# Multilinear Functions

I have a question regarding the properties of a multilinear function. This is for a linear algebra class. I know that for a multilinear function,

$f(cv_1,v_2,\dots,v_n)=c⋅f(v_1,v_2,\dots,v_n)$

Does this imply

$f(cv_1,dv_2,\dot,v_n)=c⋅d⋅f(v_1,v_2,\dots,v_n)$?

It is for a question involving a multilinear function $f:\mathbb{R}^2\times\mathbb{R}^2\times\mathbb{R}^2\to \mathbb{R}$. I am given eight values of $f$, each of which is composed of a combination three unit vectors. They are:

$f((1,0),(1,0),(1,0))=e$

$f((1,0),(1,0),(0,1))=\sqrt{7}$

$f((1,0),(0,1),(1,0))=0$

$f((1,0),(0,1),(0,1))=2$

$f((0,1),(1,0),(1,0))=\sqrt{5}$

$f((0,1),(1,0),(0,1))=0$

$f((0,1),(0,1),(1,0))=\pi$

$f((0,1),(0,1),(0,1))=3$

Then I am asked to compute for different values of f. For instance,

$f((2,3),(-1,1),(7,4))$

How do I approach this question to solve for the value of $f$?

-
I edited your question in $\LaTeX$. Please check if everything is correct. –  Git Gud Jan 20 '13 at 22:36
Besides the property you listed above for scalar product, you'll also need the additive property of multilinear functions in order to solve this problem. –  Ted Jan 20 '13 at 22:59
In a multilinear function, each of the seperate variables are linear. For example $f(cx, by, aw)= cf(x)+ bf(y)+af(w)$, where the property that $f(a+b)=f(a)+f(b)$ of a linear map was used along with the other property. –  Jaivir Baweja Jan 21 '13 at 14:47

as it stands: no, because linearity in first variable (with rest variables fixed) doesn't imply linearity in second variable.

But, $f$ being multilinear means it is linear in each variable (when the rest are fixed), so if we also assume that, the $c\cdot d\cdot$ line is correct.

For the actual question, I couldn't find simpler solution for now, than to introduce 8 undeterminates, say $a_1,..,a_8$, the coefficients of the appropriate "matrices" where $f$ is given. So that $a_1$ belongs to the matrix $((1,0),(1,0),(1,0))$, that is, we need $$a_1((1,0),(1,0),(1,0)) +a_2((1,0),(1,0),(0,1))+... = ((2,3),(−1,1),(7,4))$$ that is, \begin{align} a_1+a_2+a_3+a_4 &= 2 \\ a_5+a_6+a_7+a_8 &=3 \\ a_1+a_2+a_5+a_6 & = -1 \\ \dots \end{align} Then you can solve it using Gaussian elimination, for example. Hmm.. having 6 equations for 8 variables. But it usually means that you can choose 2 variables freely in the solution...

-

If I interpret your question correctly, then the clue is in $f:\mathbb{R}^2\times\mathbb{R}^2\times\mathbb{R}^2\to\mathbb{R}$, which seems to imply that the function is trilinear, that is, in three inputs of two dimensions each. In that case,

\begin{align*}f((2,3),(-1,1),(7,4)) =&2\cdot -1\cdot 7\cdot f((1,0),(1,0),(1,0)) \\+&2\cdot -1\cdot 4\cdot f((1,0),(1,0),(0,1)) \\+&2\cdot 1\cdot 7\cdot f((1,0),(0,1),(1,0)) \\+&2\cdot 1\cdot 4\cdot f((1,0),(0,1),(0,1)) \\+&3\cdot -1\cdot 7\cdot f((0,1),(1,0),(1,0)) \\+&3\cdot -1\cdot 4\cdot f((0,1),(1,0),(0,1)) \\+&3\cdot 1\cdot 7\cdot f((0,1),(0,1),(1,0)) \\+&3\cdot 1\cdot 4\cdot f((0,1),(0,1),(0,1)) \end{align*}

I'll leave you to plug in the values, I'm too lazy.

This might be easier to see if you realize that a bilinear function $f(x,y)=x'Ay$ (where $A$ is a matrix) works the same way.

-
Why was this downvoted? –  Pedro Tamaroff Apr 12 '13 at 23:58
I have no idea. I thought I got the interpretation of the question correct, but I didn't push it. –  John Moeller Apr 13 '13 at 1:56

A multilinear function $f: V_1 \times \ldots \times V_n \to W$, where all the $V_i$ and $W$ are vector spaces is a function such that if we fix $v_1, \ldots v_{i-1}, v_{i+1}, \ldots v_n$, the function $\phi(v_i) := f(v_1, \ldots, v_i, \ldots, v_n)$ is linear in $v_i$. Linearity means: $$\forall v,w \in V_i, a \in \mathbb{R}: \phi(a v + w) = a \phi(v) + \phi(w)$$

Now, since we know that your function is multilinear, and we know it's value for all possible combinations of linearly independent vectors in $V_1, V_2$ and $V_3$, we can express every input $(v_1, v_2, v_3)$ as linear combinations of the unit vectors in the various spaces and then use the multilinearity to calculate the correct value.

For example: $$((2,3),(-1,1),(7,4)) = (2(1,0)+3(0,1), -1(1,0)+(1,0), 7(1,0)+4(0,1)).$$ I hope this clears it up. ;)

-