# Geometric intuition for why an outer product of two vectors makes a correlation matrix? [closed]

I understand that the outer product of two vectors, say representing two detrended time series, can represent a cross-correlation (well covariance) matrix.

I also know that the inverse of a correlation matrix represents the partial correlations between two variables. Geometrically, I know that the partial correlation between two variables is the angle formed by the projection of their residuals when regressed against all other variables onto the surface perpendicular to all other variables.

I'm wondering how these two relate. That is, I know the interpretation of the inverse of matrix (partial correlation) but not the matrix or its construction.

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Your reference to "outer product" does not accord with my understanding of this operation, which I believe is a conventional one. The rank of any outer product is at most one, which would produce a highly degenerate matrix: that's not what one expects of a covariance. Could you indicate what your "outer product" operation is? –  whuber Sep 14 '12 at 22:27
am sure he refers to the sum of all the outer products of $x_i$ with itself over $i$. That is just the sum of cross-products ... –  kjetil b halvorsen Sep 14 '12 at 23:09
By outer product, I mean $\mathbf{x}\mathbf{y}^T$. I thought that for two random vectors, $\mathbf{x}$ and $\mathbf{y}$ the covariance matrix would be $E\left(\mathbf{xy^T}\right) = E\left(\mathbf{x}\right) E\left(\mathbf{y}\right)^T$. –  mac389 Sep 15 '12 at 2:58
Maybe I don't understand. Suppose that the random vectors $X$ and $Y$ have mean vectors $u=\mathrm{E}[X]$ and $v=\mathrm{E}[Y]$, respectively. The matrix $A=u\,v^T$ can't be, in general, a covariance matrix, because, suppose that $X$ and $Y$ are such that the inner product $\langle u,v\rangle<0$. Then, we have $\langle u,Au\rangle=u^TAu=u^Tu\,v^Tu=||u||^2 \langle v,u\rangle<0$, and $A$ is not positive definite. –  Zen Sep 16 '12 at 2:23