Let $V$ a finite inner product space, $dim V = n \geq 3$. Let $w_1,w_2 \in V$ such that: $<w_1,w_2>=0$, $||w_1||=||w_2||=1$ where $||w||$ is the norm of a vector $w$.
The inner product space is defined $<v,u>=(v_1u_1,v_2u_2 + ... + v_nu_n)$.
Let $T:V \to V$ be a linear transformation, self-adjoint such that:
$$T_v=-v+2<v,w_1>w_1+2<v,w_2>w_2$$
Find the Jordan Normal Form of T.
What I've done so far
I had to prove that T is self-adjoint and to do so I have defined an orthonormal basis $B=sp \left \{ w_1,w_2,...,w_n \right \}$. From $B$ definition:
$T(w_1)=-w_1+2<w_1,w_1>w_1 + 2<w_1,w_2>w_2=-w_1+2w_1+0=w_1$ (Remember the basis is orthonormal).
$T(w_2)=w_2$
$T(w_3)=-w_3$
$T(w_4)=-w_4$
and continuing...
$T(w_n)=-w_n$
So what I have to do (and correct me if I'm wrong) is to find the representing matrix of $T$ with respect to the basis $B$. I don't know how to do it. I think it's way too simple to say that it is simply:
$$[T]_B=\begin{bmatrix}1 & 0 & 0 & 0 & ...& 0 \\ 0 & 1 & 0 & 0 & ... & 0 \\ 0 & 0 & -1 &0 & ... & 0 \\ 0 & 0 & 0 & -1 &... & 0 \end {bmatrix}$$ (continue this matrix until $n$, don't know how to latex it).
How do I complete my task to find Jordan Normal Form?