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Let's say we have a positive-definite symmetric matrix $M$ with individual elements sample from a distribution containing an expected value of $0$.

$$M = Q \Lambda Q^{T}$$ $$E \left[ M\right] = 0 = E \left[Q \Lambda Q^{T} \right]$$

Now since we know that $M$ is symmetric, it is positive definite therefore it's eigenvalues are all greater than zero, therefore, we can say

$$0 \leq E\left[\Lambda\right]$$

How do we show that $ E \left[Q \right] = 0$. Is it as simple as stating $$E \left[ M\right]=E \left[Q \Lambda Q^{T} \right] = E \left[Q \right] * E \left[\Lambda \right]* E \left[Q^T \right]$$

And since the $\Lambda$ expectation is greater than 0, the only way this expression is true is $E \left[Q \right]$ is 0.

Is this a valid statement? Do we have to write the covariances? And if so how?

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    Comments: 1. If $M$ is positive definite, this means that all its eigenvalues are greater than zero (not one, as you wrote). In any case, the inequality $E[\Lambda]\leq 1$ comes from nowhere (the eigenvalues of a positive definite matrix need not be smaller than one). 2. Your last equality only holds true if $\Lambda$ and $Q$ are independent sets of variables, which in general is not true. 3. What do you exactly mean by $E[A]$, where $A$ is a matrix? – 2017-02-18
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    @PierpaoloVivo My bad, that should say eigenvalues are greater than zero. And I don't know why i put $\leq 1$. By expected value, I'm referring to the mean of all the elements in the matrix – 2017-02-20

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