i get:
v^{H}M = v^{H}M^{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?).
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!