# Chapter 21. Basis

# 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 \(V\) be a vector space over a field \(F\). A list of vectors

$$
B = (v_1,v_2,\ldots,v_n)
$$

is called a basis of \(V\) if:

| Property | Meaning |
|---|---|
| Spanning | \(\operatorname{span}(v_1,\ldots,v_n)=V\) |
| Linear independence | \(c_1v_1+\cdots+c_nv_n=0\) implies \(c_1=\cdots=c_n=0\) |

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 \(\mathbb{R}^n\)

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

$$
e_1 =
\begin{bmatrix}
1\\
0\\
\vdots\\
0
\end{bmatrix},
\quad
e_2 =
\begin{bmatrix}
0\\
1\\
\vdots\\
0
\end{bmatrix},
\quad
\ldots,
\quad
e_n =
\begin{bmatrix}
0\\
0\\
\vdots\\
1
\end{bmatrix}.
$$

Each vector has one component equal to \(1\) and all other components equal to \(0\).

Every vector

$$
x =
\begin{bmatrix}
x_1\\
x_2\\
\vdots\\
x_n
\end{bmatrix}
$$

can be written as

$$
x = x_1e_1+x_2e_2+\cdots+x_ne_n.
$$

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

The same expression also proves uniqueness. If

$$
c_1e_1+\cdots+c_ne_n=0,
$$

then

$$
\begin{bmatrix}
c_1\\
c_2\\
\vdots\\
c_n
\end{bmatrix} =
\begin{bmatrix}
0\\
0\\
\vdots\\
0
\end{bmatrix},
$$

so

$$
c_1=c_2=\cdots=c_n=0.
$$

Therefore the standard basis is linearly independent.

## 21.3 Basis and Coordinates

A basis allows vectors to be represented by coordinates.

Let

$$
B=(v_1,\ldots,v_n)
$$

be a basis of \(V\). Since \(B\) spans \(V\), every vector \(v\in V\) can be written as

$$
v=c_1v_1+\cdots+c_nv_n.
$$

Since \(B\) is linearly independent, the scalars \(c_1,\ldots,c_n\) are unique.

The coordinate vector of \(v\) with respect to \(B\) is

$$
[v]_B =
\begin{bmatrix}
c_1\\
\vdots\\
c_n
\end{bmatrix}.
$$

The vector \(v\) belongs to the original space. The coordinate vector \([v]_B\) belongs to \(F^n\). 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

$$
v=c_1v_1+\cdots+c_nv_n
$$

and also

$$
v=d_1v_1+\cdots+d_nv_n.
$$

Subtract the two equations:

$$
0=(c_1-d_1)v_1+\cdots+(c_n-d_n)v_n.
$$

Since \(v_1,\ldots,v_n\) are linearly independent,

$$
c_1-d_1=0,\quad \ldots,\quad c_n-d_n=0.
$$

Therefore

$$
c_i=d_i
$$

for every \(i\).

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

## 21.5 A Nonstandard Basis of \(\mathbb{R}^2\)

Consider

$$
v_1=
\begin{bmatrix}
1\\
1
\end{bmatrix},
\qquad
v_2=
\begin{bmatrix}
1\\
-1
\end{bmatrix}.
$$

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

A general linear combination is

$$
av_1+bv_2 =
a
\begin{bmatrix}
1\\
1
\end{bmatrix}
+
b
\begin{bmatrix}
1\\
-1
\end{bmatrix} =
\begin{bmatrix}
a+b\\
a-b
\end{bmatrix}.
$$

To represent an arbitrary vector

$$
\begin{bmatrix}
x\\
y
\end{bmatrix},
$$

we solve

$$
a+b=x,
\qquad
a-b=y.
$$

Adding gives

$$
2a=x+y,
$$

so

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

Subtracting gives

$$
2b=x-y,
$$

so

$$
b=\frac{x-y}{2}.
$$

These scalars exist for every \(x,y\in\mathbb{R}\). Hence \(B\) spans \(\mathbb{R}^2\).

Also, if

$$
av_1+bv_2=0,
$$

then

$$
a+b=0,
\qquad
a-b=0.
$$

Adding gives \(2a=0\), so \(a=0\). Then \(b=0\). Thus \(B\) is linearly independent.

Therefore \(B\) is a basis.

## 21.6 Coordinates in a Nonstandard Basis

Using the basis

$$
B=
\left(
\begin{bmatrix}
1\\
1
\end{bmatrix},
\begin{bmatrix}
1\\
-1
\end{bmatrix}
\right),
$$

find the coordinates of

$$
x=
\begin{bmatrix}
7\\
3
\end{bmatrix}.
$$

We need

$$
x=av_1+bv_2.
$$

Thus

$$
a+b=7,
\qquad
a-b=3.
$$

Adding gives

$$
2a=10,
$$

so

$$
a=5.
$$

Then

$$
5+b=7,
$$

so

$$
b=2.
$$

Therefore

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

This means

$$
\begin{bmatrix}
7\\
3
\end{bmatrix} =
5
\begin{bmatrix}
1\\
1
\end{bmatrix}
+
2
\begin{bmatrix}
1\\
-1
\end{bmatrix}.
$$

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

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

Solve for \(z\):

$$
z=-x-y.
$$

Then every vector in \(W\) has the form

$$
\begin{bmatrix}
x\\
y\\
-z
\end{bmatrix}
$$

with \(z\) replaced correctly as

$$
\begin{bmatrix}
x\\
y\\
-x-y
\end{bmatrix}.
$$

Separate the free parameters:

$$
\begin{bmatrix}
x\\
y\\
-x-y
\end{bmatrix} =
x
\begin{bmatrix}
1\\
0\\
-1
\end{bmatrix}
+
y
\begin{bmatrix}
0\\
1\\
-1
\end{bmatrix}.
$$

Thus

$$
W=
\operatorname{span}
\left(
\begin{bmatrix}
1\\
0\\
-1
\end{bmatrix},
\begin{bmatrix}
0\\
1\\
-1
\end{bmatrix}
\right).
$$

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

## 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 \(\mathbb{R}^2\),

$$
S=
\left(
\begin{bmatrix}
1\\
0
\end{bmatrix},
\begin{bmatrix}
0\\
1
\end{bmatrix},
\begin{bmatrix}
1\\
1
\end{bmatrix}
\right)
$$

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

$$
\begin{bmatrix}
1\\
1
\end{bmatrix} =
\begin{bmatrix}
1\\
0
\end{bmatrix}
+
\begin{bmatrix}
0\\
1
\end{bmatrix}.
$$

Removing it leaves

$$
\left(
\begin{bmatrix}
1\\
0
\end{bmatrix},
\begin{bmatrix}
0\\
1
\end{bmatrix}
\right),
$$

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,

$$
\left(
\begin{bmatrix}
1\\
0\\
0
\end{bmatrix},
\begin{bmatrix}
0\\
1\\
0
\end{bmatrix}
\right)
$$

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

Adding

$$
\begin{bmatrix}
0\\
0\\
1
\end{bmatrix}
$$

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

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

$$
B=(v_1,\ldots,v_n)
$$

be a basis. If one vector, say \(v_j\), could be removed without changing the span, then \(v_j\) 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:

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

## 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

$$
B=(v_1,\ldots,v_n)
$$

be a basis of \(V\). Since \(B\) spans \(V\), every vector \(w\in V\) is already a linear combination of \(v_1,\ldots,v_n\).

Therefore adding \(w\) to the list creates a redundancy.

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

Therefore:

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

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,

$$
\dim(\mathbb{R}^n)=n.
$$

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

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

## 21.13 Basis of Polynomial Spaces

Let \(P_n\) be the vector space of real polynomials of degree at most \(n\).

The list

$$
(1,x,x^2,\ldots,x^n)
$$

is a basis of \(P_n\).

It spans \(P_n\) because every polynomial \(p\in P_n\) has the form

$$
p(x)=a_0+a_1x+\cdots+a_nx^n.
$$

It is linearly independent because

$$
a_0+a_1x+\cdots+a_nx^n=0
$$

as a polynomial implies

$$
a_0=a_1=\cdots=a_n=0.
$$

Therefore

$$
\dim(P_n)=n+1.
$$

## 21.14 Basis of Matrix Spaces

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

Define

$$
E_{11}=
\begin{bmatrix}
1&0\\
0&0
\end{bmatrix},
\quad
E_{12}=
\begin{bmatrix}
0&1\\
0&0
\end{bmatrix},
$$

$$
E_{21}=
\begin{bmatrix}
0&0\\
1&0
\end{bmatrix},
\quad
E_{22}=
\begin{bmatrix}
0&0\\
0&1
\end{bmatrix}.
$$

Every matrix

$$
A=
\begin{bmatrix}
a&b\\
c&d
\end{bmatrix}
$$

can be written as

$$
A=aE_{11}+bE_{12}+cE_{21}+dE_{22}.
$$

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

Thus

$$
(E_{11},E_{12},E_{21},E_{22})
$$

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

$$
\dim M_{2\times 2}(\mathbb{R})=4.
$$

More generally,

$$
\dim M_{m\times n}(\mathbb{R})=mn.
$$

## 21.15 Ordered Bases

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

The order matters for coordinates. If

$$
B=(v_1,v_2)
$$

and

$$
C=(v_2,v_1),
$$

then \(B\) and \(C\) contain the same vectors, but coordinate vectors are arranged differently.

For example, if

$$
v=3v_1+5v_2,
$$

then

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

while

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

The vector \(v\) has not changed. Only its coordinate description has changed.

## 21.16 Coordinates as an Isomorphism

Let \(B=(v_1,\ldots,v_n)\) be a basis of \(V\). The coordinate map

$$
\Phi_B:V\to F^n
$$

defined by

$$
\Phi_B(v)=[v]_B
$$

is a linear isomorphism.

It is linear because coordinates respect addition and scalar multiplication:

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

and

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

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

Thus every \(n\)-dimensional vector space over \(F\) is structurally the same as \(F^n\) after a basis is chosen.

## 21.17 Finding a Basis by Row Reduction

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

$$
A=
\begin{bmatrix}
|&|&&|\\
v_1&v_2&\cdots&v_k\\
|&|&&|
\end{bmatrix}.
$$

To find a basis for the span of these vectors, row reduce \(A\). 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

$$
A=
\begin{bmatrix}
1&2&1\\
0&1&1\\
1&3&2
\end{bmatrix}.
$$

Row reduce:

$$
\begin{bmatrix}
1&2&1\\
0&1&1\\
1&3&2
\end{bmatrix}
\to
\begin{bmatrix}
1&2&1\\
0&1&1\\
0&1&1
\end{bmatrix}
\to
\begin{bmatrix}
1&0&-1\\
0&1&1\\
0&0&0
\end{bmatrix}.
$$

The pivot columns are columns \(1\) and \(2\). Therefore a basis for the column space of \(A\) is

$$
\left(
\begin{bmatrix}
1\\
0\\
1
\end{bmatrix},
\begin{bmatrix}
2\\
1\\
3
\end{bmatrix}
\right).
$$

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

Indeed,

$$
\begin{bmatrix}
1\\
1\\
2
\end{bmatrix} = -
\begin{bmatrix}
1\\
0\\
1
\end{bmatrix}
+
\begin{bmatrix}
2\\
1\\
3
\end{bmatrix}.
$$

## 21.19 Bases Are Not Unique

A vector space usually has many bases.

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

$$
\left(
\begin{bmatrix}
1\\
0
\end{bmatrix},
\begin{bmatrix}
0\\
1
\end{bmatrix}
\right),
$$

but it also has

$$
\left(
\begin{bmatrix}
1\\
1
\end{bmatrix},
\begin{bmatrix}
1\\
-1
\end{bmatrix}
\right),
$$

and infinitely many others.

Any two nonparallel nonzero vectors in \(\mathbb{R}^2\) 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:

| Concept | Meaning |
|---|---|
| Basis | Linearly independent spanning list |
| Coordinate vector | Coefficients relative to a basis |
| Standard basis | Usual coordinate basis of \(F^n\) |
| Minimal spanning set | Spanning set with no removable vector |
| Maximal independent set | Independent set that cannot be enlarged |
| Dimension | Number of vectors in any basis |
| Ordered basis | Basis with fixed order for coordinates |
| Pivot columns | Columns 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.