Meaning of Square Root of Covariance / Precision Matrices Say $X \in \mathbb{R}^n$ is a random variable with covariance $\Sigma \in \mathbb{R}^{n\times n}$. By definition, entries of the covariance matrix are covariances: 
$$
\Sigma_{ij} = Cov( X_i,X_j).
$$
Also, it is known that entries of the precision $\Sigma^{-1}$ satisfy:
$$
\Sigma^{-1}_{ij} = Cov(X_i,X_j| \{X_k\}_{k=1}^n \backslash X_i,X_j\}),
$$
where the right hand side is the covariance of $X_i$ with $X_j$ conditioned on all other variables.
Is there a statistical interpretation to the entries of a square root of $\Sigma$ or $\Sigma^{-1}$? By square root of a square matrix $A$ I mean any matrix $M$ such that $M^tM = A$. 
An eigenvalue decomposition of said matrices does not give such entry-wise interpretation as far as I can see.
 A: I will write matrix square roots of $\Sigma$ as $\Sigma=A A^T$, to be consistent with the Cholesky decomposition which is written as $\Sigma = L L^T$ where $L$ is lowtri (lower triangular). So let $X$ be a random vector with $\DeclareMathOperator{\E}{\mathbb{E}} \DeclareMathOperator{\Cov}{\mathbb{Cov}}\DeclareMathOperator{\Var}{\mathbb{Var}} \E X=\mu$ and $\Var X =\Sigma$. Let now $Z$ be a random vector with expectation zero and unit covariance matrix. 
Note that there are (except for the scalar case) infinitely many matrix square roots. If we let $A$ be one of the, then we can find all the others as $A \mathcal{O}$ where $\mathcal{O}$ is any orthogonal matrix, that is, $\mathcal{O} \mathcal{O}^T = \mathcal{O}^T \mathcal{O} =I$.  This is known as unitary freedom of square roots. 
Let us look at some particular matrix square roots.


*

*First a symmetric square root. Use the spectral decomposition to write $\Sigma = U \Lambda U^T = U\Lambda^{1/2}(U\Lambda^{1/2})^T$. Then $\Sigma^{1/2}=U\Lambda^{1/2}$ and this can be interpreted as the PCA (principal component analysis) of $\Sigma$. 

*The Cholesky decomposition $\Sigma=L L^T$ and $L$ is lowtri. We can represent $X$ as $X=\mu + L Z$. Multiplying out to get scalar equations, we get a triangular system in $Z$, which in the time series case can be interpreted as a MA (moving average) representation. 

*The general case $A= L \mathcal{O}$, using the above we can interpret this as a MA representation after rotating $Z$. 
A: An nxn matrix can have many square roots as you mention.  However a covariance matrix must be positive semi-definite and a positive semi-definite matrix has only one square root that is also positive semi-definite.  Take a look at the wikipedia article titled "Square root of a matrix".
A: Sometimes people are interested in estimating the locations of zeros in the precision matrix for the same reason you describe above. If $M$ is your square root matrix, i.e. $M'M = \Sigma^{-1}$, then for nodes $i \neq j$
$$
\Sigma^{-1}_{i,j} = 0 \iff M_i'M_j = 0
$$ 
so I imagine that looking at the inner product between columns of your estimated square root matrix will give you some number akin to how close to conditionally independent two nodes are. Just an idea.
The square root of the covariance matrix, that's the scale. I imagine simulating a standard normal random vector, and then pre-multiplying by the square root matrix. If this matrix is lower-triangular, then I always imagine doing all the little multiplications and additions out.
There are also individual cases you can think of where certain elements of the square root matrix are just square roots of individual elements of the covariance matrix. This isn't that interesting, though, so I figure you have already thought of this.
