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Chapter 21. Basis

A basis is a list of vectors that gives a coordinate system for a vector space. It has two properties: it spans the space, and it is linearly independent. The spanning property says that every vector can be built from the basis. Linear independence says that no basis vector is redundant. Equivalently, every vector in the space has a unique expression as a linear combination of the basis vectors.

21.1 Definition

Let VV be a vector space over a field FF. A list of vectors

B=(v1,v2,…,vn) B = (v_1,v_2,\ldots,v_n)

is called a basis of VV if:

PropertyMeaning
Spanningspan⁡(v1,…,vn)=V\operatorname{span}(v_1,\ldots,v_n)=V
Linear independencec1v1+⋯+cnvn=0c_1v_1+\cdots+c_nv_n=0 implies c1=⋯=cn=0c_1=\cdots=c_n=0

Thus a basis is a linearly independent spanning list.

Both conditions are necessary. A spanning list may contain redundant vectors. A linearly independent list may fail to reach the whole space. A basis is exactly balanced: it contains enough vectors, but no unnecessary vectors.

21.2 The Standard Basis of Rn\mathbb{R}^n

The standard basis of Rn\mathbb{R}^n is

e1=[10⋮0],e2=[01⋮0],…,en=[00⋮1]. e_1 = \begin{bmatrix} 1\\ 0\\ \vdots\\ 0 \end{bmatrix}, \quad e_2 = \begin{bmatrix} 0\\ 1\\ \vdots\\ 0 \end{bmatrix}, \quad \ldots, \quad e_n = \begin{bmatrix} 0\\ 0\\ \vdots\\ 1 \end{bmatrix}.

Each vector has one component equal to 11 and all other components equal to 00.

Every vector

x=[x1x2⋮xn] x = \begin{bmatrix} x_1\\ x_2\\ \vdots\\ x_n \end{bmatrix}

can be written as

x=x1e1+x2e2+⋯+xnen. x = x_1e_1+x_2e_2+\cdots+x_ne_n.

This proves that the standard basis spans Rn\mathbb{R}^n.

The same expression also proves uniqueness. If

c1e1+⋯+cnen=0, c_1e_1+\cdots+c_ne_n=0,

then

[c1c2⋮cn]=[00⋮0], \begin{bmatrix} c_1\\ c_2\\ \vdots\\ c_n \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ \vdots\\ 0 \end{bmatrix},

so

c1=c2=⋯=cn=0. c_1=c_2=\cdots=c_n=0.

Therefore the standard basis is linearly independent.

21.3 Basis and Coordinates

A basis allows vectors to be represented by coordinates.

Let

B=(v1,…,vn) B=(v_1,\ldots,v_n)

be a basis of VV. Since BB spans VV, every vector v∈Vv\in V can be written as

v=c1v1+⋯+cnvn. v=c_1v_1+\cdots+c_nv_n.

Since BB is linearly independent, the scalars c1,…,cnc_1,\ldots,c_n are unique.

The coordinate vector of vv with respect to BB is

[v]B=[c1⋮cn]. [v]_B = \begin{bmatrix} c_1\\ \vdots\\ c_n \end{bmatrix}.

The vector vv belongs to the original space. The coordinate vector [v]B[v]_B belongs to FnF^n. A basis creates the bridge between abstract vectors and columns of scalars.

21.4 Uniqueness of Coordinates

The uniqueness of coordinates follows from linear independence.

Suppose

v=c1v1+⋯+cnvn v=c_1v_1+\cdots+c_nv_n

and also

v=d1v1+⋯+dnvn. v=d_1v_1+\cdots+d_nv_n.

Subtract the two equations:

0=(c1−d1)v1+⋯+(cn−dn)vn. 0=(c_1-d_1)v_1+\cdots+(c_n-d_n)v_n.

Since v1,…,vnv_1,\ldots,v_n are linearly independent,

c1−d1=0,…,cn−dn=0. c_1-d_1=0,\quad \ldots,\quad c_n-d_n=0.

Therefore

ci=di c_i=d_i

for every ii.

So a basis gives exactly one coordinate description for each vector.

21.5 A Nonstandard Basis of R2\mathbb{R}^2

Consider

v1=[11],v2=[1−1]. v_1= \begin{bmatrix} 1\\ 1 \end{bmatrix}, \qquad v_2= \begin{bmatrix} 1\\ -1 \end{bmatrix}.

We show that B=(v1,v2)B=(v_1,v_2) is a basis of R2\mathbb{R}^2.

A general linear combination is

av1+bv2=a[11]+b[1−1]=[a+ba−b]. av_1+bv_2 = a \begin{bmatrix} 1\\ 1 \end{bmatrix} + b \begin{bmatrix} 1\\ -1 \end{bmatrix} = \begin{bmatrix} a+b\\ a-b \end{bmatrix}.

To represent an arbitrary vector

[xy], \begin{bmatrix} x\\ y \end{bmatrix},

we solve

a+b=x,a−b=y. a+b=x, \qquad a-b=y.

Adding gives

2a=x+y, 2a=x+y,

so

a=x+y2. a=\frac{x+y}{2}.

Subtracting gives

2b=x−y, 2b=x-y,

so

b=x−y2. b=\frac{x-y}{2}.

These scalars exist for every x,y∈Rx,y\in\mathbb{R}. Hence BB spans R2\mathbb{R}^2.

Also, if

av1+bv2=0, av_1+bv_2=0,

then

a+b=0,a−b=0. a+b=0, \qquad a-b=0.

Adding gives 2a=02a=0, so a=0a=0. Then b=0b=0. Thus BB is linearly independent.

Therefore BB is a basis.

21.6 Coordinates in a Nonstandard Basis

Using the basis

B=([11],[1−1]), B= \left( \begin{bmatrix} 1\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ -1 \end{bmatrix} \right),

find the coordinates of

x=[73]. x= \begin{bmatrix} 7\\ 3 \end{bmatrix}.

We need

x=av1+bv2. x=av_1+bv_2.

Thus

a+b=7,a−b=3. a+b=7, \qquad a-b=3.

Adding gives

2a=10, 2a=10,

so

a=5. a=5.

Then

5+b=7, 5+b=7,

so

b=2. b=2.

Therefore

[x]B=[52]. [x]_B= \begin{bmatrix} 5\\ 2 \end{bmatrix}.

This means

[73]=5[11]+2[1−1]. \begin{bmatrix} 7\\ 3 \end{bmatrix} = 5 \begin{bmatrix} 1\\ 1 \end{bmatrix} + 2 \begin{bmatrix} 1\\ -1 \end{bmatrix}.

The same vector has different coordinate columns under different bases.

21.7 Basis of a Subspace

A basis can be chosen not only for a whole vector space, but also for a subspace.

Consider the plane

W={(x,y,z)∈R3:x+y+z=0}. W=\{(x,y,z)\in\mathbb{R}^3:x+y+z=0\}.

Solve for zz:

z=−x−y. z=-x-y.

Then every vector in WW has the form

[xy−z] \begin{bmatrix} x\\ y\\ -z \end{bmatrix}

with zz replaced correctly as

[xy−x−y]. \begin{bmatrix} x\\ y\\ -x-y \end{bmatrix}.

Separate the free parameters:

[xy−x−y]=x[10−1]+y[01−1]. \begin{bmatrix} x\\ y\\ -x-y \end{bmatrix} = x \begin{bmatrix} 1\\ 0\\ -1 \end{bmatrix} + y \begin{bmatrix} 0\\ 1\\ -1 \end{bmatrix}.

Thus

W=span⁡([10−1],[01−1]). W= \operatorname{span} \left( \begin{bmatrix} 1\\ 0\\ -1 \end{bmatrix}, \begin{bmatrix} 0\\ 1\\ -1 \end{bmatrix} \right).

The two spanning vectors are linearly independent, since neither is a scalar multiple of the other. Therefore they form a basis of WW.

21.8 Basis from a Spanning Set

A spanning set may contain redundant vectors. By removing redundant vectors, one can obtain a basis.

For example, in R2\mathbb{R}^2,

S=([10],[01],[11]) S= \left( \begin{bmatrix} 1\\ 0 \end{bmatrix}, \begin{bmatrix} 0\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ 1 \end{bmatrix} \right)

spans R2\mathbb{R}^2. But the third vector is redundant because

[11]=[10]+[01]. \begin{bmatrix} 1\\ 1 \end{bmatrix} = \begin{bmatrix} 1\\ 0 \end{bmatrix} + \begin{bmatrix} 0\\ 1 \end{bmatrix}.

Removing it leaves

([10],[01]), \left( \begin{bmatrix} 1\\ 0 \end{bmatrix}, \begin{bmatrix} 0\\ 1 \end{bmatrix} \right),

which is a basis.

Thus a finite spanning set can be reduced to a basis by deleting vectors that already lie in the span of the others.

21.9 Extending an Independent Set

A linearly independent set may fail to span the whole space. By adding suitable vectors, one can extend it to a basis.

For example,

([100],[010]) \left( \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}, \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix} \right)

is linearly independent in R3\mathbb{R}^3, but it does not span R3\mathbb{R}^3. It spans only the xyxy-plane.

Adding

[001] \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}

gives the standard basis of R3\mathbb{R}^3.

In finite-dimensional spaces, every linearly independent set can be extended to a basis, and every spanning set can be reduced to a basis.

21.10 Minimal Spanning Sets

A basis is a minimal spanning set.

This means that the list spans the space, but removing any vector destroys the spanning property.

Let

B=(v1,…,vn) B=(v_1,\ldots,v_n)

be a basis. If one vector, say vjv_j, could be removed without changing the span, then vjv_j would be a linear combination of the remaining vectors. That would make the original list linearly dependent.

Thus no basis vector can be removed.

Conversely, any minimal spanning set is linearly independent. If it were dependent, some vector could be removed without changing the span, contradicting minimality.

Therefore:

basis=minimal spanning set. \text{basis} = \text{minimal spanning set}.

21.11 Maximal Independent Sets

A basis is also a maximal linearly independent set.

This means that the list is linearly independent, but adding any new vector from the space makes it dependent.

Let

B=(v1,…,vn) B=(v_1,\ldots,v_n)

be a basis of VV. Since BB spans VV, every vector w∈Vw\in V is already a linear combination of v1,…,vnv_1,\ldots,v_n.

Therefore adding ww to the list creates a redundancy.

Conversely, if a linearly independent set is maximal, it must span VV. If it failed to span VV, one could choose a vector outside its span and add it while preserving linear independence.

Therefore:

basis=maximal linearly independent set. \text{basis} = \text{maximal linearly independent set}.

These equivalent descriptions are standard consequences of the basis definition.

21.12 Basis and Dimension

All bases of a finite-dimensional vector space contain the same number of vectors. This number is called the dimension of the space.

For example,

dim⁡(Rn)=n. \dim(\mathbb{R}^n)=n.

The standard basis has nn vectors, and every other basis of Rn\mathbb{R}^n also has nn vectors.

This fact is fundamental. It allows dimension to be defined without referring to a particular basis.

21.13 Basis of Polynomial Spaces

Let PnP_n be the vector space of real polynomials of degree at most nn.

The list

(1,x,x2,…,xn) (1,x,x^2,\ldots,x^n)

is a basis of PnP_n.

It spans PnP_n because every polynomial p∈Pnp\in P_n has the form

p(x)=a0+a1x+⋯+anxn. p(x)=a_0+a_1x+\cdots+a_nx^n.

It is linearly independent because

a0+a1x+⋯+anxn=0 a_0+a_1x+\cdots+a_nx^n=0

as a polynomial implies

a0=a1=⋯=an=0. a_0=a_1=\cdots=a_n=0.

Therefore

dim⁡(Pn)=n+1. \dim(P_n)=n+1.

21.14 Basis of Matrix Spaces

Let M2×2(R)M_{2\times 2}(\mathbb{R}) be the vector space of all 2×22\times 2 real matrices.

Define

E11=[1000],E12=[0100], E_{11}= \begin{bmatrix} 1&0\\ 0&0 \end{bmatrix}, \quad E_{12}= \begin{bmatrix} 0&1\\ 0&0 \end{bmatrix}, E21=[0010],E22=[0001]. E_{21}= \begin{bmatrix} 0&0\\ 1&0 \end{bmatrix}, \quad E_{22}= \begin{bmatrix} 0&0\\ 0&1 \end{bmatrix}.

Every matrix

A=[abcd] A= \begin{bmatrix} a&b\\ c&d \end{bmatrix}

can be written as

A=aE11+bE12+cE21+dE22. A=aE_{11}+bE_{12}+cE_{21}+dE_{22}.

The expression is unique because each coefficient controls a different matrix entry.

Thus

(E11,E12,E21,E22) (E_{11},E_{12},E_{21},E_{22})

is a basis of M2×2(R)M_{2\times 2}(\mathbb{R}), and

dim⁡M2×2(R)=4. \dim M_{2\times 2}(\mathbb{R})=4.

More generally,

dim⁡Mm×n(R)=mn. \dim M_{m\times n}(\mathbb{R})=mn.

21.15 Ordered Bases

A basis is often treated as an ordered list rather than an unordered set.

The order matters for coordinates. If

B=(v1,v2) B=(v_1,v_2)

and

C=(v2,v1), C=(v_2,v_1),

then BB and CC contain the same vectors, but coordinate vectors are arranged differently.

For example, if

v=3v1+5v2, v=3v_1+5v_2,

then

[v]B=[35], [v]_B= \begin{bmatrix} 3\\ 5 \end{bmatrix},

while

[v]C=[53]. [v]_C= \begin{bmatrix} 5\\ 3 \end{bmatrix}.

The vector vv has not changed. Only its coordinate description has changed.

21.16 Coordinates as an Isomorphism

Let B=(v1,…,vn)B=(v_1,\ldots,v_n) be a basis of VV. The coordinate map

ΦB:V→Fn \Phi_B:V\to F^n

defined by

ΦB(v)=[v]B \Phi_B(v)=[v]_B

is a linear isomorphism.

It is linear because coordinates respect addition and scalar multiplication:

[u+v]B=[u]B+[v]B, [u+v]_B=[u]_B+[v]_B,

and

[cv]B=c[v]B. [cv]_B=c[v]_B.

It is one-to-one because coordinates are unique. It is onto because every coordinate column in FnF^n corresponds to a linear combination of basis vectors.

Thus every nn-dimensional vector space over FF is structurally the same as FnF^n after a basis is chosen.

21.17 Finding a Basis by Row Reduction

Suppose vectors in Rm\mathbb{R}^m are placed as columns of a matrix

A=[∣∣∣v1v2⋯vk∣∣∣]. A= \begin{bmatrix} |&|&&|\\ v_1&v_2&\cdots&v_k\\ |&|&&| \end{bmatrix}.

To find a basis for the span of these vectors, row reduce AA. The pivot columns of the original matrix form a basis for the column space.

The phrase original matrix is important. Row operations change the columns themselves, but they preserve the linear dependence relations among columns. Therefore the pivot positions identify which original vectors are needed.

21.18 Example: Basis for a Column Space

Let

A=[121011132]. A= \begin{bmatrix} 1&2&1\\ 0&1&1\\ 1&3&2 \end{bmatrix}.

Row reduce:

[121011132]→[121011011]→[10−1011000]. \begin{bmatrix} 1&2&1\\ 0&1&1\\ 1&3&2 \end{bmatrix} \to \begin{bmatrix} 1&2&1\\ 0&1&1\\ 0&1&1 \end{bmatrix} \to \begin{bmatrix} 1&0&-1\\ 0&1&1\\ 0&0&0 \end{bmatrix}.

The pivot columns are columns 11 and 22. Therefore a basis for the column space of AA is

([101],[213]). \left( \begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix}, \begin{bmatrix} 2\\ 1\\ 3 \end{bmatrix} \right).

The third column is redundant because it lies in the span of the first two columns.

Indeed,

[112]=−[101]+[213]. \begin{bmatrix} 1\\ 1\\ 2 \end{bmatrix} = - \begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix} + \begin{bmatrix} 2\\ 1\\ 3 \end{bmatrix}.

21.19 Bases Are Not Unique

A vector space usually has many bases.

For example, R2\mathbb{R}^2 has the standard basis

([10],[01]), \left( \begin{bmatrix} 1\\ 0 \end{bmatrix}, \begin{bmatrix} 0\\ 1 \end{bmatrix} \right),

but it also has

([11],[1−1]), \left( \begin{bmatrix} 1\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ -1 \end{bmatrix} \right),

and infinitely many others.

Any two nonparallel nonzero vectors in R2\mathbb{R}^2 form a basis.

Although bases are not unique, the number of vectors in a basis is unique. That number is the dimension.

21.20 Summary

A basis is a linearly independent spanning list. It gives a coordinate system for a vector space.

The key ideas are:

ConceptMeaning
BasisLinearly independent spanning list
Coordinate vectorCoefficients relative to a basis
Standard basisUsual coordinate basis of FnF^n
Minimal spanning setSpanning set with no removable vector
Maximal independent setIndependent set that cannot be enlarged
DimensionNumber of vectors in any basis
Ordered basisBasis with fixed order for coordinates
Pivot columnsColumns that form a basis for a column space

A basis is the point where span and independence meet. Span gives enough vectors to describe the whole space. Independence removes redundancy. Together they produce unique coordinates, and unique coordinates make calculation possible.