Linear Algebra II (pdf, 500 kB)

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Proof. We leave the (by now standard) proof that the given maps are linear as
an exercise. It remains to check that they are inverses of each other. Call the
second map γV,W . So let φ : V × W → F be a bilinear form. Then γV,W (φL)
sends (v, w) to (φL(v))(w) = φ(v, w), so γV,W ◦ βV,W is the identity. Conversely,
let f ∈ Hom(V, W ∗ ), and let φ = γV,W (f). Then for v ∈ V , φ(v) sends w to
(φL(v))(w) = φ(v, w) = (f(v))(w), so φL(v) = f(v) for all v ∈ V , hence φL = f.
This shows that βV,W ◦ γV,W is also the identity map.
If V = W , we write βV : Bil(V ) → Hom(V, V ∗ ) for this isomorphism.
8.8. Example. Let V now be finite-dimensional. We see that a non-degenerate
bilinear form φ on V allows us to identify V with V ∗ via the isomorphism φL.
On the other hand, if we fix a basis v1, . . . , vn, we also obtain an isomorphism
ι : V → V ∗ by sending vj to v∗ j , where v∗ 1, . . . , v∗ n is the dual basis of V ∗ . What
is the bilinear form corresponding to this map? We have, for v = n j=1 λjvj,
w = n j=1 µjvj,
φ(v, w) = ι(v) (w) =
n =
j=1
λjv ∗ j
n ι
n
k=1
j=1
µkvk
λjvj
=
n
n
j,k=1
k=1
µkvk
λiµk v ∗ j (vk) =
n
λiµkδjk =
j,k=1
n
j=1
λjµj .
This is just the standard dot product if we identify V with F n using the given
basis; it is a symmetric bilinear form on V.
8.9. Corollary. Let V be a finite-dimensional vector space, and let φ be a nondegenerate
bilinear form on V. Then every linear form ψ ∈ V ∗ is represented as
ψ(w) = φ(v, w) for a unique v ∈ V.
Proof. The equality ψ = φ(v, ·) means that ψ = φL(v). The claim now follows
from the fact that φL is an isomorphism.
8.10. Example. Let V be the real vector space of polynomials of degree at most 2.
Then
φ : (p, q) ↦−→
1
0
p(x)q(x) dx
is a bilinear form on V. It is non-degenerate since for p = 0, we have φ(p, p) > 0.
Evaluation at zero p ↦→ p(0) defines a linear form on V, which by Cor. 8.9 must
be representable in the form p(0) = φ(q, p) for some q ∈ V. To find q, we have to
solve a linear system:
φ(a0 + a1x + a2x 2 , b0 + b1x + b2x 2 )
= a0b0 + 1
2 (a0b1 + a1b0) + 1
3 (a0b2 + a1b1 + a2b0) + 1
4 (a1b2 + a2b2) + 1
5 a2b2 ,
and we want to find a0, a1, a2 such that this is always equal to b0. This leads to
a0 + 1
2 a1 + 1
3 a2 = 1 ,
1
2 a0 + 1
3 a1 + 1
4 a2 = 0 ,
1
3 a0 + 1
4 a1 + 1
5 a2 = 0