# Multilinear transformations being determined by their values on basis elements

The following is stated in the book Analysis on Manifolds by James Munkres

Just as is the case with linear transformations, a multilinear transformation is entirely determined once one knows its values on basis elements. That we now prove.

And then he gives the following lemma.

Lemma 26.2 Let $a_1, \dots, a_n$ be a basis for $V$. If $f, g : V^k \to \mathbb{R}$ are $k$-tensors on $V$ and if $$f\left(a_{i_1}, \dots, a_{i_k}\right) = g\left(a_{i_1}, \dots, a_{i_k}\right)$$ for every $k$-tuple $I = (i_1, \dots, i_k)$ of integers from the set $\{1, \dots, n\}$ then $f=g$.

Now what I don't understand is why to we even need the following in the above lemma

"for every $k$-tuple $I = (i_1, \dots, i_k)$ of integers from the set $\{1, \dots, n\}$"

Because $V^k$ has dimension $k \cdot n$ since $V$ has dimension $n$, and has as basis elements \begin{align*}&(a_1, 0, \dots, 0), \dots, &(a_n,0, \dots, 0), \\ &(0, a_1, \dots, 0), \dots, &(0, a_n, \dots, 0), \\ \ \ \ &. &. \\ \ \ \ &. &. \\ \ \ \ &. &. \\ &(0, 0, \dots, a_1), \dots, &(0, 0, \dots, a_n) \end{align*}

So if $\mathcal{B}$ was the set of basis elements of $V^k$ above then I'd say that the following proposed lemma would make more sense

Proposed Lemma: Let $V$ be a vector space of dimension $n$. If $f, g : V^k \to \mathbb{R}$ are $k$-tensors on $V$ and if $$f(\alpha) = g(\alpha)$$ for every $\alpha \in \mathcal{B}$ where $\mathcal{B}$ is a basis for $V^k$, then $f= g$

Furthermore the same part of Lemma 26.2

"for every $k$-tuple $I = (i_1, \dots, i_k)$ of integers from the set $\{1, \dots, n\}$"

Taken literally gives $n^k$ possible $k$-tuples, which would correspond to checking to see if the values of $f$ and $g$ agree on $n^k$ basis elements, which confuses me since $\dim(V^k) = kn$

I'm sure that Lemma 26.2 must be correct and I'm just making some error somewhere, if so could someone please point out what that error is.

• take a look at how this construction looks for $V=\Bbb R^1$ or $V=\Bbb R^2$ or $V=\Bbb R^3$ with $k=1$ or $=2$ or $k=3$ – janmarqz Jul 20 '18 at 18:33
• @janmarqz Okay so take the case $V = \mathbb{R}^2$ and $k=2$, suppose $f, g : V^2 = \mathbb{R}^4 \to \mathbb{R}$ are two $2$-tensors, by the above theorem we need to show $$f(e_1, e_1) = g(e_1, e_1);\\ f(e_1, e_2) = g(e_1, e_2);\\ f(e_2, e_1) = g(e_2, e_1); \\ f(e_2, e_2) = g(e_2, e_2);$$ for basis elements $e_1 = (1, 0)$ and $e_2 = (0, 1)$ of $\mathbb{R}^2$. But for example $(e_1, e_1) = (1, 0, 1, 0)$ is not a basis element of $\mathbb{R}^4$ – Perturbative Jul 20 '18 at 19:48
• So I take it that Munkres is referring to basis elements of $V$ as opposed to $V^k$ – Perturbative Jul 20 '18 at 19:58
• for bilinear maps $\Bbb R^2\times\Bbb R^2\to\Bbb R$ (which is a vector space) you only need the 4 basic "vectors": $$\left(\begin{array}{cc}1&0\\0&0\end{array}\right)$$ $$\left(\begin{array}{cc}0&1\\0&0\end{array}\right)$$ $$\left(\begin{array}{cc}0&0\\1&0\end{array}\right)$$ $$\left(\begin{array}{cc}0&0\\0&1\end{array}\right)$$ – janmarqz Jul 20 '18 at 20:23
• @janmarqz But that's not what the theorem above asserts.. – Perturbative Jul 20 '18 at 21:29

The dimension of $\underbrace{V\otimes\dots\otimes V}_{k \text{ times}} = \bigotimes^k V$ is $n^k$, even though the dimension of $V^k$ is $kn$. We're talking about multilinear maps on $V^k$, not linear maps. (Note that $n=k=2$ is a bad example to pick, since then $kn = n^k$. :))
EDIT: I should comment that the vector space of multilinear maps on $V^k$ is isomorphic to $\big({}\bigotimes^k V\big)^*$, but dimensions are the same.
• Thanks for your answer Ted! Just one follow up question, you're saying then that Munkres is talking about basis elements from $\bigotimes^k V$ and not $V^k$, correct? – Perturbative Jul 22 '18 at 13:33
• Right. That's why we're considering the $n^k$ $k$-tuples $(a_{i_1},\dots,a_{i_k})$. – Ted Shifrin Jul 22 '18 at 14:57