## Halmos Section 23: Bilinear Forms

**Point of post:** This post will document the solutions to Halmos’s section 23 on Bilinear forms, the content of which is explained (not quoted) here. Some interesting results in this post is the, very very perfunctory, discussion of quadratic forms and a how a symmetric bilinear form on a field of characteristic greater than two is completely determined by it’s associated quadratic form.

*Remark:* As usual, I use somewhat different notation than Halmos. To see this, and examples, etc. see the above link.

1.

**Problem: **

**a) **If , then there exists scalars such that

**b) **If then there exists scalars such that

**Proof:**

**a) **This follows immediately from my discussion in the above link by taking

**b) **This is true in a much more general setting as we shall see in the next section, but for now it suffices to note that for any we have that

Thus, note that if is a linear functional on that the above implies that

and clearly this representation is unique. So, taking and gives the desired result.

2.

**Problem:** A bilinear form is *degenerate* if, as a function of one of its two arguments, it vanishes identically for some non-zero value of its other argument, otherwise it is *non-degenerate*

**a) **Give an example of a degenerate linear form (not identically zero) on

**b) **Give an example of a non-degenerate bilinear form on .

**Proof: **Define

Evidently is linear (this really isn’t that hard to check), but notice that

**b) **Consider (as defined in the last post). Then, is a bilinear form on but notice that

and

Thus, if we fixed the first entry of , then the only way it can be zero for all second entries is if it’s first entry is zero. Considering that interchanging the columns (switching the first and second entries) only changed the sign of the conclusion follows.

3.

**Problem:** If , if and if is defined by , prove that . Is it true that if is non-degenerate, then every linear functional on can be obtained this way?

**Proof: **Evidently such a is a linear functional since

Now, consider the map

clearly this is linear since

Also note that if is non-degenerate then

and by assumption this implies that . Thus, is injective. It follows that

but if this function cannot be an isomorphism, and thus cannot be surjective. For a concrete example consider to be over and to over itself. Then, by the above analysis any non-degenerate function cannot induce a surjection . But, let’s show that for a specific example. Consider

Then, we note that

but, clearly is a linear functional and not of the above form.

4.

**Problem: **Suppose that for each , the function is defined by

**a) **

**b) **

**c) **

**d) **

**Proof:**

**a) **This is evidently.

Since the rest follows by symmetry.

**b)** This is not since

**c) **This is since

And the other direction follows by symmetry

**d) **This is, the case that is linear in the first entry is trivial (done above basically) and the second follows since

**a) **This is non-degenerate since if one fixed then if for all , then in particular

**c) **This is degenerate since for all

**d) **Similarly noting that for all we see that is degenerate.

5.

**Problem:** Does there exist a vector space and a bilinear form such that but for all ?

**Proof:** Yes. Let and take . Then clearly this is not the zero form since

but,

6.

**Problem: **

**a) **A bilinear form on is *symmetric *if for all . A *quadratic form *on is a function by taking for some . Prove that if , then every symmetric bilinear form is uniquely determined by the corresponding quadratic form. What happens if ?

**b) **Can a non-symmetric bilinear form define the same quadratic linear form as a symmetric one?

**Proof:**

**a) **As is common in field theoretic literature denote . Note then that since we (and it evidently does not equal one, by the assumption on fields that ) see that and so has an inverse . But, notice that (remembering that we assumed that is symmetric)

\

and thus from prior discussion we may conclude that

Thus, if was such that and symmetric we note that for every we have that

Evidently a problem can occur if since will not have a multiplicative inverse.

**b) **Yes. What about the zero form and ? The first is clearly symmetric since

for all . Also, is non-symmetric since

That said, if are the quadratic forms for respectively then we see (as was shown earlier)

for any .

**References:**

1. Halmos, Paul R. *Finite-dimensional Vector Spaces,*. New York: Springer-Verlag, 1974. Print.

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