Show $\det \left[T\right]_\beta=-1$, for any basis $\beta$ when $Tx=x-2(x,u)u$, $u$ unit vector Let $u$ be a unit vector in an $n$ dimensional inner product space $V$.
Define the orthogonal operator
$$
Tx= x - 2 (x,u)u,
$$
where $x \in V$. Show
$$
\det A = -1,
$$
whenever $A$ is a matrix representation of $T$.
I can show that $\det A=\pm1$
$$
1=\det(I)=\det(A^tA)=(\det A)^2.
$$
Then I guess I should extend $u$ to a basis $\beta$ for $V$
$$
\beta=\{u,v_1,v_2,\ldots,v_{n-1}\}.
$$
I don't see what to do after this.

To complement rschwieb's answer, he is giving you way to determine a specific basis to perform your calculation in easily.  Then recall that if $A$ and $A'$ are two matrices that represent the same linear transformation in two different bases, then $A' = P^{-1} A P$ where $P$ is a change of basis matrix.  In particular, this implies that $\det(A') = \det(A)$ (why?).
