Suppose $Q^T$ equals $Q^{-1}$ (transpose equals inverse, so $Q^TQ^{-1} = 1$).
a)Show that the columns $q_l, ... , q_n$ are unit vectors: $||q_1||^2 = 1$.
b)Show that every two columns of $Q$ are perpendicular: $q_1^{T}q_2 = 0$.
This problem is connected with the definition of dot product: multiplication of pairwised columns and rows.
So, a)by the definition of dot product if $i = j$, then $\sum_{i=1}^n q_i^2 = 1$
b) The dot product isn't defined for i does't equal to j. So, when i does't equal to j, then $q_1^{T}q_2 = 0$.
Please correct my logic, because i am not sure if i am on the right way.
Since $Q^TQ=I$, we have $(Q^TQ)_{ij}= \begin{cases} 1 &\ i=j\\ 0 &\text{otherwise.} \end{cases}$
Also, from matrix multiplication, note that $(Q^TQ)_{ij}=(q_i \cdot q_j)$. Hence, \begin{equation} (q_i \cdot q_j)=\begin{cases} 1 &\ i=j\\ 0 &\text{otherwise.} \end{cases} \end{equation} This completes both proofs.