Page 381 - 35Linear Algebra

G.4 Vector Spaces 381

We are half-way done, now we need to consider the rules for scalar multipli-
cation. Notice, that we multiply vectors by scalars (i.e. numbers) but do NOT
multiply a vectors by vectors.
(·i) Multiplicative closure: Again, we are checking that an operation does
not produce vectors outside the vector space. For a scalar a ∈ R, we

x 1 2
require that a lies in R . First we compute using our component-
x 2
wise rule for scalars times vectors:

x 1 ax 1
a = .
x 2 ax 2
Since products of real numbers ax 1 and ax 2 are again real numbers we see
2
this is indeed inside R .
(·ii) Multiplicative distributivity: The equation we need to check is

x 1 ? x 1 x 1
(a + b) = a + b .
x 2 x 2 x 2
Once again this is a simple LHS=RHS proof using properties of the real
numbers. Starting on the left we have

x 1 (a + b)x 1 ax 1 + bx 1
(a + b) = =
x 2 (a + b)x 2 ax 2 + bx 2

ax 1 bx 1 x 1 x 1
= + = a + b ,
ax 2 bx 2 x 2 x 2
as required.

(·iii) Additive distributivity: This time we need to check the equation The
equation we need to check is

x 1 y 1 ? x 1 y 1
a + = a + a ,
x 2 y 2 x 2 y 2
i.e., one scalar but two different vectors. The method is by now becoming
familiar

x 1 y 1 x 1 + y 1 a(x 1 + y 1 )
a + = a =
x 2 y 2 x 2 + y 2 a(x 2 + y 2 )

ax 1 + ay 1 ax 1 ay 1 x 1 y 1
= = + = a + a ,
ax 2 + ay 2 ax 2 ay 2 x 2 y 2
again as required.

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