1
Given a set of vectors, how to determine if there are any vectors that
are linear combinations of other vectors?
Idea:
in {
u
1
,
…
,
u
i
,
…
,
u
k
}, if is
u
i
a linear combination of other
vectors, then there exists scalars
c
1
,
…
,
c
i
,
…
,
c
k
,
not all zero
,
such that
c
1
u
1
+
"
+
c
i
u
i
+
"
+
c
k
u
k
=
0
. (at least
c
i
= 1
)
(
L.D.
)
(
L.I.
)
Property: Any finite set
S
= {
0
,
u
1
,
u
2
,
…
,
u
k
} that contains the zero
vector is L.D., since 1·
0
+ 0·
u
1
+ 0·
u
2
+
"
+0
·
u
k
=
0
.
Condition for L.D.:
⇔
Example:
L.D. or L.I.? and which element(s), if
any, can be linearly combined by others?
A
= [
"
]
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the augmented matrix of
A
x
=
0
is
reduced row
echelon form
⇒
the general solution of
A
x
=
0
is
setting
x
3
= 1 leads to
Conclusion:
S
is L.D., and the last vector is not a linear combination
of others.
In general, the set is L.I. if and only if there is not any free variable.
always zero
column,
redundant.
(g)
There is a pivot position in each column of
A
.
Proof
(a)
⇔
(f): by definition, as noted.
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 Spring '09
 Fong
 UK, Row echelon form, Corollaries, ciui

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