### Dual

##### 2. Dual basis

A basis of a vector space can be used to construct an isomorphism, via the coordinates, between the vector space and the euclidean space. The basis can also be used to identify the dual space with the euclidean space.

Let {b1, b2, ..., bn} be a basis of V. Then

f → (f(b1), f(b2), ..., f(bn)), V*Rn

is an isomorphism. In particular, we have dimV = dimV*.

Proof Denote the transformation by E(f) = (f(b1), f(b2), ..., f(bn)). The following verifies E(f + g) = E(f) + E(g).

E(f + g)
= ((f + g)(b1), (f + g)(b2), ..., (f + g))(bn))
= (f(b1) + g(b1), f(b2) + g(b2), ..., f(bn) + g(bn)) (definition of the addition f + g)
= (f(b1), f(b2), ..., f(bn)) + (g(b1), g(b2), ..., g(bn))
= E(f) + E(g).

We may verify E(cf) = cE(f) in the similar way. Therefore E is linear.

Next, we construct an inverse for E. For any a = (a1, a2, ..., an) ∈ Rn, we construct a linear function on V via the formula

fa(c1b1 + c2b2 + ... + cnbn) = a1c1 + a2c2 + ... + ancn.

Strictly speaking, we need to explain that fa is well-defined and linear. This can be proved directly, or can be seen from the fact that fa is a composition

v = c1b1 + c2b2 + ... + cnbn → (c1, c2, ..., cn) → a1c1 + a2c2 + ... + ancn,

where the first → is the coordinates with respect to the basis {b1, b2, ..., bn}, and the second → is a well-known linear function on the euclidean space.

By the way fa is constructed, it is easy to verify that the transformation afa, RnV* is the inverse of E.

In the special case V = Rn and {b1, b2, ..., bn} is the standard basis, the isomorphism f → (f(b1), f(b2), ..., f(bn)) is simply the isomorphism ιn: a1x1 + a2x2 + ... + anxn → (a1, a2, ..., an) constructed early on.

Under the isomorphism f → (f(b1), f(b2), ..., f(bn)), V*Rn, the standard basis on the right side corresponds to a basis {β1, β2, ... , βn} of V*. In other words, we should have (βi(b1), βi(b2), ..., βi(bn)) = ei. This leads to the following concept.

The dual basis {β1, β2, ... , βn} of a basis {b1, b2, ..., bn} of V consists of linear functions satisfying

 βi(bj) = { 1, if i = j 0, if i ≠ j

Example The linear functions

ε1(x1, x2, x3) = x1,
ε2(x1, x2, x3) = x2,
ε3(x1, x2, x3) = x3,

form a basis of (R3)*, which is in fact the dual basis of the standard basis of R3.

The linear functions

α(a + bt + ct2 + dt3) = a,
β(a + bt + ct2 + dt3) = b,
γ(a + bt + ct2 + dt3) = c,
δ(a + bt + ct2 + dt3) = d,

form the dual basis of the basis {1, t, t2, t3} of P3.

Example In an earlier example, we had a basis B = {(3, 1, 2), (1, -1, 2), (-1, 1, 1)} of R3. Let B* = {a11x1 + a12x2 + a13x3, a21x1 + a22x2 + a23x3, a31x1 + a32x2 + a33x3} be its dual basis. Then the definition of the dual basis can be reformulated exactly as

 [ a11 a12 a13 ][ 3 1 -1 ] = [ 1 0 0 ]. a21 a22 a23 1 -1 1 0 1 0 a31 a32 a33 2 2 1 0 0 1

Then by this example, we get

 [ a11 a12 a13 ] = [ 3 1 -1 ]-1 = [ 1/4 1/4 0 ]. a21 a22 a23 1 -1 1 -1/12 -5/12 1/3 a31 a32 a33 2 2 1 -1/3 1/3 1/3

We conclude that the dual basis is B* = {(1/4)x1 + (1/4)x2, -(1/12)x1 - (5/12)x2 + (1/3)x3, -(1/3)x1 + (1/3)x2 + (1/3)x3}.