why is this vector multiplication move valid?

Question

This is dealing with properties of unitary matrices and this is proving that

eigenvectors corresponding to different eigenvalues are orthonormal.

start with $Ux = \lambda_1 x$ and $Uy = \lambda_2 y$ and take inner products...

$$ x^H y = (Ux)^H(Uy) = (\lambda_1x)^H(\lambda_2 y) = \overline{\lambda_1} \lambda_2 x^Hy $$

I can follow this up until the last step where I guess we unpack the $(\lambda_1x)^H(\lambda_2y)$ into $x^H \lambda_1^H \lambda_2y$ but then I can't see how we can make a valid move of the vectors and move from the conjugate transpose to just the conjugate

Answer 1

Since $\lambda_{1}$ and $\lambda_{2}$ are scalars, not vectors, transpose doesn't "have an effect" on them.

Answer 2

$\lambda_i\in\Bbb C\implies(\lambda_1 x)^H=\overline{(\lambda_1 x)^T}=\overline{\lambda_1 x^T}=\bar\lambda_1 x^H$.