My book says that:
(a) $A$ is orthogonal if and only if the row vectors of $A$ form an orthonormal basis of $\mathbb{R}^n$ under the euclidean inner product; and
(b) $A$ is orthogonal if and only if the column vectors of $A$ form an orthonormal basis of $\mathbb{R}^n$ under the euclidean inner product.
And then they proof (a):
Let $\text{r}_1,...,\text{r}_n$ denote the row vectors of A. Then
$$AA^T= \begin{bmatrix} \text{r}_1 \\ \vdots \\ \text{r}_n \end{bmatrix} \begin{bmatrix} \text{r}_1 & \cdots & \text{r}_n \end{bmatrix}= \begin{bmatrix} \text{r}_1 \cdot \text{r}_1 & \cdots & \text{r}_1 \cdot \text{r}_n \\ \vdots & &\vdots \\ \text{r}_n \cdot \text{r}_1 & \cdots & \text{r}_n \cdot \text{r}_n \end{bmatrix}. $$
It follows that $AA^T=I$ iff
$$\text{r}_i \cdot \text{r}_j= \begin{cases} 1 & \text{if} \ i=j \\ 0 & \text{if} \ i\ne j \end{cases} $$
iff $\text{r}_1,...,\text{r}_n$ are orthonormal.
Then they say that the proof of (b) is almost identical.
Can anyone help I am little bit confused how to show it actually .
Hint: $AA^T=I$ if and only if $A^TA=I$. So $A$ is orthogonal if and only if $A^T$ is orthogonal.