Question from the topic "Linear Algebra"

Question

Consider the vector space V = R+ × R+ × R+, where R+ is the set of all positive real numbers with

addition and scalar multiplication defined by

(x1, y1, z1) ⊕ (x2, y2, z2) = (x1x2, y1y2, z1z2) and

αⵙ(x, y, z) = (x^α, y^α, z^α), α ∈ R.

(a) Find the dimension of the vector space V .

(b) Find a basis of V which contains the element (e, e, 1).

Answer 1

Solution:

Choose u = (10, 1, 1), v = (1, 10, 1), w = (1, 1, 10).

(1) Show that {u, v, w} spans V:

Suppose (a, b, c) ∈ V be any vector. We have

(a, b, c) = (a, 1, 1) ⊕ (1, b, 1) ⊕ (1, 1, c)

= (10^(log a), 1, 1) ⊕ (1, 10^(log b), 1) ⊕ (1, 1, 10^(log c))

= (log a) ¤ (10, 1, 1) ⊕ (log b) ¤ (1, 10, 1) ⊕ (log c) ¤ (1, 1, 10)

= (log a) ¤ u ⊕ (log b) ¤ v ⊕ (log c) ¤ w

This shows that {u, v, w} spans V.

(2) Show that u, v, w are linearly independent:

Let a ¤ u + b ¤ v +c ¤ w = 0. Notice 0 = (1, 1, 1) in V. Then

a ¤ (10, 1, 1) ⊕ b ¤ (1, 10, 1) ⊕ c ¤ (1, 1, 10) = (1, 1, 1)

(10^a, 1, 1,) ⊕ (1, 10^b, 1) ⊕ (1, 1, 10^c) = (1, 1, 1)

(10^a, 10^b, 10^c) = (1, 1, 1)

So 10^a = 1, 10^b =1, and 10^c = 1; or a = b = c = 0.

This shows that u, v, w, are linearly independent.

From (1) and (2) it follows that {u, v, w} is a basis of V, so dim V =3.

(b) Choose u = (e, e, 1). v = (e, 1, 1) and w = (1, 1, e).

We show that u, v, and w are linearly independent.

Let a ¤ u ⊕ b ¤ v ⊕ c ¤ w = (1, 1, 1). Then

a ¤ (e, e, 1) ⊕ b ¤ (1, e, 1) ⊕ c ¤ (1, 1, e) = (1, 1, 1)

(e^a, e^a, 1) ⊕ (1, e^b, 1) ⊕ (1, 1, e^c) = (1, 1, 1)

(e^a, e^(a+b), e^c) = (1, 1, 1)

So e^a = 1, e^(a+b) = 1, e^c = 1; or a = b = c =0.

This shows that u, v, and w are linearly independent.

Because V has dimension 3, {u, v, w} is a basis of V. ◊

Answer 2

One way to work out the basis for V is to consider (p,q,r)∈R³, and use row-reduction to derive the basis

(p,q,r) = ((q+r–p)/2)(0,1,1) + ((r+p–q)/2)(1,0,1) + ((p+q–r)/2)(1,1,0).

Then observe that under the map

s = (p,q,r) = (ln x,ln y,ln z)

from V or

v = (x,y,z) = (e^p,e^q,e^r)

to V, the algebra transforms like

(v₁⊕v₂)_i = v_1iv_2i = exp(s_1i+s_2i)

and

(αⵙv)_i = exp(αs_i).

Combining this observation with the basis for R³ yields the basis for V:

(x,y,z) = ((ln√(yz/x))ⵙ(1,e,e)) ⊕ ((ln√(zx/y))ⵙ(e,1,e)) ⊕ ((ln√(xy/z))ⵙ(e,e,1)).