For Problem DD-3: Suppose $v _ 1,\dots, v _ n$ are orthonormal. Suppose $r _ 1 v _ 1 + \dots + r _ n v _ 2 =$   0. We must show that $r _ 1 = \dots = r _ n = 0$. Take the dot product of both sides with $v _ 1$. Since the dot product is linear in each of its two arguments (with the other held fixed), we get $r _ 1 v _ 1 \cdot v _ 1 + r _ 2 v _ 1 \cdot v _ 2 + \dots + r _ n v _ 1 \cdot v _ n = 0$. Since $v _ i \cdot v _ j = \delta _ {ij}$, this says $r _ 1 = 0$. Similarly, dotting both sides with each $v _ i$ in turn, we get $r _ i = 0$ for all $i$, so the vectors are linearly independent.

For Problem DD-4:

In a matrix product $AB$, the entries are the dot products of the rows of $A$ with the columns of $B$. In $P ^ t P$, the rows of $P ^ t$ are the same as the columns of $P$, transposed. If the columns of $P$ are $v _ 1,\dots, v _ n$, then, the entries of $P ^ t P$ are the numbers $v _ i \cdot v _ j$. Therefore $P _ t P = I$ when $v _ i \cdot v _ j = I _ {ij} = \delta _ {ij}$ (the Kronecker delta symbol), which says the columns of $P$ are orthonormal.

For Problem DD-5: It's an orthogonal matrix (orthonormal columns), so just take the transpose.

For Problem DD-6:

$P ^ t P = I$ implies $\det(P ^ t P)= \det I = 1$. Since determinants are compatible with multiplication, $\det {P ^ t} \det P =$. Since $\det (P ^ t) = \det P$, we get $(\det P) ^ 2 = 1$. The solutions of $x ^ 2 = 1$ are $x = \pm 1$ so $\det P = \pm 1$.