Eigenvalues of matrices satisfying the completeness relation Suppose I have a finite collection of square complex matrices, denoted by $\{M_1, \dots, M_n\}$, such that it satisfies the following completeness relation:
$$\sum_{i=1}^n M_i^*M_i = I$$
If $\lambda$ is an eigenvalue of $M_i^*M_i$, for some $i$, then I would like to show that:
$$|\lambda| \leq 1$$
I suspect that there exists a rather straightforward proof, but I have not been able to come up with one myself, and I couldn't find one online either. Could someone give me a push in the right direction?
 A: NOTE. I previously answered a different question, and showed that the eigenvalues of the $M_i$'s are in the unit disk. This proof is now below the line. 
Here is the modification showing that the eigenvalues of $M_i^*M_i$ are in the unit interval. 
Let $v$ be an eigenvector for $M_1^*M_1$ with eigenvalue $\lambda$ and norm $1$. Then 
$$ \lambda =  (M_1^*M_1 v,v)=(M_1v,M_1 v) \ge 0 \Rightarrow \lambda \ge 0.$$  
By the assumption: 
\begin{align*} 1&= (v,v)=(Iv,v) \\&= 
\sum_{i=1}^n (M_i^*M_i  v, v)\\
&=(M_1^* M_1v,v)+\sum_{i=2}^n (M_i v,M_i v)\\
& \ge \lambda 
\end{align*} 

The assumption is equivalent to 
$$(*)\quad \sum_{i=1}^n (M_i v , M_i u)=(u,v).$$
Suppose that $\phi$ is an eigenvector for $M_1$ with eigenvalue $\lambda$, and without loss of generality $(\phi,\phi)=1$. Then letting $u=v=\phi$ in $(*)$ you obtain 
$$ \lambda \overline{\lambda} + \sum_{i=2}^n (M_i \phi,M_i \phi) = 1.$$ 
But by positivity of inner product, $(M_i \phi, M_i \phi)\ge 0$ for each $i=2,\dots,n$. 
$$\Rightarrow \lambda \overline {\lambda} \le 1$$
A: The claim is certainly true if $n=1$. If $n>1$, then $I-M_1^*M_1$ is a sum of ositive definite matrices, and hence it is positive semidefinite. Therefore all eigenvalues of $I-M_1^*M_1$ are non-negative and so all eigenvalues of $M_1^*M_1$ are at most one.
