Why does an orthogonal matrix have a transpose that equals its inverse? Wikipedia says the following: 



How does it follow from the fact that an orthogonal matrix whose columns are orthonormal that the transpose of the matrix is its inverse?
 A: Let's consider what happens when we multiply $Q^TQ$.
The $ij$ entry with $i\neq j$ is the dot product of row $i$ of $Q^T$ (i.e., column $i$ of $Q$) with column $j$ of $Q$. Since the columns of $Q$ are orthogonal, this is $0$.
The $ii$ entry is the dot product of column $i$ of $Q$ with itself, which is always $1$ because the columns of $Q$ are normal.
So $Q^TQ$ has $1$s down the diagonal and $0$s elsewhere; i.e., it is the identity matrix.
Thus $Q^T$ is the inverse of $Q$.
A: Assume that the columns of $Q$ are orthonormal.
$$(Q^TQ)_{ij} = \sum_{k=1}^n Q^T_{ik}Q_{kj} = \sum_{k=1}^n Q_{ki}Q_{kj} = \delta_{ij}$$
because the last sum is the inner product of $i$-th and $j$-th columns of $Q$.
Hence $Q^TQ = I$, so $Q$ is invertible and $Q^{-1} = Q^T$.
Therefore $$Q^TQ = QQ^T = I$$
A: It can be seen on $2\times 2$ case: if $Q=\begin{pmatrix}a&c\\b&d\end{pmatrix}$ has orthonormal columns, i.e. $a^2+b^2=1, c^2+d^2=1, ac+bd=0$, calculate $Q^\top Q$:
$$Q^\top Q=\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}a&c\\b&d\end{pmatrix}=\begin{pmatrix}a^2+b^2&ac+bd\\ac+bd&b^2+d^2\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I$$
and, conversely, $Q^\top Q=I$ implies columns of $Q$ are orthonormal. The proof in the general case is identical, but a bit harder to write, so I will let you get the intuition from the $2\times 2$ case and try to write the general proof yourself.
A similar statement can be proven for rows, and with replacing $Q^\top Q$ with $QQ^\top$.
Thus, we have the equivalences:
columns of $Q$ are orthonormal $\Longleftrightarrow Q^\top Q=I \Longleftrightarrow QQ^\top=I \Longleftrightarrow$ rows of $Q$ are orthonormal
A: If $\{e_i\}_{i=1}^N$ is an orthonormal basis, then, for all $1 \le i, j \le N$, $e_i \circ e_j = \delta_{i,j}=
\begin{cases} 
   1 & \text{If $i = j$} \\
   0 & \text{If $i \ne j$} \\
\end{cases}$.
This leads directly to $Q^TQ= QQ^T = I$
