# 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.

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also just to add some intuition, T is a reflection through the hyperplane that is orthogonal to $u$. Thus, $u$ is quite obviously an eigenvector of $A$ since $Tu = -u$ given the geometric definition of a reflection. $u$ is called the root of the reflection $T$. As you might have guessed, each reflection corresponds to 2 roots $\pm u$. They don't have to be of unit length as well, but in this case it's convenient to have them in unit length. – Enzo Apr 23 '13 at 14:35

1. Check that $u$ is always an eigenvector for this transformation corresponding to eigenvalue $-1$.
2. Check that any vector perpendicular to $u$ is an eigenvector for the eigenvalue $+1$.
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?).