Trying to prove
Reference: https://arxiv.org/abs/1407.1244
I was trying to prove a claim in above paper. To summarize, they first define a Hermitian matrix $H(\alpha)$ which depends on a paramter $\alpha$. Since this is Hermitian matrix, all eigenvalues are real, so they first order eigenvalues in increasing order $\lambda_1 \leq \lambda_2, ...$ and define projection operator $P$ onto the eigenspace with eigenvalues $\lambda_1, \lambda_2, ... \lambda_n$. Multiplicity on $\lambda_n$ is 1, so the projection is well defined.
Then they define a quantity $\gamma$ such that (equation 4)
$$ \gamma = n - Tr(P \tilde{P}) = n - \sum_{i,j}^n|M_{ij}|^2 $$
where $P$ and $\tilde{P}$ are the two projection for two different $\alpha$, $P=\sum_i^n{|\psi_i\rangle \langle \psi_i|}$ and $M_{ij} = \langle \psi_i | \tilde{\psi_j} \rangle$ (equation 5). Here $\psi_i$ ($\psi_j$) are the ith (jth) eigenvector for $H(\alpha)$ ($H(\tilde{\alpha})$)
This is the definitions, and in the line 35 in page 2, there is a claim which says that "since a singular M would imply $\gamma \geq 1$.
I am trying to prove this one.
Attemped solution
We can define a matrix $X$ as
$$ X = I - MM^{\dagger}. $$ Then $$ \gamma = Tr(X) = n - Tr(M M^{\dagger}) = n - \sum_i{|\mu_i|^2} $$ Here $\mu$ refers the eigenvalues of $M$.
If $M $ is singular matrix, $\det(M)=0$ and $\det(M) = \prod_i{\mu_i}$ implies that at least one of the eigenvalues of $M$ is zero. If $P = \tilde{P}$, $M$ is identity matrix, so all $\mu=1$. If $\tilde{P} = (1-P)$, $M$ = 0. Considering this exteme case, I guessed that the modulus of all the eigenvalues for matrix $M $ could be less than one. Then since at least one of the eigenvalues is zero, we have
$$ \sum_i{|\mu_i|^2} \leq n-1 $$ so can prove $\gamma \geq 1$
Then my question is for the matrix $M$ defined above (or $MM^{\dagger}$), is it true that the modulus of eigenvalue is less than one?
Original post
I keep below the original post only to relate it to the received comments. You don't need to read this part.
Let $M=M(\lambda)$ be a diagonalizable $n \times n$ matrix which depends on a paramter $\lambda$.
Then assume that for distinct paramters $\lambda_1$ and $ \lambda_2$, we construct a matrix $P$ such that
$$ P_{ij} = \langle u_i^{\lambda_1} |v_j^{\lambda_2}\rangle $$
where $u_i^{\lambda_1}$ and $v_j^{\lambda_1}$ are the normalized eigenvectors for given paramter $\lambda$.
Also I assume that the multiplicity of all eigenvalues is 1, real number so I can order the eigenvectors by comparing their eigenvalues in increasing order.
For example, if $\lambda_1 = \lambda_2$,
$$ P_{ij} = \delta_{ij}. $$
Also, if all of the eigenvectors for two $\lambda$s have no overlap, then
$$ P_{ij} = 0 $$
So for these two extreme cases, we have the modulus of all eigenvalues of matrix $P$ is less than one. My question is, can we say that for arbitrary paramter $\lambda_1$ and $\lambda_2$, the modulus of all the eigenvalues of P is less than one?