Linear Algebra II (pdf, 500 kB)

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Applying this to d ′ and n ′ , we see that d and d ′ commute, and that n and n ′
commute. We can write
d − d ′ = n ′ − n ;
then the right hand side is nilpotent (for this we need that n and n ′ commute!). By
Thm. 5.9, the left hand side is diagonalizable, so we can assume it is represented
by a diagonal matrix. But the only nilpotent diagonal matrix is the zero matrix,
therefore d − d ′ = n ′ − n = 0, i.e., d ′ = d and n ′ = n.
6. The Dual Vector Space
6.1. Defintion. Let V be an F -vector space. A linear form or linear functional
on V is a linear map φ : V → F .
The dual vector space of V is V ∗ = Hom(V, F ), the vector space of all linear forms
on V .
Recall how the vector space structure on V ∗ = Hom(V, F ) is defined: for φ, ψ ∈ V ∗
and λ, µ ∈ F , we have, for v ∈ V ,
(λφ + µψ)(v) = λφ(v) + µψ(v) .
6.2. Example. Consider the standard example V = F n . Then the coordinate
maps
are linear forms on V .
The following result is important.
pj : (x1, . . . , xn) ↦−→ xj
6.3. Proposition and Definition. Let V be a finite-dimensional vector space
with basis v1, . . . , vn. Then V ∗ has a unique basis v∗ 1, . . . , v∗ n such that
v ∗
1 if i = j
i (vj) = δij =
0 if i = j .
This basis v ∗ 1, . . . , v ∗ n of V ∗ is called the dual basis of v1, . . . , vn or the basis dual to
v1, . . . , vn.
Proof. Since linear maps are uniquely determined by their images on a basis, there
certainly exist unique linear forms v ∗ i ∈ V ∗ with v ∗ i (vj) = δij. We have to show that
they form a basis of V ∗ . First, it is easy to see that they are linearly independent,
by applying a linear combination to the basis vectors vj:
0 = (λ1v ∗ 1 + · · · + λnv ∗ n)(vj) = λ1δ1j + · · · + λnδnj = λj .
It remains to show that the v ∗ i generate V ∗ . So let φ ∈ V ∗ . Then
φ = φ(v1)v ∗ 1 + · · · + φ(vn)v ∗ n ,
since both sides take the same values on the basis v1, . . . , vn.