+b_{ij} = a_{ji}^t+b_{ji}^t = d_{ij}$$
which completes the proof.
For the proof of $C)$ let $A B = C$ and $A^t B^t = D$. Similar to the previous argument we will show that $c_{ji}^t = d_{ij}$ for every $c_{ji}^t$ and $d_{ij}$ in $C^t$ or $D$. Notice that $c_{ji}^t = c_{ij} = \sum_{k=1}^n a_{ik} b_{kj}$ and $d_{ij} = \sum_{k=1}^n b_{jk}^t a_{ki}^t$, due to matrix multiplication. Since by matrix transposition we also have $a_{ik} = a_{ki}^t$ as well as $b_{kj} = b_{jk}^t$ we come to the conclusion that
$$c_{ji}^t = c_{ij} = \sum_{k=1}^n a_{ik} b_{kj} = \sum_{k=1}^n b_{jk}^t a_{ki}^t = d_{ij}$$
as required.
Corollary: Per definition, a matrix is symmetric, if for any matrix $B^t=B$ and skew-symmetric if $B^t=-B$. We now can immediately deduce, by $A)$ and $C)$, that
$$(B B^t)^t=(B^t)^t B^t = B B^t$$
which shows, that $BB^t$ is symmetric and, using $B)$, that
$$(B - B^t)^t = B^t - (B^t)^t =  B^t - B = - (B - B^t)$$
so $B - B^t$ is skew-symmetric. $\blacksquare$
