Find out if a operator is self-adjoint or orthogonal

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 $$==0$$ and $$=1/2$$.

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:

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

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:

$$=$$

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

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

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

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

A self-adjoint operator $$T$$ satisfies $$\left=\left\quad\forall x,y\in V$$ 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 $$\left=\left=\sum_{i=1}^3\sum_{j=1}^3c_id_j\left=\alpha c_1d_1+\beta c_2d_2+\beta\frac{1}{2}c_2d_3+\gamma\frac{1}{2}c_3d_2+\gamma c_3d_3$$ and $$\left=\left<\sum_{i=1}^3c_iv_i,T\sum_{i=1}^3d_iv_i\right>=\sum_{i=1}^3\sum_{j=1}^3c_id_j\left=\alpha c_1d_1+\beta c_2d_2+\gamma\frac{1}{2}c_2d_3+\beta\frac{1}{2}c_3d_2+\gamma c_3d_3$$ Thus $$\left=\left\Leftrightarrow \gamma=\beta$$