Is the matrix used to prove independence of $\bar Y $ and $S^2$ for a random normal sample orthogonal? I'm looking at a proof that $\bar Y $ and $S^2$ for a sample of $n$ random normal variables are independent using two different sources. Both sources makes use of identical matrices linked in the images below.
In the first source, the matrix is defined as $A$ but the author states that $A'A = A^TA =I$. In the second source, the matrix is defined as $B^T$ (so $B^T = A$), but the author also states that $B^T B = I$.
If I denote this matrix as $A$, is $A$ orthogonal or does $A^T A = AA^T = I$? Otherwise the two sources seem to contradict.
From Statistical Models by Davison
From Mathematical Statistics by Wackerly
 A: The matrix is orthogonal as it consists of orthogonal unit row and column vectors. The norm of the first row is$$\sum_{i=1}^n\left(\frac1{\sqrt n}\right)^2=1$$The norm of $j^\text{th}(j>1)$ row is$$\sum_{i=1}^{j-1}\left(\frac1{\sqrt{j(j-1)}}\right)^2+\left[\frac{1-j}{\sqrt{j(j-1)}}\right]^2=1$$The norm of the $i^\text{th}$ column is given by$$\left(\frac1{\sqrt{n}}\right)^2+\left[\frac{1-i}{\sqrt{i(i-1)}}\right]^2+\sum_{j=i+1}^n\left(\frac1{\sqrt{j(j-1)}}\right)^2=1$$

 Use the fact $\frac1{j(j-1)}=\frac1j-\frac1{j-1}$.


The dot product of $i^\text{th},j^\text{th}(j>i)$ columns is$$\frac1{\sqrt n}\cdot\frac1{\sqrt n}+\frac{1-j}{\sqrt{j(j-1)}}\cdot\frac1{\sqrt{j(j-1)}}+\sum_{k=j+1}^n\left(\frac1{\sqrt{k(k-1)}}\right)^2=0$$
The dot product of the first and $j^\text{th}(j\ne1)$ rows is$$\frac1{\sqrt n}\left[\sum_{i=1}^{j-1}\frac1{\sqrt{j(j-1)}}+\frac{1-j}{\sqrt{j(j-1)}}\right]=0$$
The dot product of $i^\text{th},j^\text{th}(i\ne j,j>i>1)$ rows is$$\frac1{\sqrt{j(j-1)}}\left[\sum_{k=1}^{i-1}\frac1{\sqrt{i(i-1)}}+\frac{1-i}{\sqrt{i(i-1)}}\right]=0$$
