For a matrix $O$ containing columns which are an orthonormal basis for a column space, why does $O^{\prime}O = I$?

Question

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 $O^{\prime}O = I$?

Answer 1

Note that, by the definition of matrix multiplication, the $i,j$ entry of $O'O$ is the inner product (dot-product) of the columns $o_i$ and $o_j$.

By the definition of an orthonormal basis, the conclusion follows.