Another example of avector space: Km×nthespace 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 {M∈K3×3MT=
−M}is asubspace of K3×3and ﬁnd its dimension.
Solution:ToshowWis asubspace weshould verify these twoconditions:
1. If A, B∈Wthen A+B∈W
2. If λ∈K,A∈Wthen λA ∈W.
Indeed: 1. If AT=−Aand BT=−B,then (A+B)T=AT+BT=
−A−B=−(A+B), so A+B∈W
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
=
−a−b−c
−d−e−f
−g−h−k
}.
In other words W={
0bc
−b0f
−c−f0
}={bE1+cE2+fE3},where E1=
0 1 0
−100
0 0 0
,
E2=
0 0 1
0 0 0
−100
,E3=
000
001
0−1 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
−c−f0
=
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
−1−1 0
,
010
−1 0 −1
010
,
0 1 0
−101
0−1 0
}
is abasis for W.
Solution:Weshowﬁrst that vectors in Bare linearly independent.
If a
0 1 1
−101
−1−1 0
+b
010
−1 0 −1
010
+c
0 1 0
−101
0−1 0
=0, then a+b+c=0,a=
0,a−b+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 athreedimensional
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 3m 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.