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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?

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2 Answers 2

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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 $i=2,\dots,n$.

$$\Rightarrow \lambda \overline {\lambda} \le 1$$

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  • $\begingroup$ Here you assume that $\phi$ is an eigenvector for $M_1$. Is that equivalent to saying that $\phi$ is an eigenvector for $M_1^*M_1$? $\endgroup$
    – arriopolis
    Commented Jun 8, 2016 at 22:50
  • $\begingroup$ To clarify: take the matrix $M_1 = \left[\begin{array}{cc} 0 & 1 \\ 0 & 0\end{array}\right]$ and $\phi = \left[\begin{array}{c} 0 \\ 1 \end{array}\right]$. Then $\phi$ is not an eigenvector for $M_1$, but it is for $M_1^*M_1$. So does this prove that $|\lambda| \leq 1$ for any eigenvalue of $M_1^*M_1$, or only for those that have simultaneous eigenvectors with $M_1$? $\endgroup$
    – arriopolis
    Commented Jun 8, 2016 at 22:58
  • $\begingroup$ That's not what we're saying (because, as you observed, the claim is not true). By definition, $(M^*u,v) = (u,Mv)$ for any $u,v$. In particular if $u=Mv$, then $ (M^*Mv,v) = (Mv,Mv) \ge 0$. Now if $v$ is an eigenvector for $M$ with eigenvalue $\lambda$, then $(M^*M v,v) =(Mv,Mv) = (\lambda v ,\lambda v) = |\lambda|^2 (v,v)$. $\endgroup$
    – Fnacool
    Commented Jun 8, 2016 at 23:21
  • $\begingroup$ I see, but then this proof isn't complete, as it does not necessarily take into account all eigenvalues of $M_1^*M_1$, right? By "the claim is not true", do you mean the statement I'm trying to prove? If so, could you give a counterexample? $\endgroup$
    – arriopolis
    Commented Jun 8, 2016 at 23:37
  • $\begingroup$ Sorry, I didn't read the question carefully. Edited the answer accordingly. $\endgroup$
    – Fnacool
    Commented Jun 9, 2016 at 2:48
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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.

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