[Math] Show that $A$ is orthogonal iff the column vectors of $A$ form an orthonormal basis

Question

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 .

Answer 1

Hint: $AA^T=I$ if and only if $A^TA=I$. So $A$ is orthogonal if and only if $A^T$ is orthogonal.