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