Orthogonal matrix with single $0$ entry To my surprise there is an orthogonal matrix dimension $3 \times 3$ with a single $0$ entry as  it was shown in this   answer.   
Moreover it was possible to identify the pattern for matrix entries to satisfy this condition i.e.
$$Q=\begin{bmatrix} 0 & -a & -b \\  a   &  -b^2 &  ab \\ b & ab & -a^2\end{bmatrix}$$
when $a,b$ are constrained by $a^2+b^2=1$.
For orthogonal matrix dimension $2 \times 2$ single $0$ entry is not possible.
Here my new questions: 


*

*Is it possible a single $0$ entry for orthogonal matrix dimension $4 
   \times 4$ ?

*Can we  also provide for this case some more general formula? (to be more specific in a similar form as it was shown above for 3d)

*Is the method listed above the only one for orthogonal matrices dimension $3 
   \times 3$ or can we generate such matrix also with some other substantially different algorithm?
 A: I can answer your first question, but the other two seem less reasonable to answer concisely.
It's possible to find such a matrix for any $n\ge 3$, and there are tons of them. Let's just choose the first column to be $w'=(0,1,\dots,1)^T$ and $w = w'/||w'||$. We are thus looking to complete $w$ to an orthonormal basis $w,v_1,\dots,v_{n-1}$ with no coordinate of any $v_i$ equal to $0$. The space of vectors orthogonal to $w$ is the space 
$$W = \{(x_1,\dots,x_n)^T : x_2+\dots+x_n = 0\},$$
and we want to find an orthonormal basis of this space with no zero coordinates. That is, we want to find an orthonormal basis of this space which avoids the $n$ hyperplanes
$$A_i = \{(x_1,\dots,x_n)^T : x_i = 0\}.$$
The key use of having $n\ge 3$ is so that $W \cap A_i$ is $(n-2)$-dimensional for each $i$. When $n=2$, for example, $W \cap A_2$ is $1$-dimensional. Now, for each $i$, let $B_i = W \cap A_i$, an $(n-2)$-dimensional subspace of the $(n-1)$-dimensional space $W$. For each $i$, let $C_i = B_i^\perp \cap W$, a $1$-dimensional subspace of $W$ (and $1\le n-2$, remember). Now, since all the $B_i$ and $C_i$ have dimension smaller than that of $W$, their union cannot be $W$. Pick a unit vector $v_1$ in $W$ but not any $B_i$ or $C_i$ (you can pick it with norm $1$ because if $v$ is any vector in $W$ not in their union, then $\lambda v$ is also not in their union for $\lambda \ne 0$: if $\lambda v$ were in their union, then it would be in one of them in particular, and then so would $v=(1/\lambda)\lambda v$). Now let $W_1 = W\cap \{v_1\}^\perp$, which is $(n-2)$-dimensional. And for each $i$, let $B_i^1 = B_i \cap W_1$, which is $(n-3)$-dimensional, since we picked $v_1$ not in any $C_i$. Now we're looking for $n-2$ unit-norm vectors in $W_1$ not in any of the $B_i^1$, and we can iterate.
A: As so far there was no answer which could provide an  algebraic form of orthogonal matrix with single $0$ entry.   
However   looking a few days at the 3d form I was able   to find the missing pattern   at the end yesterday.
The pattern is amazingly simple.
Let $   v=\begin{bmatrix} a_1 & a_2 & \dots & a_n  \end{bmatrix}^T $ be n-dimensional unit vector ($v^Tv=1$).  
Now we can construct a block matrix dimension ${(n+1) \times (n+1)}$ based only on $v$ vector:
$$Q_{(n+1) \times (n+1)}  =\begin{bmatrix} 0_{1 \times 1}   & -v^T_{1 \times n}   \\  v_{n \times 1}   &  vv^T-I_{n \times n}  \end{bmatrix}$$
For this matrix:
$$Q^T  =\begin{bmatrix}   0  &  v^T    \\  -v   &  vv^T-I   \end{bmatrix}$$
One can check that really $QQ^T=I$,   hence if only unit vector $v$ has non-zero components then we have orthogonal matrix with single zero entry.
