How can I prove following density matrices have same eigenvalues? I have the following two density operators, the paper I am reading says that these two operators have same eigenvalues

$$\rho^i = \frac{1}{3} ( |0\rangle \langle 0 | +|1\rangle \langle 1 |+|2\rangle \langle 2 |+a|0\rangle \langle 1 |+a|1\rangle \langle 0 |+c|1\rangle \langle 2 |+c|2\rangle \langle 1
 |),$$
$$\rho^f = \frac{1}{3} ( |0\rangle \langle 0 | +|1\rangle \langle 1 |+|2\rangle \langle 2 |+ax|0\rangle \langle 1 |+ax^*|1\rangle \langle 0 |+cy|1\rangle \langle 2 |+cy^*|2\rangle \langle 1
 |).$$
Here $a,c$ are real numbers and $x,y$ are unimodular complex numbers. Also $\{|0\rangle ,|1\rangle , |2\rangle \} $ are three orthonormal vectors of a 3-D Hilbert space.
My approach : its apparent that the operators are hermitian and thus have a spectral decomposition in terms of orthonormal vectors. Lets say one of the vectors in spectral decomposition of first operator is $$|\psi \rangle = A|0 \rangle + B|1\rangle,$$ then for the second operator the corresponding one would be $$|\psi \rangle = Ax^{\frac{1}{2}}|0 \rangle +(x^*)^{\frac{1}{2}} B|1\rangle$$ ( $A,B,C,D$ are some complex constants ). Similarly another vector in spectral decomposition of first operator will be of form $$|\psi \rangle = C|1 \rangle + D|2\rangle$$ and its counterpart in second operators spectral decomposition will be $$|\psi \rangle = y^{\frac{1}{2}}C|1 \rangle + (y^*)^{\frac{1}{2}}D|2\rangle.$$ Thus keeping eigen values same.
Doubt : But my approach seems to be an intuitive and vague proof, is there a way where it is evident that they have same eigenvalues in a simple manner? 
 A: Hints:


*

*What happens to the density operator and its eigenvalues under a change of the orthonormal basisvectors $|0\rangle $, $|1\rangle $, $|2\rangle$ by phase factors?

*More generally, what happens to the density operator and its eigenvalues under a unitary transformation $\rho\longrightarrow U^{\dagger}\rho U$?
A: 
I have the following two density operators, the paper I am reading
  says that these two operators have same eigenvalues 
$$\rho^i = \frac{1}{3} ( |0\rangle \langle 0 | +|1\rangle \langle 1
> |+|2\rangle \langle 2 |+a|0\rangle \langle 1 |+a|1\rangle \langle 0
> |+c|1\rangle \langle 2 |+c|2\rangle \langle 1  |),$$
$$\rho^f = \frac{1}{3} ( |0\rangle \langle 0 | +|1\rangle \langle 1
> |+|2\rangle \langle 2 |-ax|0\rangle \langle 1 |+ax^*|1\rangle \langle
> 0 |+cy|1\rangle \langle 2 |+cy^*|2\rangle \langle 1  |).$$

They don't. 
Did you mean to write "+ax" instead of "-ax" in the second matrix?
If so, then they do... in either case (if you meant to write "+ax") the eigenvalue equation is:
$$
(1-\lambda)^3=(1-\lambda)(a^2+c^2)
$$
giving 
$$
\lambda=1\;,\;1\pm\sqrt{a^2+c^2}
$$
