Eigenvalues and eigenvectors in hermitian matrix

Given 3 vectors in C^3

e1=(i,1,1) , e2= (i,0,-1), e3= (i,0,1)

and A a linear operator in C^3 with these vectors as eigenvectors, corresponding eigenvalues a,b,c respectively. Let be A hermitian and b=0 and a=w, where w is a real parameter, find c.


Thanks in advance

Eigenvalues of a hermitian matrix in C^n are all real.
Eigenvectors of a hermitian matrix can always be chosen to be orthogonal. Thus, for any hermitian matrix, there is always an orthonormal set of eigenvectors that span C^n.

So if any two eigenvectors are non-orthogonal, they must either have the same eigenvalue, or one or both eigenvalues must be 0.

Of the 3 given eigenvectors, e2 is orthogonal to each of the others, but e1 is not orthogonal to e3:

e1•e3 = i•i* + 1•0* + 1•1* = i•(-i) + 1•0 + 1•1 = 1 + 0 + 1 = 2

Further, those 3 vectors are linearly independent, since their determinant ≠ 0:

| i 1. 1 |
| i 0 -1 | = -2i
| i 0. 1 |

Thus, c is either 0 or w.

Thanks back!
I felt my answer was unfinished, but I don't know whether any more can be done with it.

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