Examples, Subspaces

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Published on 23 Jun 2011
School
UTSG
Department
Mathematics
Course
MAT224H1
Professor
Another example of avector space: Km×n-thespace of m×nmatrices
with entries from the field K.This is avector space over Kwith the addition
being the usual matrix addition and scalar multiplication being the entry-
wise multiplication byscalars: λ((aij )) =((λaij )).
Problem:Provethat the set of antisymmetric matrices {MK3×3|MT=
M}is asubspace of K3×3and find its dimension.
Solution:ToshowWis asubspace weshould verify these twoconditions:
1. If A, BWthen A+BW
2. If λK,AWthen λA W.
Indeed: 1. If AT=Aand BT=B,then (A+B)T=AT+BT=
AB=(A+B), so A+BW
2. If AT=A,then (λA)T=λAT=λA so λA W
Tofind the dimension, wewill find aspanning set and showthat it is
linearly independent(or eliminate the dependentelements from it, if it turns
out to belinearly dependent).
According to the definition of W,W={
abc
def
ghk
|
adg
beh
cfk
=
abc
def
ghk
}.
In other words W={
0bc
b0f
cf0
}={bE1+cE2+fE3},where E1=
0 1 0
100
0 0 0
,
E2=
0 0 1
0 0 0
100
,E3=
000
001
01 0
.
So E={E1,E2,E3}is aspanning set for W.In particular dim(W)3.
Toshowthe other inequality,weshould showthat Eis linearly independent.
Indeed, if bE1+cE2+fE3=0, then
0bc
b0f
cf0
=
000
000
000
and so b=c=
f=0. This shows that Eis linearly independent, completing the proof that
dim W=3.
Problem:Showthat if 1+16=0in K,then the set
B={
0 1 1
101
11 0
,
010
1 0 1
010
,
0 1 0
101
01 0
}
is abasis for W.
Solution:Weshowfirst that vectors in Bare linearly independent.
If a
0 1 1
101
11 0
+b
010
1 0 1
010
+c
0 1 0
101
01 0
=0, then a+b+c=0,a=
0,ab+c=0, so a=0and b+c=0,b+c=0or a=0,2b=0,2c=0.
Since 1+16=0, wecan divide by2, so a=b=c=0.
Since these are three linearly independentvectors in athree-dimensional
space, they must beabasis.
1
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Document Summary

M} is a subspace of k3 3 and nd its dimension. Problem: prove that the set of antisymmetric matrices {m k3 3|m t = Solution: to show w is a subspace we should verify these two conditions: if a, b w then a + b w, if k,a w then a w . If at = a and bt = b, then (a + b)t = at + bt = A b = (a + b), so a + b w: if at = a, then ( a)t = at = a so a w. To nd the dimension, we will nd a spanning set and show that it is linearly independent (or eliminate the dependent elements from it, if it turns out to be linearly dependent). According to the de nition of w , w = {(cid:16)a b c d e f. 1 0 0(cid:17), e3 = (cid:16)0 0 0.