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When we handle with some dynamic input-output mappings, there occurs a question as follows:

Let $M$ be a random matrix, of which each element contains random terms. Consider the two expectation values $E\{MM^{T}\}$ and $E\{M^{T}M\}$. Are their largest eigenvalues equal? If so, how to prove it.

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  • $\begingroup$ $MM^T$ and $M^TM$ have the same set of nonzero eigenvalues. $\endgroup$
    – Carlo Beenakker
    Commented Oct 10, 2018 at 10:12
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    $\begingroup$ Thanks for your reply @CarloBeenakker. I know this fact. But I am confused that, if we take expectation on them. $\endgroup$
    – Dude-Ray
    Commented Oct 10, 2018 at 10:16

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Having noted that $MM^T$ and $M^TM$ have the same eigenvalues, you seek the eigenvalues of the expectation values of each matrix. These can be very different, take for example $M$ of dimension $n\times 1$, and let each element of $M_{i}$ fluctuate independently between $M_i=+1$ and $M_i=-1$. Then $E[M^TM]$ has a single eigenvalue equal to $n$ while $E[MM^T]$ has $n$ eigenvalues equal to 1.

The sum of the eigenvalues of $E[M^TM]$ and $E[MM^T]$ is the same, which follows from the fact that, on the one hand, the trace of $MM^T$ equals the trace of $M^TM$, and on the other hand taking the trace commutes with taking the expectation value.

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  • $\begingroup$ Good answer and Thanks. @Carlo Beenakker. Would you please check another question of mine: mathoverflow.net/questions/312449/… $\endgroup$
    – Dude-Ray
    Commented Oct 10, 2018 at 14:47
  • $\begingroup$ Indeed, there is some link between this question and another one on the induced norm. For a stochastic dynamic system, we can always find a map that makes $\varrho=\mathcal{T}_{a}\tau$, and here I think $\mathcal{T}_{a}$ can be viewed as $M$ in this question. However, if the result on the induced norm of a stochastic operator and its adjoint operator ( in that question) is correct, we may deduce $E[M^{\rm T}M]$ and $E[MM^{\rm T}]$ have the same largest eigenvalue. I do not know where goes wrong in that question. If possible, please check it. Thanks once again. @Carlo Beenakker $\endgroup$
    – Dude-Ray
    Commented Oct 10, 2018 at 14:55

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