The N(I+V) must be bigger than 4 in complex matrices? In this problem, all matrices are $n*n$ with complex entries. Let $U$ and $V$ be matrices such that $UV \not = VU$. Assume that $U$ is diagonalizable and commutes with $VUV^{-1}$
(a).For $\lambda , \mu \in \mathbb C$ , let
$E_{\lambda , \mu}=\{x \in \mathbb C^n|Ux= \lambda x ,VUV^{-1}x=\mu x \}.$
Show that there exist couples $(\lambda_1,\mu_1) \not = (\lambda_2,\mu_2)$,satisfying $\lambda_i \not = \mu_i$ and $E_{\lambda_i , \mu_i} \not = 0$ for $ i = 1,2.$
(b).For a matrix $A$, we define $N(A):=tr(A^*A),$ where $A^*$ is the conjugate transpose of $A$. Assume that $U$ and $V$ are unitary (namely,$U^*U=V^*V$ is the identity matrix).Deduce that $N(1+V) \ge 4$.
 A: (a). Let $\lambda_1,\ldots,\lambda_r$ be the eigenvalues of $U$ (and of $VUV^{-1}$) and set $N_j = \ker(U-\lambda_i I_n)$. Then
$$
\mathbb C^n = N_1\,\oplus\,\ldots\,\oplus\,N_r.
$$
Now, each $N_j$ is invariant with respect to $VUV^{-1}$. Hence, each $N_j$ has a basis $\{x_1^j,\ldots,x_{r_j}^j\}$ consisting of eigenvectors of $VUV^{-1}$. Hence, we have $VUV^{-1}x_k^j = \mu_{kj}x_k^j$ for all $j$ and all $k$. Note that $VUV^{-1}$ has the same eigenvalues as $U$ with the same multiplicities, respectively. If $\mu_{kj} = \lambda_j$ for all $j=1,\ldots,r$ and all $k=1,\ldots,r_j$, then $VUV^{-1} = U$, which contradicts $UV\neq VU$. Therefore, there exists some $j$ and $k_j\in\{1,\ldots,r_j\}$ such that $\mu_{k_j,j} = \lambda_i$ for some $i\neq j$. But this also means that there must be some $\ell\neq j$ such that $\mu_{k_\ell,\ell} = \lambda_j$ with some $k_\ell\in\{1,\ldots,r_\ell\}$. Hence, we have $Ux_{k_j}^j = \lambda_jx_{k_j}^j$, $VUV^{-1}x_{k_j}^j = \lambda_ix_{k_j}^j$, and $Ux_{k_\ell,\ell} = \lambda_\ell x_{k_\ell}^\ell$ and $VUV^{-1}x_{k_\ell}^\ell = \lambda_jx_{k_\ell}^\ell$. So, $E_{\lambda_j,\lambda_i}\neq\{0\}$ and $E_{\lambda_\ell,\lambda_j}\neq\{0\}$ with $\lambda_i\neq\lambda_j$, $\lambda_j\neq\lambda_\ell$, and thus also $(\lambda_j,\lambda_i)\neq (\lambda_\ell,\lambda_j)$.
(b). We have
\begin{align}
N(I+V)
&= tr((I+V^*)(I+V)) = tr(2I+V^*+V)\\
&= 2n + \overline{tr V} + tr V\\
&= 2n+2\operatorname{Re}(tr V).
\end{align}
So, we have to prove that $-\operatorname{Re}tr V\le n-2$. Note that $(U,V)$ satisfies the assumptions, then also $(U,-V)$ does. So, the claim is $|\operatorname{Re}tr V|\le n-2$.
I can only show this if the eigenvalues of $U$ are distinct. Then $U = WDW^*$ with a unitary $W$ and a diagonal matrix $D$ with $n$ distinct diagonal entries $\lambda_1,\ldots,\lambda_n$. Consider $V_0 = W^*VW$. Then it is easy to see that $D$ commutes with $V_0DV_0^*$ and $tr V_0 = tr V$. We therefore may assume that $U = D$ is diagonal. If $Ux = \lambda x$, then $VDV^*(Vx) = \lambda(Vx)$. Hence, $V\ker(U-\lambda I) = \ker(VDV^*-\lambda I)$. Hence, $Ve_j\in\ker(VDV^*-\lambda_j I)$. But as we saw above, $\ker(VUV^*-\lambda_j I) = \ker(U-\lambda_k)$ for some $k$. So, $Ve_j = \mu_je_{\pi(j)}$, where $\mu_j\in\mathbb C$ and $\pi$ is a permutation of $\{1,\ldots,n\}$. We have $|\mu_j|^2 = \|\mu e_{\pi(j)}\|^2 = \|Ve_j\|^2 = (Ve_j)^*(Ve_j) = e_j^*V^*Ve_j = \|e_j\|^2=1$. So $V$ is a matrix with exactly one entry $\mu_j$ of modulus $1$ in each column and each row. Since $V$ is not diagonal (otherwise $UV = VU$), there are at least two zero entries on the diagonal. This implies $|\operatorname{Re}tr V|\le n-2$.
