Relationship between columns of a unitary matrix

Question

$U=[u_1\quad u_2\quad u_3]$ is a 3 x 3 unitary matrix and $u_i\in \mathbb R^3, i=1,2,3$. It seems that $$u_2u_1^T - u_1u_2^T$$ is equal to $[u_3]_\times$, i.e. the skew-symmetric matrix of $u_3$ (up to a positive and negative sign). I write programs to test it and it seems to be correct. I have tried to expand all the values to prove it but in vain. Any ideas?

Answer 1

I assume that by $[u_3]_{\times}$, you are referring to the matrix defined so that $$ [u_3]_{\times}v = u_3 \times v $$ for all vectors $v \in \Bbb R^3$.

Your statement is correct. In particular, we have $[u_3]_{\times} = u_2u_1^T - u_1u_2^T$ whenever $U$ is orthogonal (i.e. real and unitary) and the following equivalent conditions holds:

• $(u_1,u_2,u_3)$ forms a right-handed coordinate system,
• $u_3 = u_1 \times u_2$,
• $\det(U) = 1$.

A proof of this result is given in the link I provide above (see the "proof by substitution").

An alternative proof is as follows: note that the cross-product under rotations. That is, if $U$ is orthogonal with $\det(U) = 1$, then we have $(Ux) \times (Uy) = U(x \times y)$. Let $e_1,e_2,e_3$ denote the standard basis. We have $$ u_2 u_1^T - u_1u_2^T = U[e_2 e_1^T - e_1e_2^T]U^T. $$ It now suffices to show that $[e_3]_{\times} = e_2 e_1^T - e_1e_2^T$. Once this is established, we have $$ \begin{align} [u_2 u_1^T - u_1u_2^T]v &= U[e_2 e_1^T - e_1e_2^T]U^T v = U[e_3]_{\times}(U^T v) \\ & = U(e_3 \times U^Tv) = (U e_3) \times (UU^Tv) = u_3 \times v, \end{align} $$ as desired.