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Basis and Dimension
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Finite-dimensional spacesRecall we defined a finite-dimensional space as one that
is spanned by a finite number of vectors.
Today we will see how questions about a finite-dimensionalvector space can be translated into questions about
Euclidean space.
Then we can use all the techniques we have developed
for Euclidean space to solve them.
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BasisLet be a vector space. Vectors are a finite
basis for if they are linearly independent and
v1, . . . , vrV
V
Note: Every vector can be written as a linear
combination of in a unique way.
span({v1, . . . , vr}) = V .
v ∈ V
1, . . . , r
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Example
Consider the vector space of degree polynomials.2 ≤ 2
A basis for this space is given by .{1, x, x2}
We saw yesterday that these functions are linearly
independent and span .2
More generally, a basis for the vector space ofpolynomials of degree is given by .
P n
≤ n {1, x, x2
, . . . , xn}
This is the canonical basis for .n
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Example
Consider the vector space of degree polynomials.2 ≤ 2
Another basis for this space is given by .{1 + x, x, x2− x}
These functions span because their span contains
the functions .
P 2
1 = 1 + x + (−1)x
x = x
1, x, x2
x2= x
2− x+ x
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Example
Consider the vector space of degree polynomials.2 ≤ 2
Another basis for this space is given by .{1 + x, x, x2− x}
These functions are linearly independent. Suppose
a ·
1 + (a + b−
c)x + c · x2 = 0
a(1 + x) + bx + c(x2 − x) = 0.
.
By linear independence of this means1, x, x2
a = 0, a+ b− c = 0, c = 0 a = 0, b = 0, c = 0
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ExampleConsider the vector space of 2-by-3 matrices with real
entries.
A basis for this space is given by the matrices
1 0 0
0 0 0 , 0 1 0
0 0 0 , 0 0 1
0 0 0 ,
0 0 0
1 0 0
,
0 0 0
0 1 0
,
0 0 0
0 0 1
.
This is the canonical basis for M (2, 3).
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Construction of a basis
Theorem: Every finite-dimensional vector space has afinite basis.
V
Proof: Let be such that .v1, . . . , vr span({v1, . . . , vr}) = V
We know that such vectors exist as is finite-dimensional.V
If are linearly independent, then we have found a
basis.
v1, . . . , vr
Otherwise, some can be expressed as a linear
combination of the others.i
vi = a1v1 + . . .+ ai−1vi−1 + ai+1vi+1 + . . .+ arvr
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Construction of a basis
Theorem: Every finite-dimensional vector space has afinite basis.
V
Otherwise, some can be expressed as a linearcombination of the others.i
vi = a1v1 + . . .+ ai−1vi−1 + ai+1vi+1 + . . .+ arvr
Proof:
This means
span({v1, . . . , vr}) = span({v1, . . . , vi−1, vi+1, . . . , vr})
We can “throw away” without changing the span.i
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Construction of a basis
Theorem: Every finite-dimensional vector space has afinite basis.
V
Proof:
We can “throw away” without changing the span.i
If the vectors are linearly
independent, then we are done.1, . . . , i−1, i+1, . . . , r
Otherwise, we repeat this process.
This process must terminate, and when it does, we have
found a basis.
the “throw away” theorem
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CoordinatesWe discussed the coordinates of a vector with respect to
a given basis in Euclidean space.
Let be a basis for .B = {(1, 0, 0), (1, 1, 0), (1, 1, 1)} R
3
(0,−1, 1) = 1 · (1, 0, 0) + (−2) · (1, 1, 0) + 1 · (1, 1, 1)
Example:
The coordinates of with respect to thebasis are
v = (0,−
1, 1)
(v)B = (1,−2, 1).
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CoordinatesWe can do the same thing in a general finite-dimensional
vector space .V
Fix some basis for . We have already seen
such a basis exists.
v1, . . . , vr V
Every vector can be written as a unique linear
combination of the basis vectors .u ∈
v1, . . . , vr
u = a1v1 + · · ·+ arvr
(a1, . . . , ar) ∈ Rr
The coordinate vector of with respect to the
basis is .1, . . . , r
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ExampleConsider the vector space of degree polynomials.P 2 ≤ 2
A basis for this space is given by .{1 + x, x, x2− x}
What is the coordinate vector of with respect to thisbasis?
x
2
It is because(0, 1, 1)
x2 = 0 · (1 + x) + 1 · x + 1 · (x2 − x).
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Coordinate MappingFix some basis for .1, . . . , r V
u = a1v1 + · · ·+ arvr
(a1, . . . , ar) ∈ Rr
then the coordinate vector of with respect to the
basis is .v1, . . . , vr
u
The basis defines a mapping= {v1, . . . , vr}
coordB : V → Rr
where if .coordB(u) = (a1, . . . , ar) u = a1v1 + · · ·+ arvr
This is a function, because the representation is unique.
If
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Coordinate MappingThe basis defines a mappingB = {v1, . . . , vr}
coordB : V → Rr
where if .coordB(u) = (a1, . . . , ar) u = a1v1 + · · ·+ arvr
This is a function, because the representation is unique.
Claim: The coordinate mapping is one-to-one.
coordB(u1) = coordB(u2)If then .1 = 2
Claim: The coordinate mapping is onto.
for any (a1, . . . , ar) ∈ Rr
a1v1 + · · ·+ arvr ∈
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Coordinate MappingThe basis defines a mappingB = {v1, . . . , vr}
coordB : V → Rr
where if .coordB(u) = (a1, . . . , ar) u = a1v1 + · · ·+ arvr
This mapping is one-to-one and onto. It is a bijection between and .V R
r
This mapping has even more nice properties.
coordB(u + v) = coordB(u) + coordB(v)
u = a1v1 + · · · + arvr
v = b1v1 + · · · + brvru + v = (a1 + b1)v1 + · · · + (ar + br)vr
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Coordinate MappingThe basis defines a mappingB = {v1, . . . , vr}
coordB : V → Rr
where if .coordB(u) = (a1, . . . , ar) u = a1v1 + · · ·+ arvr
This mapping is a bijection between and .V Rr
§
coordB(u + v) = coordB(u) + coordB(v)§
§ coordB(c · u) = c · coordB(u) c ∈ Rfor any
u = a1v1 + · · ·+ arvrIf then c · u = c(a1v1 + · · · + arvr)
= ca1v1 + · · · + carvr
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ExampleConsider the vector space of degree polynomials.P 2 ≤ 2
Fix the basis .B = {1,x ,x2}
coordB(3− 2x + 5x2) = (3,−2, 5)
coordB(−
1 + x + x2) = (−
1, 1, 1)
(3− 2x + 5x2) + (−1 + x + x2) = 2− x + 6x2
(3,−2, 5) + (−1 + 1 + 1) = (2,−1, 6)
Adding polynomials in is just like adding vectors in .2 R3
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IsomorphismLet and be two vector spaces. A function
is an isomorphism if
V W f : V → W
is a bijection between and .V §
§
§
f W
f (u + v) = f (u) + f (v) for all u, v ∈ V
f (c · v) = c · f (v) for all c ∈ R, v ∈ V
Two spaces are isomorphic if there is an isomorphism
between them.
Isomorphic spaces are essentially the same.
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Linear combinationsLet and be two vector spaces. If is
an isomorphism then
V W f : V →W
f (c1v1 + c2v2 + · · · + crvr) = c1f (v1) + c2f (v2) + · · · + crf (vr)
This follows by progressively applying the properties of an
isomorphism.
f ((c1v1) + (c2v2 + · · · + crvr)) = f (c1v1) + f (c2v2 + · · · + c
rvr)
= c1f (v1) + f (c2v2 + · · · + crvr)
= c1f (v1) + c2f (v2) + · · · + crf (vr)
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Coordinate MappingThe basis defines a mappingB = {v1, . . . , vr}
coordB : V → Rr
where if .coordB(u) = (a1, . . . , ar) u = a1v1 + · · ·+ arvr
This mapping is a bijection between and .V Rr
§
coordB(u + v) = coordB(u) + coordB(v)§
§ coordB(c · u) = c · coordB(u) c ∈ Rfor any
If has a basis with many elements then is
isomorphic to .
V V r
Rr
Reason: The coordinate mapping is an isomorphism.
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ConsequencesExample: The vector space of polynomials of degree
is isomorphic to .2 ≤ 2
R3
An isomorphism is given by the coordinate mapping with
respect to the basis .{1, x, x2}
The coordinate mapping lets us transfer questions about
a general finite-dimensional vector space to questions about
Euclidean space.
We are already familiar with Euclidean space, and know how
to answer the questions there!
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Consequences
Theorem: Let be a vector space with zero element andbasis . Then
V B = {v1, . . . , vr}
coordB(0) = 0̄ ∈ Rr
Proof: coordB(0) = coordB(0 · 0)
= 0 · coordB(0)
= 0̄
Question: Does the same hold for any isomorphism?
Li I d d
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Linear IndependenceTheorem: Let be a finite-dimensional vector space and fix
a basis for . Then vectorsare linearly dependent if and only if
V
B =
{v1, . . . , vr} V u1, . . . , uk ∈
V
coordB(u1), . . . , coordB(uk) ∈ Rr
are linearly dependent.
Proof:
Suppose are linearly dependent, thusu1, . . . , uk ∈ V
c1u1 + · · · ckuk = 0
where some .ci 6= 0
Apply the function to both sides.coordB
coordB
(0
) = ¯0
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Theorem: Let be a finite-dimensional vector space and fix
a basis for . Then vectors
are linearly dependent if and only if
V
B = {v1, . . . , vr} V u1, . . . , uk ∈ V
coordB(u1), . . . , coordB(uk) ∈ Rr
are linearly dependent.
Proof:
c1u1 +· · ·
ckuk = 0 where some .ci 6= 0
Apply the function to both sides.coordB coordB(0) = 0̄
Simplifying the left hand side:
coordB(c1u1 + · · ·
+ ckuk) = coordB(c1u1) + coordB(c2u2 + · · ·
+ ckuk)
= c1coordB(u1) + coordB(c2u2 + · · · + ckuk)
= c1coordB(u1) + · · · + ckcoordB(uk)
This shows are lin. dependent.coordB(u1), . . . , coordB(uk)
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Theorem: Let be a finite-dimensional vector space and fix
a basis for . Then vectors
are linearly dependent if and only if
V
B = {v1, . . . , vr} V u1, . . . , uk ∈ V
coordB(u1), . . . , coordB(uk) ∈ Rr
are linearly dependent.
Proof:
Suppose are lin. dependent.coordB(u1), . . . , coordB(uk)
0̄ = c1 · coordB(u1) + · · · + ck · coordB(uk)
= coordB(c1u1) + · · · + coordB(ckuk)
= coordB(c1u1 + · · · + ckuk)
We know that and as is
one-to-one this means
coordB(0) = 0̄ coordB : V → Rr
c1u1 + · · · + ckuk = 0
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Comment
Notice that in the last proof we only used the fact thatis an isomorphism.
coordB
We did not use any specific details about the action ofcoord
B
The last theorem actually holds with respect to any
isomorphism.
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SpanTheorem: Let be a finite-dimensional vector space and fix
a basis for . Thenif and only if
V
B = {v1, . . . , vr} V v ∈ span({u1, . . . , uk})
for any .v, u1, . . . , uk ∈ V
Proof: If thenv ∈ span({u1, . . . , uk}) v = c1u1 + . . .+ ckuk
coordB(v) = coordB(c1u1 + · · ·
+ ckuk)= c1coordB(u1) + · · · + ckcoordB(uk)
and so .coordB(v) ∈ span({coordB(u1), . . . , coordB(uk)})
coordB(v) ∈ span({coordB(u1), . . . , coordB(uk)}),
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SpanTheorem: Let be a finite-dimensional vector space and fix
a basis for . Thenif and only if
V
B = {v1, . . . , vr} V v ∈ span({u1, . . . , uk})
coordB(v) ∈ span({coordB(u1), . . . , coordB(uk)})
for any .v, u1, . . . , uk ∈ V
Proof: If
coordB
(v
) = c1
coordB
(u1
) + · · ·
+ ck
coordB
(uk
)= coordB(c1u1 + · · · + ckuk)
Again as is one-to-one, this meanscoordB
v = c1u1 + · · · + ckuk
coordB(v) ∈ span({coordB(u1), . . . , coordB(uk)}),
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Dimension TheoremTheorem: Let be a finite dimensional vector space and
let and be two bases
for .
V
B1 = {u1, . . . , ur} B2 = {v1, . . . , vs}V Then .r = s
Proof: Suppose that . From the dimension theorem
in Euclidean space we know that the largest linearly
independent set in is of size .
r < s
Rr
r
Consider . As are linearly
independent, so arecoordB1
: V → Rr
coordB1(v1), . . . , coordB1
(vs)
v1, . . . , vs
∈ Rr
.
This is a contradiction as r < s.
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Dimension TheoremTheorem: Let be a finite dimensional vector space and
let and be two bases
for .
V
B1 = {u1, . . . , ur} B2 = {v1, . . . , vs}V Then .r = s
Proof: We can make an analogous argument if This
completes the proof.s < r.
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Dimension
Definition: Let be a finite-dimensional vector space. Thedimension of is the number of elements in a basis for
V
V V.
This definition makes sense by the dimension theorem.
Remark: If is a vector space of dimension then it is
isomorphic to .V d
Rd
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Example: PolynomialsConsider the vector space of polynomials of degree
at most
P n
.
A basis is given by {1, x, x2
, . . . , xn}.
The dimension of is .P n n + 1
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Example: MatricesLet be the vector space of m-by-n matrices
with real entries.
M (m,n)
A basis for this space is given by the matrices
E ij(k, `) =
(1 if k = i, ` = j
0 otherwise
where 1 ≤ i ≤ m, 1 ≤ j ≤ n.
In total, there are elements in this basis.mn
The dimension of isM (m,n) .
S b
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SubspacesTheorem: Let be a vector space of dimension with a
basis and let If
rV
U ⊆ V.B = {v1, . . . , vr},
{coordB(u) : u ∈ U } ⊆ Rr
is a subspace of then is a subspace ofRr
U V.
S =
Proof: As is a subspace of it contains .S R 0̄
Thus as this is the only vector satisfying0 ∈ U coordB(v) = 0̄.
S b
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SubspacesTheorem: Let be a vector space of dimension with a
basis and let If
rV
U ⊆ V.B = {v1, . . . , vr},
{coordB(u) : u ∈ U } ⊆ Rr
is a subspace of then is a subspace ofRr
U V.
S =
Proof: Now we see that is closed under scalar mult.U
Let and As is a subspace,
alsou ∈ U, c ∈ R s = coordB(u) ∈ S. S
c · s ∈ S.
Note that , and is theonly vector with this property as is one-to-one.
coordB(c ·
u) = c ·
coordB(u) = c ·
s c ·
u
coordB
Thus c · u ∈ U.
S b
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SubspacesTheorem: Let be a vector space of dimension with a
basis and let If
rV
U ⊆ V.B = {v1, . . . , vr},
{coordB(u) : u ∈ U } ⊆ Rr
is a subspace of then is a subspace ofRr
U V.
S =
Proof: Now we see that is closed under vector addition.U
Take and letu1, u2 ∈ U s1 = coordB(u1), s2 = coordB(u2) ∈ S.
S As is a subspace also s1 + s2 ∈ S.
Note that coordB(u1 + u2) = coordB(u1) + coordB(u2) = s1 + s2
and this is the only vector with this property (as one-to-one).
Thus u1 + u2 ∈ U.
E l
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ExampleConsider the vector space of 2-by-2 matrices.M (2, 2)
S =
a11 a12
a21 a22
: a11 + a22 = 0
Let
be the subset of of matrices whose diagonalelements sum to zero.
M (2, 2)
We could show this directly and verify that contains
the all-zero matrix and satisfies the closure conditions.S
Is a subspace?S
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Consider the vector space of 2-by-2 matrices.M (2, 2)
S =a11 a12
a21 a22
: a11 + a22 = 0
Let
be the subset of of matrices whose diagonal
elements sum to zero.
M (2, 2)
A basis for is given by the set of matricesM (2, 2)
E = 1 0
0 0 , 0 1
0 0 , 0 0
1 0 , 0 0
0 1 .
Note that
coordE
a11 a12
a12 a22
= (a11, a12, a21, a22)
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Consider the vector space of 2-by-2 matrices.M (2, 2)
S =
a11 a12
a21 a22
: a11 + a22 = 0
Let
be the subset of of matrices whose diagonal
elements sum to zero.
M (2, 2)
Consider the set .T = {coordE (A) : A ∈ S } ⊆ R4
By the previous theorem, if is a subspace then so is .T S
But is simply the null space of the matrixand therefore is a subspace.
1 0 0 1
E l ti d
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Consider the vector space of 2-by-2 matrices.M (2, 2)
S =
a11 a12
a21 a22
: a11 + a22 = 0
Let
be the subset of of matrices whose diagonal
elements sum to zero.
M (2, 2)
Example continued
What is a basis for S ?
To do this, we can first find a basis for
T = {coordE (A) : A ∈ S }
= {(a11, a12, a21, a22) : a11 + a22 = 0}
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To do this, we can first find a basis for
T = {coordE (A) : A ∈ S }
= {(a11, a12, a21, a22) : a11 + a22 = 0}
We can do this by finding the special solutions:
{(−1, 0, 0, 1), (0, 1, 0, 0), (0, 0, 1, 0)}
Now we look for vectors in which map to these special
solutions under .S
coordE
coordE
0 10 0
= (0, 1, 0, 0)
coordE
0 01 0
= (0, 0, 1, 0)
coordE
−
1 00 1
= (
−1, 0, 0, 1)
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To do this, we can first find a basis for
T = {coordE (A) : A ∈ S }
= {(a11
, a12
, a21
, a22
) : a11
+ a22
= 0}
The matrices form a basis for .
−1 0
0 1
,
0 1
0 0
,
0 0
1 0
S
They are linearly independent as their coordinate imagesare.
They span as their coordinate images span .S T
The dimension of is 3.S