Proof that eigenvalues are constants

[jk22]

i suppose matrices of the form $$A_i=\left(\begin{array}[cc] ccos(x_i)&sin(x_i)\\sin(x_i)&-cos(x_i)\end{array}\right)$$

And i consider the matrix

$$C=A1\otimes A2\otimes 1_4-A1\otimes 1_4\otimes A2$$

I would like to show that the eigenvalues of C are independent of the x
i tried with mathematica but i have no proof.

 

[robphy]

What is $1_4$?

 

[jk22]

The identity matrix of dimension 4 $$\left(\begin{array}[cccc]a1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{array}\right)$$

 

[Hawkeye18]

What do you mean by $\otimes$?

 

[jk22]

What do you mean by $\otimes$?

The usual Kronecker product or tensor product

 

[Hawkeye18]

First, your matrices $A_i$ are the reflection matrices: in particular that means that the spectrum of $A_i$ is $\{-1, 1\}$ and that $A_i$ are unitarily equivalent to $$A_0= \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right),$$ meaning that $A=U_i A_0 U_i^{-1}$, where $U_i$ are unitary operators (with real entries). Matrices $U_i$ can be easily computed, but it does not matter here.

Since the spectrum $\sigma(A\otimes B)$ is the product of $\sigma(A)$ and $\sigma (B)$ you just need to prove that the spectrum of $A_2\otimes I_4 - I_4\otimes A_2$ does not depend on $x_2$. To see that you can notice that $I_4 = I_2\otimes I_2$, so $$A_2\otimes I_4 - I_4\otimes A_2 = A_2\otimes I_2\otimes I_2 - I_2\otimes I_2\otimes A_2, $$ and the latter operator is unitarily equivalent to $$A_0\otimes I_2\otimes I_2 - I_2\otimes I_2\otimes A_0$$ (can you see why?)

So the spectrum does not depend on $x_2$.

 

[jk22]

To show this I think we multiply left by $$U\otimes U\otimes U$$ and right by the inverses of U ?

Thanks a lot.

 

[Hawkeye18]

If by $U$ you mean $U_2$, then yes, you are correct. You can also use $U_2\otimes I_2\otimes U_2$.