Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What does bilinear really mean? Everytime I heard the word, I think it should be "linear in 2 ways?"

For example, from the definition of inner product (taken from Appendix A of "Wavelets For Computer Graphics" by Stollnitz):

An inner product on a vector space V is any map from $ V \times V $ to $\mathbb{R}$ that is:

  • Symmetric $ \langle u | v \rangle = \langle v | u \rangle $
  • Bilinear $ \langle au + bv | w \rangle = a \langle u | w \rangle + b \langle v | w \rangle $
  • Positive definite $ \langle u | u \rangle > 0 $ for all $ u \ne 0 $

But how is bilinearity "linear in 2 ways", (if bilinear really does mean $2\times$ linear!)

share|cite|improve this question
Addition ($A(x,y) = x + y$) is linear. Multiplication ($M(x,y) = xy$) is bilinear, not linear. – KCd Nov 25 '11 at 22:53
Addition for positive numbers is bilinear? – bobobobo Nov 25 '11 at 23:10
@bobobobo: no. For fixed $a$ the function $x \mapsto x+a$ is linear iff $a = 0$. – Qiaochu Yuan Nov 25 '11 at 23:24
up vote 13 down vote accepted

It means linear on the left:

$$\langle au + bv, w \rangle = a \langle u, w \rangle + b \langle v, w \rangle$$

and on the right:

$$\langle u, av + bw \rangle = a \langle u, v \rangle + b \langle u, w \rangle.$$

By symmetry, linearity on the left implies linearity on the right, so that's why right-linearity isn't explicitly mentioned (although strictly speaking it should be).

share|cite|improve this answer
Therefore the "bilinearity" should be a special case of "multilinearity" (say, a tensor) ? – Vim May 12 '15 at 12:20

By currying, you can think of a function of two variables, $f\colon X\times Y \to Z$, as a composition of two functions: first, given $x\in X$, you have a function that sends $x$ to the function $f_x\colon Y\to Z$; and then this function is evaluated at a $y\in Y$ to give $f_x(y) = f(x,y)$.

When $X$, $Y$, and $Z$ are vector spaces, the set of functions from $Y$ to $Z$ (written $Z^Y$) is also a vector space, so the notion of linearity makes sense for the set of all such functions.

The function $f$ is "bilinear" if and only if both the map $X\to Z^Y$ and the maps $Y\to Z$ that we get are linear. That is, if and only if $f_{x+\alpha x'} = f_x+\alpha f_{x'}$ for all $x,x'\in X$, and scalar $\alpha$; and for each $x\in X$ the map $f_x\colon Y\to Z$ is linear. So $f$ itself is obtained by working with two linear functions.

(Symmetrically, you can think of $f$ as given first by a function that takes $y\in Y$ to a function $g_y\colon X\to Z$, and then this function is evaluated at $x\in X$ via $g_y(x) = f(x,y)$. Again, $f$ is linear if and only if the function $g\colon Y\to Z^X$ is linear, and the functions $g_y\colon X\to Z$ are each linear.)

Alternatively, notice that $f$ is a function of two variables. Bilinearity is precisely the condition "linear in each of the variables separately". So you have a function which is linear in two distinct ways: in the first variable, and in the second variable.

(Just wait until you get to the notion of sesquilinear funtions; trying to interpret "one-and-a-half linear" will give you a headache).

Also remember that not all nomenclature is necessarily accurate-as-to-intuition, and that the meaning of some words has sometimes changed from the time in which the nomenclature because fixed and present day.

share|cite|improve this answer
+ for the link to currying – hAcKnRoCk Apr 26 '13 at 17:48

Well, the definition of "bilinear" you have there is really the definition of "linear in the first variable." When you combine this with the symmetry axiom above, you get "linear in both the first and second variables," which is what "bilinear" means.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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