Page 380 - 35Linear Algebra

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This again relies on the underlying real numbers which for any x, y ∈ R
obey
x + y = y + x .

This fact underlies the middle step of the following computation

x 1 y 1 x 1 + y 1 y 1 + x 1 y 1 x 1
+ = = = + ,
x 2 y 2 x 2 + y 2 y 2 + x 2 y 2 x 2
which demonstrates what we wished to show.
(+iii) Additive Associativity: This shows that we needn’t specify with paren-
theses which order we intend to add triples of vectors because their
sums will agree for either choice. What we have to check is

x 1 y 1 z 1 ? x 1 y 1 z 1
+ + = + + .
x 2 y 2 z 2 x 2 y 2 z 2
Again this relies on the underlying associativity of real numbers:
(x + y) + z = x + (y + z) .

The computation required is

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

x 1 + (y 1 + z 1 ) x 1 y 1 + z 1 x 1 y 1 z 1
= = + = + + .
x 2 + (y 2 + z 2 ) y 1 y 2 + z 2 x 2 y 2 z 2
~
(iv) Zero: There needs to exist a vector 0 that works the way we would expect
zero to behave, i.e.

x 1 x 1
~
+ 0 = .
y 1 y 1
It is easy to find, the answer is

~ 0 = 0 .
0
You can easily check that when this vector is added to any vector, the
result is unchanged.

x 1
(+v) Additive Inverse: We need to check that when we have , there is
x 2
~
another vector that can be added to it so the sum is 0. (Note that it
~
is important to first figure out what 0 is here!) The answer for the

x 1 −x 1
additive inverse of is because
x 2 −x 2

0
x 1 −x 1 x 1 − x 1 ~
+ = = = 0 .
x 2 −x 2 x 2 − x 2 0
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