Is the Covariance Matrix a dyad? [closed]

Given a random vector $X\in\mathbb{R}^n$ with random scalar entries $X_i$, why isn't the covariance matrix $\Sigma$ of the random vector $X$ a dyad? In other words, why isn't the rank of the covariance matrix trivially 1?

closed as unclear what you're asking by Sycorax, Silverfish, gung♦, John, Sean EasterFeb 27 '16 at 3:56

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• I'm not sure I understand what this question is premised on - why would this be trivial? Without knowing this, it's difficult to see what you would like to be explained to you – Silverfish Feb 26 '16 at 21:06

For the covariance matrix to have rank 1 all its columns must be the linear combinations of the first column. There's no reason why would random entries $X_i$ be perfectly correlated to each other unless you specifically made them to be correlated.
• Your language may be confusing, because "perfectly correlated" would ordinarily be read as meaning all off-diagonal elements of the correlation matrix are $\pm 1$, which seems to be quite a different criterion than the covariance matrix having rank 1. I think you may be implicitly invoking a theorem here :-). – whuber Feb 26 '16 at 22:24
• Let's prove your theorem. Any rank-1 matrix has entries $(x_iy_j)$ for some vectors $(x_i)$ and $(y_j)$. If it's a covariance matrix, the corresponding correlation matrix has entries $(\pm\sqrt{|x_i y_j / (x_j y_i)|})$. Since that matrix is symmetric, squaring its entries and comparing to the transpose shows $$\frac{x_iy_j}{x_jy_i}=\frac{x_jy_i}{x_iy_j}$$ for all $i,j$. Since one side is the inverse of the other, each must be $\pm 1$. Therefore all the entries of the correlation matrix must be $\pm 1$, QED. – whuber Feb 26 '16 at 22:56