Theorem: let $\{o_i\}_{i \in \{1, 2, \dots, r\}}$ be an orthonormal basis for the column space of a matrix $X$ and let $O = \begin{bmatrix}o_1 & o_2 & \cdots & o_r\end{bmatrix}$. Then $OO^{\prime} = \sum\limits_{i=1}^{r}o_io^{\prime}_i$ is the perpendicular projection operator onto $C(X)$.
In the proof, they say $OO^{\prime}OO^{\prime} = OIO^{\prime}$, $I$ being the identity matrix. How do I know that $O^{\prime}O = I$?