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 $(U \otimes V)'$?

  • 1
    $\begingroup$ 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. $\endgroup$
    – Ivo Terek
    Dec 23, 2021 at 2:40

1 Answer 1


You're not reading carefully. Here's the relevant section of the text:


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.


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