Definition Let $A = \left[ a_{ij} \right]_{m \times n}$. The transpose $A^T$ of $A$ is the $n \times m$ matrix $A^T = \left[ \hat{a_{ij}} \right]$ where $\hat{a_{ij}} = a_{ji}$.
$A$ is symmetric if $A^T = A$
$A$ is skew-symmetric if $A^T = -A$
$A$ is ortoghonal if $A$ is invertible with inverse $A^{-1} = A^T$
Note that in all three cases $A$ is required to be square.
Theorem If $P$ is a permutation matrix, then $P$ is orthogonal (and so invertible).
Proof Let $P = \left[ p_{ij} \right]$ be a permutation matrix.
For all $i$, let $k(i)$ be such that $p_{ik} = 1$ if $k = k(i)$ and $p_{ik} = 0$ otherwise. Then
\begin{equation}
\left[ PP^T \right]_{ij} = \sum_{k = 1}^n{p_{ik}\hat{p_{kj}}} = \sum_{k = 1}^n{p_{ik}p_{jk}} = \cdots + 0 + p_{ik(i)} p_{jk(i)} + 0 + \cdots = 1 p_{jk(i)} = p_{jk(i)}
\end{equation}
Now if $i = j$, then $p_{jk(i)} = p_{ik(i)} = 1$. Otherwise $k(i) \not= k(j)$, because $P$ is a permutation matrix and so $p_{jk(i)} = 0$. Hence
\begin{equation}
\left[ PP^T \right]_{ij} =
\begin{cases}
1 & \text{if } i = j \\
0 & \text{otherwise}
\end{cases}
\end{equation}
and $PP^T = I_n$. Then also $P^TP = I_n$, since $P^T$ is also a permutation matrix and $P$ is its transpose, so we can apply the same argument as before again.
Proposition Let $A, B$ be $m \times n$ matrices, $C$ an $n \times p$ matrix and $\lambda \in \mathbb{R}$. Then
$(A^T)^T = A$
$(A + B)^T = A^T + B^T$
$(\lambda A)^T = \lambda A^T$
$(BC)^T = C^T B^T$
2 Vector spaces
Theorem Let $V$ be a vector space over $\mathbb{R}$. For all $\lambda \in \mathbb{R}$ and $v \in V$
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Matej Balog