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.

  • $\begingroup$ 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 $k=3$ $\endgroup$ – janmarqz Jul 20 '18 at 18:33
  • $\begingroup$ @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 $\mathbb{R}^4$ $\endgroup$ – Perturbative Jul 20 '18 at 19:48
  • $\begingroup$ So I take it that Munkres is referring to basis elements of $V$ as opposed to $V^k$ $\endgroup$ – Perturbative Jul 20 '18 at 19:58
  • $\begingroup$ 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": $$\left(\begin{array}{cc}1&0\\0&0\end{array}\right)$$ $$\left(\begin{array}{cc}0&1\\0&0\end{array}\right)$$ $$\left(\begin{array}{cc}0&0\\1&0\end{array}\right)$$ $$\left(\begin{array}{cc}0&0\\0&1\end{array}\right)$$ $\endgroup$ – janmarqz Jul 20 '18 at 20:23
  • $\begingroup$ @janmarqz But that's not what the theorem above asserts.. $\endgroup$ – 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.

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  • $\begingroup$ 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? $\endgroup$ – Perturbative Jul 22 '18 at 13:33
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    $\begingroup$ Right. That's why we're considering the $n^k$ $k$-tuples $(a_{i_1},\dots,a_{i_k})$. $\endgroup$ – Ted Shifrin Jul 22 '18 at 14:57
  • $\begingroup$ Okay thanks again Ted! :) $\endgroup$ – Perturbative Jul 22 '18 at 16:58

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