Dimension of spaces of bi/linear maps

For $V$ a finite dimensional vector space over a field $\mathbb{K}$, I have encountered the claim that $$\dim(\mathrm{Hom}(V,V)) = \dim(\mathrm{Hom}(V \times V, \mathbb{K}))$$

where $\mathrm{Hom}(V,V)$ denote the vector spaces, respectively, of all linear maps from $V$ to $V$ and all bilinear maps from $V\times V$ to the ground field $\mathbb{K}$. I'm sure I'm overlooking something elementary, but I don't see this.

There is a theorem that, in general, for any finite-dimensional vector spaces $V$ and $W$ that $$\dim(\mathrm{Hom}(V,W)) = \dim(V)\dim(W)$$

But, $\dim(V \times W) = \dim(V) + \dim(W)$ and therefore $$\dim(\mathrm{Hom}(V \times V, \mathbb{K})) = (\dim(V) + \dim(V))\cdot \dim(K) = 2\dim(V)\cdot 1$$ which is obviously not equal to $\dim(\mathrm{Hom}(V,V)) = \dim(V)\cdot\dim(V)$

Where is my mistake?

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A bilinear map from $V \times V$ to $\mathbb{K}$ is not the same thing as a linear map from $V \times V$ to $\mathbb{K}$. You have computed the dimension of the latter. –  Chris Eagle May 30 '12 at 16:01
Ok, so I guess I need to know how to calculate the dimension of the vector space of bilinear (n-linear?) maps to the ground field. Can you suggest a reference that discusses this? –  AFX May 30 '12 at 16:09
@AFX Here's a googlebooks link that looks very helpful. –  rschwieb May 30 '12 at 16:12
@AFX: You really should stop using $\mathrm{Hom}$ for both linear maps and bilinear maps. That's what got you into trouble in the first place. Assuming the first $\mathrm{Hom}$ refers to bilinear, what I'm saying is that those two spaces are not only isomorphic, but even naturally isomorphic. –  Chris Eagle May 30 '12 at 16:21
@AFX I seen Bilin(-,-) and Bihom(-,-) both used, and your suggestion is OK as long as you alert the reader to what it means. –  rschwieb May 30 '12 at 16:52
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I would not write $\operatorname{Hom}(V \times V, \mathbb K)$ for the space of bilinear maps, since there is nothing to distinguish this from your old notation for the space of linear maps. I've seen $L^2(V, V; \mathbb K)$ used, but $\operatorname{Bilin}(V, V; \mathbb K)$ has the advantage of being obvious. In any case, it never hurts to specify your notation.

Now to calculate the dimension. Let $\{e_1, \ldots, e_n\}$ be a basis for $V$, and let $\{f_i\}$ be the corresponding dual basis. Then I claim that the set of $n^2$ bilinear maps $F_{ij}(x, y) = f_i(x)f_j(y) \qquad i, j = 1, \ldots, n$ is a basis for $L^2(V, V; \mathbb K)$. To remember this fact, it might help to recall how bilinear forms correspond to matrices after one has chosen a basis.

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Chris Eagle has already comment-answered the main problem, but I'm also going to add that the set $Bilin_{\mathbb{K}}(V\times V,\mathbb{K})$ is another way of writing $V\otimes_{\mathbb{K}}V$, and this latter guy has dimension $dim(V)^2$. This is the interpretation of tensors as multilinear functionals.
EDIT: As Chris was nice enough to remind me, $Bilin_{\mathbb{K}}(V\times V,\mathbb{K})$ is actually naturally identified with the dual module $(V\otimes_{\mathbb{K}}V)^\ast$ rather than just $V\otimes_{\mathbb{K}}V$. But since finite dimensional vector spaces are isomorphic to their duals, the statement about dimensions is still OK.
You still see this first definition of $V \otimes V$ in most differential geometry books, I'm afraid. –  Dylan Moreland May 30 '12 at 16:52