Abstract Linear Algebra Inner Product

Question

Let $u\in\mathbb{R}^n$ be a vector such that $\|u\|=1$ (for the usual inner product). Prove that there exists an $n\times n$ orthogonal matrix whose first row is $u$.

Answer 1

If you have vector $\vec u=(a_1,a_2,\dots,a_n)$, then obviously the vectors $$ \begin{align*} \vec v_1=&(a_2,-a_1,0,\dots,0)\\ \vec v_2=&(0,a_3,-a_2,0,\dots,0)\\ &\vdots\\ \vec v_{n-1}=&(0,\dots,0,a_{n-1},-a_n) \end{align*} $$ are perpendicular to $\vec u$, i.e. $\langle \vec u, \vec v_i \rangle=0$.

Do you know how you can use these vectors to get an orthogonal matrix? (Hint: What lengths do rows of an orthogonal matrix have?)

Notice that you still need do tweaking if one of the vectors $\vec v_1,\dots,\vec v_{n-1}$ is zero, i.e., if the given vector $\vec u$ contains two consecutive zeroes somewhere.

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It was pointed out in comments, that you could also use the fact that every set of orthonormal vectors can be extended to an orthonormal basis. This can be done using Gram-Schmidt process.

However, for the case where you only have one vector, there is also a simple way without relying on the above result.