Find out if a operator is self-adjoint or orthogonal

Question

Let $V$ be a $\mathbb{R}$ inner product space, and $B=\left \{v_1, v_2, v_3 \right \}$ basis of $V$, with $||v_i||=1$ $\forall i=1,2,3$, and $<v_1, v_2>=<v_1, v_3>=0$ and $<v_2, v_3>=1/2$.

Let $T$ be an operator in $V$ such that $_B(T)_B=\begin{pmatrix} \alpha&0&0\\ 0&\beta&0 \\0&0&\gamma \end{pmatrix}$ (associated matrix of $T$).

Find the correct option:

1. $T$ is self-adjoint $\forall$ $\alpha$, $\beta$, $\gamma \in \mathbb{R}$.
2. If $\alpha=0$ then there are $\beta$, $\gamma$ such that $T$ is orthogonal.
3. If $\alpha=1$ then $T$ is orthogonal iff $|\beta|=|\gamma|=1$.
4. $T$ is self-adjoint iff $\beta=\gamma$. (correct answer)
5. Given $\alpha \in \mathbb{R}$, there are $\beta$, $\gamma \in \mathbb{R}$, $\beta \neq \gamma$ such that $T$ is self-adjoint.

What I've been doing:

For the self-adjoint questions, I thought of applying the change of basis theorem, so I could get the associated matrix in an orthonormal basis.

So, using Gram-Schmidt:

• $u_1=v_1$
• $u_2=v_2$
• $u_3=v_3-\frac{<v_3, u_2>}{<u_2, u_2>}u_2=v_3-\frac{1}{2}u_2$

Now I know that $B'=\left \{ u_1, u_2, v_3-\frac{1}{2}u_2\right \}$ is an orthogonal basis, but not orthonormal.

Now my problem comes when trying to find the norm of the last vector:

$<v_3-\frac{1}{2}u_2, v_3-\frac{1}{2}u_2>=$

$=<v_3, v_3-\frac{1}{2}u_2>+<-\frac{1}{2}u_2, v_3-\frac{1}{2}u_2>=$

$=<v_3, v_3>+<v_3, -\frac{1}{2}u_2>+<-\frac{1}{2}u_2, v_3>+<-\frac{1}{2}u_2, -\frac{1}{2}u_2>=$

$=1+<v_3, -\frac{1}{2}u_2>+<-\frac{1}{2}u_2, v_3>+<-\frac{1}{2}u_2, -\frac{1}{2}u_2>$.

And I really don't know what to do with all of this, nor if I'm going the correct way.

Answer 1

Regarding all the questions concerning self-adjointness:

A self-adjoint operator $T$ satisfies $$\left<Tx,y\right>=\left<x,Ty\right>\quad\forall x,y\in V$$ For two vectors given in the basis $B$ $$x=\sum_{i=1}^3c_iv_i,\quad y=\sum_{i=1}^3d_iv_i,$$ we have $$\left<Tx,y\right>=\left<T\sum_{i=1}^3c_iv_i,\sum_{i=1}^3d_iv_i\right>=\sum_{i=1}^3\sum_{j=1}^3c_id_j\left<Tv_i,v_j\right>=\alpha c_1d_1+\beta c_2d_2+\beta\frac{1}{2}c_2d_3+\gamma\frac{1}{2}c_3d_2+\gamma c_3d_3$$ and $$\left<x,Ty\right>=\left<\sum_{i=1}^3c_iv_i,T\sum_{i=1}^3d_iv_i\right>=\sum_{i=1}^3\sum_{j=1}^3c_id_j\left<v_i,Tv_j\right>=\alpha c_1d_1+\beta c_2d_2+\gamma\frac{1}{2}c_2d_3+\beta\frac{1}{2}c_3d_2+\gamma c_3d_3$$ Thus $$\left<Tx,y\right>=\left<x,Ty\right>\Leftrightarrow \gamma=\beta$$