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The example Im using comes from statistics but its not limited there. This is a question about bilinear and quadratic forms.

This might sound like a dumb question. I can readily see from its properties that variance is not a bilinear form. The variance of a sum is not the sum of the variances in the general case. I dont need it proven to me. Im just kind of surprised that it isnt bilinear. Here's why:

We know that $\mathrm{Var}(X) = \mathrm{Cov}(X,X)$, correct? Thus, variance is just a special case of covariance. But covariance is a bilinear form. Somehow variance is not.

Im having a logical error somewhere. Can someone reconcile?

It has occurred to me that the variance is a single argument function while the covariance is two argument. Naturally bilinear forms require two arguments, but a variance can easily be interpreted as a two-argument covariance. I thought that the number of arguments might have something to do with it, but it seems a rather arbitrary thing. How can the abstract properties of functions be determined by an artifact of man-made notation?

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If $B(X,Y)$ is a bilinear form, then $Q(X):=B(X,X)$ is a quadratic form. This isn't particular to variance and covariance. Take the dot product of vectors, for example: $\|x\|^2=x\cdot x$ is not a bilinear form (it isn't even a function of two variables, which is a prerequisite to be bilinear - this is not just "man-made notation"), but a quadratic form.

One can convert any symmetric bilinear form to a quadratic form, and also back again using the polarization identity $Q(X+Y)=Q(X)+2B(X,Y)+Q(Y)$.

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  • $\begingroup$ That last paragraph, although very useful, also has a limitation: you can do this only if the field in question does not have characteristic two. If it does, you can't go from $Q$ to $B$ because it involves division by ... uh... $2$. $\endgroup$ – John Hughes Oct 13 '19 at 17:18
  • $\begingroup$ I never meant to imply that it was only true for variance and covariance. It was just the example I came across. I do appreciate the answer. I didnt realize all bilinear functions became quadratic when the argument is repeated. $\endgroup$ – SquishyRhode Oct 13 '19 at 17:19

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