# how does semi-inner product and symmetric positive semi-definite bilinear form are different?

Given vector space $V$ over scalar field $\mathbb R$, I wonder if two definitions "semi-inner product" and "symmetric positive semi-definite bilinear form" are actually equivalent. The definition of "semi-inner product" is given in the textbook " a course in functional analysis" by Conway. I can not find exact definition of second one, but the meaning is pretty straightforward, and often used by many people. If they are different, can you find an example which belongs to one but not the other?

• In this case, they are actually the same. – Luiz Cordeiro May 13 '14 at 5:01

A semi-inner product on a vector space $V$ over the field $\mathbb{F} \in \{\mathbb{R}, \mathbb{C}\}$ is a map $V\times V \to \mathbb{F}$ that is linear in the first variable, conjugate symmetric, and positive semi-definite.

So a semi-inner product on a real vector space $V$ is linear in the first variable, and by symmetry (no conjugation occurs), linear in the second variable. Therefore, a semi-inner product on a real vector space is a bilinear, symmetric, positive semi-definite form on $V$. The story for complex vector spaces however is quite different.

On a complex vector space $V$, the only bilinear, symmetric, positive semi-definite form is the zero map. To see this, suppose $B$ is such a form, then

$$B(v, v) = B(-i(iv), -i(iv)) = (-i)^2B(iv,iv) = -B(iv,iv) \leq 0$$

so $B(v,v) = 0$. Now

$$0 = B(u+v, u+v) = B(u,u)+B(u,v)+B(v,u)+B(v,v) = 2B(u,v)$$

so $B(u, v) = 0$. That is, $B$ is the zero map.

So, every bilinear, symmetric, positive semi-definite form on a complex vector space is a semi-inner product, but not a particularly interesting one. Of course, the converse doesn't hold; not every semi-inner product on a complex vector space is a bilinear, symmetric, positive semi-definite form because there are non-trivial semi-inner products.