Recipe

Evaluate a bilinear form and check its properties

Use when: you're given a two-argument formula and need its value at a specific pair, or need to confirm it's genuinely bilinear and check whether it's symmetric

Goal: the form's value at any given pair, plus confidence in whether it's linear in each argument and whether it's symmetric

Taught in Bilinear forms

Quick recipe

Evaluate the form at a specific pair
Test linearity in the first argument
Test linearity in the second argument
Check whether the form is symmetric

1
Substitute each vector's components directly into the formula, one term at a time.

Example

For $B(\mathbf u,\mathbf v)=2u_1v_1+u_1v_2+3u_2v_2$ , evaluate $B((1,2),(3,1))$ : $2(1)(3)+(1)(1)+3(2)(1)=6+1+6=13$ .

Why: a bilinear form is completely determined by its formula, so substituting components computes it the same way any function gets evaluated.
2
Fix $\mathbf v$ , pick two vectors $\mathbf u_1,\mathbf u_2$ , and confirm $B(\mathbf u_1+\mathbf u_2,\mathbf v)$ equals $B(\mathbf u_1,\mathbf v)+B(\mathbf u_2,\mathbf v)$ .

Example

With $\mathbf u_1=(1,0)$ , $\mathbf u_2=(0,1)$ , $\mathbf v=(2,1)$ : $B((1,1),(2,1))=8$ , while $B((1,0),(2,1))+B((0,1),(2,1))=5+3=8$ .
- the two sides match — the "linear in the first slot" half of bilinearity holds for this pair.
- the two sides disagree — the expression fails to be bilinear, no matter how it looked at first glance.
Why: bilinearity in the first slot requires this equality for every choice of vectors, so a single mismatched pair is already a genuine counterexample.
3
Repeat the same additivity test with the roles reversed, fixing $\mathbf u$ this time and splitting $\mathbf v$ instead.

Example

With $\mathbf u=(1,1)$ , $\mathbf v_1=(1,0)$ , $\mathbf v_2=(0,1)$ : $B((1,1),(1,1))=6$ , while $B((1,1),(1,0))+B((1,1),(0,1))=2+4=6$ .

Why: bilinearity is two separate promises, one per argument, so confirming the first slot says nothing about the second.
4
Compute $B(\mathbf u,\mathbf v)$ and $B(\mathbf v,\mathbf u)$ for a pair where the two values could plausibly differ, then compare them.

Example

$B((1,0),(0,1))=1$ , but $B((0,1),(1,0))=0$ for the same $B$ above, so the two orders genuinely disagree.
- the two values agree — $B$ is symmetric on this pair, worth checking more pairs before trusting it everywhere.
- the two values disagree — $B$ is a genuine asymmetric bilinear form, still a completely legitimate one.
Why: bilinearity alone never guarantees $B(\mathbf u,\mathbf v)=B(\mathbf v,\mathbf u)$ ; that's an extra property studied on its own later in this chapter.

Common mistakes

Assuming a formula is bilinear just because it looks similar to the dot product, without testing additivity in either slot.
Testing linearity with only one pair of vectors and treating that as proof, rather than as one piece of supporting evidence.
Concluding a form is symmetric from a single matching pair instead of checking the general formula.
Mixing up which argument is "first" when writing $B(\mathbf u,\mathbf v)$ versus $B(\mathbf v,\mathbf u)$ .