Hermitian Eigenvalues

Let v be the column vector [1-i ; -3 ; -2i ; 0] and suppose v is an eigenvector of a Hermitian matrix M with eigenvalue -3. Calculate v^H M

What I did-

We know by definition Mv = -3v

Now, M = M^H because M is Hermitian. Then,

M^Hv = (v^H M)^H = conj(v^H M).

So, v^H M = conj(-3v)

Thoughts?

Thanks!

i get:

v^HM = v^HM^H = (Mv)^H = (-3v)^H = (-3)(1+i,-3,2i,0) = (-3-3i,9,-6i,0) (row-vector).

(since we're already using _^H for the conjugate-transpose, why not keep using it?).

Ah. Does my logic work going the other way?

So you started from v^H M ....does what I did starting from Mv make sense too?