Given $u_1$, find orthogonal matrix whose first column is $u_1$?

Question

Let $u_1 \in \mathbb{R}^n$, how can we construct an orthogonal matrix $U \in \mathbb{R}^{n \times n}$ whose first column is $u_1$? $$ U = \begin{bmatrix} u_1 & \hat{U} \end{bmatrix} $$

My try: I know that orthogonal complement of $u_1$ has $n-1$ vector and applying G-S procedure gives us $\hat{U}$, but is there any intuitive way of finding $\hat{U}$ form $u_1$?

Answer 1

Let $u_1=(a_1,a_2,\ldots, a_n)$. Take the set of all solutions of $a_1x_1+a_2x_2+\cdots +a_nx_n=0$. Take a basis for the set of solutions (they form a vector space of dimension $n-1$). One easy basis for this is:

$v_2=(a_2, -a_1,0,0,\ldots, 0)$

$v_3=(a_3, 0, -a_1,0,0,\ldots, 0)$

$\vdots$

$v_n=(a_n,0,,0,\ldots, 0, -a_1)$.

Now for $\{u_1, v_2,v_3, \ldots, v_n\}$ apply Gram-Schmidt and get an orthonormal basis. When these basis vectors are made columns of a matrix that will provide the orthogonal matrix you are looking for.

Answer 2

You could take a reflection mapping the vector $(1,0,\ldots,0)^t$ to $u_1$.