Adjoint problem eigenvalue Here is a problem I am stuck on even though I suspect the solutions is fairly straightforward:
Show that if $\lambda$ is an eigenvalue of the problem
$$
Ax = \lambda x
$$
for some square matrix $A$ then it is also an eigenvalue for the adjoint problem
$$
A^* y = \lambda y\,,
$$
where * indicates the conjugate transpose.
 A: For $X\in M_n(\Bbb C)$ we have 
$$
\det(X)^*=\det(X^*)\tag{1}
$$
Can you prove this? 
Taking $X=A-\lambda I$ in (1) gives
$$
\det(A-\lambda I)^*=\det((A-\lambda I)^*)=\det(A^*-\lambda^* I^*)=\det(A^*-\lambda^* I)
$$
This shows that $\lambda$ is a root of the characteristic polynomial of $A$ if and only if $\lambda^*$ is a root of the characteristic polynomial of $A^*$. This proves that the eigenvalues of $A$ are exactly the complex-conjugates of the eigenvalues of $A^*$. In particular, $A$ and $A^*$ share their real eigenvalues.
Now, note that $A$ and $A^*$ might have different eigenvalues. For example, the eigenvalues of 
$$
A=
\begin{bmatrix}
i & 0\\ 0 & 1
\end{bmatrix}
$$
are $\lambda_1=i$ and $\lambda_2=1$ while the eigenvalues of 
$$
A^*=
\begin{bmatrix}
-i&0 \\ 0 & 1
\end{bmatrix}
$$
are $\lambda_1^*=-i$ and $\lambda_2^*=1$.
A: Suppose $\lambda$ is an eigenvalue of a given square matrix $A$, and $x$ satisfies $Ax=\lambda x$. Then for any vector $y$,
$$\langle y,Ax\rangle=\langle y,\lambda x\rangle = \langle\lambda y, x\rangle = \lambda \langle y,x\rangle.$$
The adjoint of $A$, denoted $A^*$, is defined to satisfy
$$\langle y,Ax\rangle = \langle A^* y,x\rangle,~~~\forall x,y.$$
Hence, if $\lambda$ is an eigenvalue of a given square matrix $A$, and $x$ satisfies $Ax=\lambda x$, then
$$\forall x:\langle A^* y,x\rangle = \langle\lambda y, x\rangle\\
\implies A^*y=\lambda y.$$
