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Linear Independence
Linear Independence.
Definition. Let v ,v ,...,v ∈ Rn be a set of vectors.
1 2 k
• The vectors are linearly dependent if there exist scalars λ1,λ2,...,λk ∈ R, not
all zero, such that
λ v +λ v +···+λ v =0.
1 1 2 2 k k
• The vectors are linearly independent if λ v + ··· + λ v = 0 for scalars
1 1 k k
λ1,...,λk implies that
λ1 = λ2 = ··· = λk = 0. N
Example1.
1. • Thevectors
1 0 0
0 1 0
e = ,e = ,...,e =
1 . 2 . n .
. . .
. . .
0 0 1
in Rn are linearly independent.
• Toshowthis, consider any scalars λ ,...,λ such that λ e + λ e ··· +
1 n 1 1 2 2
λ e =0,i.e.suchthat
n n
1 0 0 λ1 0
0 1 0 λ2 0
λ +λ +···+λ = =
1 . 2 . n . . .
. . . . .
. . . . .
0 0 1 λn 0
• Thelast equation implies that λ1 = λ2 = ··· = λn = 0.
2. • Thevectors
1 4 7
w = 2 ,w = 5 , andw = 8
1 2 n
3 6 9
in R3 are linearly dependent, since w1 − 2w2 + w3 = 0.
• Howdidwecomeupwiththis? Startwiththeequation in the definition of
linear dependence
1 4 7 0
λ 2 +λ 5 +λ 8 = 0
1 2 3
3 6 9 0
and solve this system for possible values of λ ,λ ,λ .
1 2 3
2. • Wecanwritethis system in matrix form as
1 4 7 λ1 0
2 5 8 λ2 = 0 .
3 6 9 λ3 0
• The coefficient matrix A is simply the matrix with columns w ,w ,w
1 2 3
which we can write as A = (w1w2w3). If we reduce A to row echelon
form we get
1 4 7
0 −3 −6 .
0 0 0
• We have found that rankA is not maximal, and so the system is singular
and has a nonzero solution (many in fact).
• Thefollowing theorem generalizes the last example.
Theorem1(S&B11.1). Vectorsv ,...,v ∈ Rn arelinearly dependent iff the linear
1 k
system
λ
1
λ
A 2 =0,
.
.
.
λk
where A = (v v ···v ), has a nonzero solution (λ ,...,λ ).
1 2 k 1 k
• The following theorem gives us a way of checking that a set of n vectors in Rn
are linearly independent.
Theorem2(S&B11.2). Vectorsv1,...,vn ∈ Rn are linearly independent iff
det(v v ···v ) 6= 0.
1 2 n
Theorem3(S&B11.3). Ifk > n,anysetofk vectors in Rn is linearly dependent.
Proof. Let v1,...,vk be k vectors in Rn with k > n. By theorem 1, this set of vectors
is linearly dependent if and only if the system
λ1
λ2
Aλ=A =0
.
.
.
λk
where A = (v v ···v ), has a nonzero solution (λ ,...,λ ).
1 2 k 1 k
• But, since the vectors belong to Rn, the matrix A has n rows and k columns with
k > n.
2
• Since A has more columns than rows, there must be some column, say column
j, in the row echelon forms of A without a pivot.
• But then λj is a free variable, and since Aλ = 0 has a solution, it must have
infinitely many solutions all but one of which is nonzero.
• Hencethevectors are linearly dependent.
• Instead of the above, we could have used fact 7.11(a)(i) of S&B which says
that if there are more unknowns than equations, the system Aλ = 0 must have
infinitely many solutions.
SpanningSets.
• Asubset of V of the vector space Rn is a subspace of Rn if it is
– closed under addition (the sum of any two elements in V is also in V ), and
– closed under scalar multiplication (any scalar multiple of an element in V
is also in V ).
Example2.
– Theset{(1,1,1)}isnotasubspaceofR3,becauseforexample(0,0,0) ∈/
{(1,1,1)}.
– ThesetV = L[(1,1,1)] = {λ(1,1,1) | λ ∈ R} is a subspace of R3 as you
can check.
Definition. Let v ,...,v ∈ Rn be a set of vectors and let V ⊆ Rn be a subspace.
1 k
• Theset of all linear combinations of these vectors
L[v ,...,v ] = {λ v +···+λ v |λ ,...,λ ∈ R}
1 k 1 1 k k 1 k
is called the set generated or spanned by v ,...,v .
1 k
• Equivalently, the span of {v ,...,v } is the set L[v ,...,v ] (sometimes
1 k 1 k
written span(v ,...,v )).
1 k
• If V =L[v ,...,v ],
1 k
wesayv ,...,v spanV. N
1 k
• A set of vectors spans (or is a spanning set of) a subspace V of Rn if we
can write any element in V as a linear combination of the vectors.
Example3.
1. The n-dimensional Euclidean space is spanned by the vectors e1,...,en of our
earlier example. Take any vector a = (a ,...,a ) ∈ Rn. Then a = a e +
1 n 1 1
· · · + a e .
n n
2. There are many sets of vectors that span the same space. Each of the following
sets spans the space R2.
3
(a) 1 0 1 0 56
0 , 1 . (c) 0 , 1 , −24 .
−1 0 6 31
(b) 0 , 1 . (d) −6 , 31 .
Theorem4(S&B11.4). Letv ,...v ∈ Rn beasetofvectors. LetA = (v v ···v )
1 k 1 2 k
andletb ∈ Rn beavector. ThenbliesinthespaceL[v ...,v ]spannedbyv ,...,v
1 k 1 k
iff the system Aλ = b has a solution λ.
• The first corollary to this theorem provides a way of checking whether or not a
set of vectors spans all of Rn, while the second gives the minimum number of
vectors needed in a set spanning Rn.
Corollary1(S&B11.5). Letv ,...v ∈ Rnbeasetofvectors. LetA = (v v ···v ).
1 k 1 2 k
Then v ,...,v span Rn iff the system of equations Ax = b has a solution x for every
1 k
b ∈ Rn.
Corollary2(S&B11.6). AsetofvectorsthatspansRn mustcontainatleastnvectors.
Basis and Dimension.
• Once we have a spanning set of vectors, we can always make the spanning set
larger by including more vectors.
• We want to find the smallest set of vectors that spans a subset V of the n-
dimensional Euclidean space.
Definition. Let v ,...v ∈ V be a set of vectors and let V ⊆ Rn. Then v ,...v
1 k 1 k
forms a basis of V if
1. v ,...v span V, and
1 k
2. v ,...v are linearly independent. N
1 k
Example4. 1. Fromourpreviousexamples,wecanseethattheunitvectorse1,...,en
form a basis of Rn. This natural basis, is called the canonical basis of Rn.
2
2. In a previous example (2), we listed four spanning sets of R . The set (2c) is not
linearly independent, since it contains more than two vectors. The collections
(2a), (2b), and (2d) are linearly independent and so each forms a basis for R2.
Theorem5(S&B11.7). EverybasisofRn containsexactly n vectors.
• Using theorems 1, 2 and corollary 1, together with the fact that a square matrix
is nonsingular iff its determinant is nonzero gives the following theorem.
Theorem6(S&B11.8). Letv ,...,v ∈ RnbeasetofvectorsandletA = (v v ···v ).
1 n 1 2 n
Thenthe following statements are equivalent.
4
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