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 $Q(X+Y)=Q(X)+2B(X,Y)+Q(Y)$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.