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?
Best Answer
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.