Linear Algebra

82 Chapter 2. Vector Spaces
For the third condition, that scalar multiplication distributes from the left over
vector addition, the check is also straightforward.
� � � � � � � � � � � �
v1 w1 r(v1 + w1) rv1 + rw1 v1 w1
r · ( + )=
=
= r · + r ·
v2 w2 r(v2 + w2) rv2 + rw2 v2 w2
The fourth
� � � � � � � �
v1 (rs)v1 r(sv1)
v1
(rs) · = = = r · (s · )
v2 (rs)v2 r(sv2)
v2
and fifth conditions are also easy.
� � � � � �
v1 1v1 v1
1 · = =
v2 1v2 v2
In a similar way, each R n is a vector space with the usual operations of
vector addition and scalar multiplication. (In R 1 , we usually do not write the
members as column vectors, i.e., we usually do not write ‘(π)’. Instead we just
write ‘π’.)
1.4 Example This subset of R3 that is a plane through the origin
⎛ ⎞
x
P = { ⎝y⎠
z
� � x + y + z =0}
is a vector space if ‘+’ and ‘·’ are interpreted in this way.
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
x1 x2 x1 + x2
x rx
⎝y1⎠
+ ⎝y2⎠
= ⎝y1
+ y2 ⎠ r · ⎝y⎠
= ⎝ry⎠
z rz
z1
z2
z1 + z2
The addition and scalar multiplication operations here are just the ones of R3 ,
reused on its subset P .WesayPinherits these operations from R3 .Hereisa
typical addition in P .
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 −1 0
⎝ 1 ⎠ + ⎝ 0 ⎠ = ⎝ 1 ⎠
−2 1 −1
This illustrates that P is closed under addition. We’ve added two vectors from
P — that is, with the property that the sum of their three entries is zero —
and we’ve gotten a vector also in P . Of course, this example of closure is not
a proof of closure. To prove that P is closed under addition, take two elements
of P
⎛
⎝ x1
⎞ ⎛
⎠ , ⎝ x2
⎞
⎠
y1
z1
y2
z2