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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,

$$f(c\vec{v}_1, \vec{v}_2,\ldots,\vec{v}_n)=c \cdot f(\vec{v}_1, \vec{v}_2,\ldots,\vec{v}_n)$$

Does this imply

$$f(c\vec{v}_1, d\vec{v}_2,\ldots,\vec{v}_n)=c\cdot d \cdot f(\vec{v}_1, \vec{v}_2,\ldots,\vec{v}_n)?$$

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. For instance, $$ f\left ( \begin{bmatrix} 1\\ 0 \end{bmatrix} , \begin{bmatrix} 1\\ 0 \end{bmatrix} , \begin{bmatrix} 1\\ 0 \end{bmatrix} \right ) =e $$

and

$$ f\left ( \begin{bmatrix} 0\\ 1 \end{bmatrix} , \begin{bmatrix} 0\\ 1 \end{bmatrix} , \begin{bmatrix} 0\\ 1 \end{bmatrix} \right ) =3 $$

or, $f(\vec{e}_1,\vec{e}_1,\vec{e}_1)=e$. Then I am asked to compute for different values of $f$. For instance,

$$ f\left ( \begin{bmatrix} 1\\ 2 \end{bmatrix} , \begin{bmatrix} 1\\ 3 \end{bmatrix} , \begin{bmatrix} 1\\ 5 \end{bmatrix} \right ) $$

In this case I used

$$f(\vec{e}_1 + 2\vec{e}_2,\vec{e}_1 + 3\vec{e}_2,\vec{e}_1 + 5\vec{e}_2) = f(\vec{e}_1 ,\vec{e}_1,\vec{e}_1) + f(2\vec{e}_2 ,3\vec{e}_2,5\vec{e}_2)= f(\vec{e}_1 ,\vec{e}_1,\vec{e}_1) + 2 \cdot 3 \cdot 5 \cdot f(\vec{e}_2 ,\vec{e}_2,\vec{e}_2)$$

Which, using the given values, equals $(2)(3)(5)(3) + e = 90+e$. Is this okay?

Michael Hardy
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Delightful_Richard
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1 Answers1

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Yes. Multilinear means $f$ is linear in each term. So in the 3 variable case.

$$f(cx + x', y, z) = cf(x, y, z) + f(x', y, z)$$ $$f(x, dy + y', z) = df(x, y, z) + f(x, y', z)$$ $$f(x, y, ez + z') = ef(x, y, z) + f(x, y, z')$$

So then if you have $f(2x, 3y, 5z)$,

$$f(2x, 3y, 5z) = 2f(x, 3y, 5z) = 2\cdot 3f(x, y, 5z) = 2\cdot 3\cdot 5f(x, y, z)$$.

zrbecker
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