If $P$ is an orthogonal matrix its columns form an orthonormal set.

Question

If $P$ is an $n\times{n}$ orthogonal matrix prove that the columns of $P$ form an orthonormal set in $R^n$. So I know that a matrix is orthogonal if $AA^T$=$I$ but I'm not sure how that would help me here. Any help is appreciated.

Answer 1

If $C_1,\ldots,C_n$ are the columns of $A$: $$A^tA=\begin{bmatrix}C_1^t\\{C_2^t}\\ \vdots\\{C_n^t}\end{bmatrix}\begin{bmatrix}C_1,C_2,\ldots,C_n\end{bmatrix}=\begin{bmatrix} C_1^tC_1 & C_1^tC_2 & \ldots & C_1^tC_n\\ C_2^tC_1 &C_2^tC_2 & \ldots & C_2^tC_n \\ \vdots&&&\vdots \\ C_n^tC_1 & C_n^tC_2 &\ldots & C_n^tC_n\end{bmatrix}$$ $$=\begin{bmatrix} \langle C_1,C_1\rangle & \langle C_1,C_2\rangle & \ldots & \langle C_1,C_n\rangle\\ \langle C_2,C_1\rangle & \langle C_2,C_2\rangle & \ldots & \langle C_2,C_n\rangle \\ \vdots&&&\vdots \\ \langle C_n,C_1\rangle & \langle C_n,C_2\rangle &\ldots & \langle C_n,C_n\rangle\end{bmatrix}.$$ As a consequence, $$A\text{ is orthogonal }\Leftrightarrow A^tA=I\Leftrightarrow \left \{ \begin{matrix} \langle C_i,C_j\rangle=0& i\neq j\\\langle C_i,C_i\rangle=1 & \forall i=1,\ldots,n,\end{matrix}\right.$$

Answer 2

Let's calculate $AA^T$. Notice that $(AA^T)_{ij}= \sum_{k=1}^n A_{ik}A^T_{kj}=\sum_{k=1}^n A_{ik}A_{jk}=\left\langle A_i,A_j\right\rangle$ where $A_i$ denotes the $i$-th row of $A$. Hence $\left\langle A_i,A_j\right\rangle=\delta_{ij}$ since $AA^T=I$. Now transpose and the same holds for the columns.

Answer 3

If $A A^{\top} = I$, then $A^{\top} = A^{-1}$, so that $A^{\top} A = I$.

Now note that the $(i, j)$-entry of $A^{\top} A$ is the scalar products of the $i$-th row of $A^{\top}$ (that is, the $i$-th column of $A$) by the $j$-th column of $A$.

Note the general fact that a single matrix equality compactly and conveniently encodes several scalar identities.