# On bilinear and quadratic forms: Why isnt the variance a bilinear form?

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?

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 $$Q(X+Y)=Q(X)+2B(X,Y)+Q(Y)$$.
• 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... $2$. – John Hughes Oct 13 '19 at 17:18