Hermitian operator's eigenvectors are not orthonormal and not complete

Question

I am dealing with the following operator: $$B=\begin{pmatrix} 2 & 0 & 0 \\ 0 & 0 & -3i \\ 0 & 3i & 0 \\ \end{pmatrix}.$$

This matrix is Hermitian. I need to find eigenvectors, check that they are orthonormal and complete.

I got these eigenvalues and eigenvectors:

for v1=2,

b1= \begin{pmatrix} x \\ y \\ (3/2) iy \\ \end{pmatrix}

for v1=3,

b2= \begin{pmatrix} 0 \\ y \\ iy \\ \end{pmatrix}

for v1=-3,

b3= \begin{pmatrix} 0 \\ y \\ -iy \\ \end{pmatrix}

However they are neither orthonormal nor complete and it seems impossible to find appropriate coefficients.

I have spent 2 days on this task. So far, no solution has been found. This is the only exercise I have not managed to solve, so, I guess, the algorithms I use are correct.

Answer 1

for $\lambda=2$, $a=(1,0,0)$

for $\lambda=-3$, $b=(0,i,1)$

for $\lambda=3$, $c=(0,-i,1)$

Note that $\color{red}{\langle U,V\rangle=U\bar V^T}$ so $\langle c,b\rangle=c\bar b^T=0.0+i.i+1.1=-1+1=0$...

(I think your pro was in definition of inner product)