# Bilinear forms - Paul Halmos

In the section about Bilinear forms (23) in Halmos's Finite dimensional vector spaces. He talks about $$w(x,y) = [x,y]$$ being a bilinear form in $$U \oplus U'$$. I understand that $$[x,y] = x(y) = y(x)$$ But the next leap is something I couldn't follow, as in $$w(x,y)$$ being a bilinear form on arbitrary vector spaces $$U$$ and $$V$$ and $$w(x,y) = u(x)v(x)$$.

Why should the linear functional on $$U \oplus V$$ be a product of linear functionals on the respective subspaces? I am not sure where to start to prove that.

Add to the fact that, in the following tensor product section he claims that $$w$$ is an element of $$(U \otimes V)'$$. Does that mean that $$(U \oplus V)'$$ is a subspace of $$(U \otimes V)'$$?

• It's not necessarily a product of linear functionals. We can only say that it equals a sum of products of linear functionals. The set of products of linear functionals (with obvious operations) is not a vector space, while the space of sums of products is. Dec 23, 2021 at 2:40

Halmos is only saying here that if $$u(x)$$ is a linear form on $$U$$ and $$v(y)$$ is a linear form on $$V$$, then $$w(x,y)=u(x)v(y)$$ is a bilinear form on $$U\oplus V$$. He's not saying that every bilinear form on $$U\oplus V$$ has this form. He's just providing an example, not making a statement about all bilinear forms.
The fact that $$w$$ is in $$(U\otimes V)'$$ follows immediately from Halmos' definition of the tensor product of $$U$$ and $$V$$ as the dual of the space of bilinear forms on $$U\oplus V$$.
You're also not writing carefully. You wrote $$w(x,y)=u(x)v(x)$$ which is wrong, and "linear functional on $$U\oplus V$$" which is also wrong. It's important to pay attention to details when doing math.