6.5 Basis and dimension

A basis is where the chapter comes together. Earlier notes taught you how to build vectors from other vectors, how to test whether a set is a subspace, and how to recognize linear dependence. This note combines those ideas into one question:

When do we have exactly enough vectors to describe a space without keeping any redundant directions?

Intuition first: enough directions, but not extra baggage

If a set of vectors spans a space, then it gives you enough directions to build every vector in that space.

If the set is also linearly independent, then none of those directions is wasted. No vector in the set can already be made from the others.

So a basis is the "just right" situation:

enough vectors to reach every target in the space;
not so many vectors that one of them is unnecessary.

That is why a basis is often the most useful coordinate system for a space. Once you choose a basis, every vector can be described in terms of those basis vectors.

Definition

Basis

A set of vectors $\{u_1, u_2, \ldots, u_m\}$ is a basis for a vector space $V$ if both conditions hold:

$u_1, u_2, \ldots, u_m$ are linearly independent.
$Span\{u_1, u_2, \ldots, u_m\} = V$ .

The two conditions do different jobs.

The spanning condition says the set is large enough.
The independence condition says the set is not larger than necessary.

If you forget either condition, you can misidentify a basis.

Pause and vary the coefficients in a spanning example. Watch how changing the generators changes the set of vectors you can reach.

Read and try

Build one vector from a span

The live explorer lets you vary coefficients and watch the resulting vector move inside the span.

(1, 0)

(0, 1)

Result

αu + βv = (1, 0)

Every output vector is built from the horizontal and vertical directions.

The standard basis is the model example

The first basis most students meet is the standard basis of $R^3$ :

e_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \quad e_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \quad e_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}.

These vectors are important because each one isolates one coordinate direction.

Worked example

Why the standard basis of $R^3$ really is a basis

To show that $\{e_1, e_2, e_3\}$ is a basis for $R^3$ , you must check the two basis conditions.

First, they span $R^3$ . If

x = \begin{bmatrix} a \\ b \\ c \end{bmatrix},

then

x = a e_1 + b e_2 + c e_3.

So every vector in $R^3$ can be built from $e_1, e_2, e_3$ .

Second, they are linearly independent. If

\alpha_1 e_1 + \alpha_2 e_2 + \alpha_3 e_3 = 0,

then comparing coordinates gives

\alpha_1 = 0,\quad \alpha_2 = 0,\quad \alpha_3 = 0.

So the only linear relation is the trivial one.

Since both conditions hold, $\{e_1, e_2, e_3\}$ is a basis for $R^3$ .

This example also explains why basis coordinates are useful. Once the basis is fixed, the coefficients a, b, and c tell you exactly how to rebuild the vector.

Why the number of basis vectors matters

The notes stress a fact that is easy to say and very important to use:

Theorem

Every basis of the same space has the same size

If $B_1$ and $B_2$ are both bases of the same vector space $V$ , then $B_1$ and $B_2$ contain the same number of vectors.

This statement is what makes dimension well-defined. Without it, the phrase "the number of vectors in a basis" would depend on which basis you picked, and dimension would not be a stable idea.

The reason behind the theorem is the following counting principle.

Theorem

Too many vectors in a spanned space force dependence

Suppose $\{v_1,\ldots,v_m\}$ spans a vector space $V$ . Then any collection of more than m vectors in $V$ is linearly dependent.

Here is the idea. If $u_1,\ldots,u_n$ are vectors in $V$ with $n > m$ , then each $u_j$ can be written as a linear combination of $v_1,\ldots,v_m$ . A linear relation

c_1u_1+\cdots+c_nu_n=0

then becomes a homogeneous system with m equations and n unknowns. Since there are more unknowns than equations, a non-trivial solution exists. That non-trivial solution gives a non-trivial linear relation among the $u_j$ , so the $u_j$ are dependent.

This also explains why two different bases of the same space must have the same size. If one basis had more vectors than the other, the larger one would be a linearly dependent set inside a space spanned by the smaller one, contradicting the definition of basis.

Dimension is a count of independent directions

Definition

Dimension

The dimension of a vector space $V$ , written dim(V), is the number of vectors in any basis of $V$ .

Dimension tells you how many genuinely independent directions the space has.

$dim(R^m) = m$
$dim(M_{mn}) = mn$
$dim(P_n) = n + 1$
the space of all real polynomials has infinite dimension, because no finite list can span all powers $1,x,x^2,\ldots$ .

The zero space is the exceptional case:

dim(\{0\}) = 0.

That makes sense because the zero space has no nonzero direction at all.

A basis does not have to look like the standard one

A basis is not required to use the standard coordinate vectors. Many different sets can serve as a basis for the same space, as long as they span the space and stay linearly independent.

Worked example

A basis for a plane inside $R^3$

Let

W = \left\{ \begin{bmatrix} x \\ y \\ 0 \end{bmatrix} : x, y \in R \right\}.

This is the plane $z = 0$ inside $R^3$ .

Consider

u_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \qquad u_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}.

Every vector in $W$ has the form

\begin{bmatrix} x \\ y \\ 0 \end{bmatrix} = x u_1 + y u_2,

so $u_1$ and $u_2$ span $W$ .

They are also linearly independent, because

\alpha_1 u_1 + \alpha_2 u_2 = 0

forces $\alpha_1 = \alpha_2 = 0$ .

Therefore $\{u_1, u_2\}$ is a basis for $W$ , and

dim(W) = 2.

So dimension does not count how many entries a vector happens to have. It counts how many independent directions the space itself contains.

How dimension becomes a shortcut

Once you know the dimension of a space, many basis questions become shorter.

If $dim(V) = m$ , then:

any m linearly independent vectors in $V$ automatically form a basis;
any m vectors in $V$ that span $V$ automatically form a basis;
fewer than m vectors cannot span $V$ ;
more than m vectors cannot all remain linearly independent.

This is why dimension is powerful. It turns a long two-part basis test into a one-part check when the number of vectors already matches the dimension.

A useful way to package the shortcut is the following three-way criterion. Suppose $u_1,\ldots,u_n$ are vectors in a subspace $W$ . Consider the three statements:

$dim(W)=n$ .
$u_1,\ldots,u_n$ are linearly independent.
$Span\{u_1,\ldots,u_n\}=W$ .

If any two of these statements are true, then the third is automatically true, and $\{u_1,\ldots,u_n\}$ is a basis for $W$ .

This is often the cleanest way to write a solution. Instead of proving both independence and spanning from scratch, you first identify the dimension and then prove whichever of independence or spanning is easier.

Worked example

Using dimension to avoid a second calculation in $R^3$

Let

u_1=\begin{bmatrix}1\\2\\2\end{bmatrix},\quad u_2=\begin{bmatrix}2\\3\\4\end{bmatrix},\quad u_3=\begin{bmatrix}3\\8\\7\end{bmatrix}.

We want to show that these three vectors form a basis for $R^3$ .

First test linear independence. The equation

c_1u_1+c_2u_2+c_3u_3=0

has coefficient matrix

\begin{bmatrix} 1 & 2 & 3\\ 2 & 3 & 8\\ 2 & 4 & 7 \end{bmatrix}.

Row reduction gives an echelon form with a pivot in every column, for example

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

Therefore the homogeneous equation has only the trivial solution, and $u_1,u_2,u_3$ are linearly independent.

Since $dim(R^3)=3$ , three linearly independent vectors in $R^3$ automatically span $R^3$ . Hence the vectors form a basis. The point is that the spanning part is not ignored; it is supplied by the dimension theorem.

Try a few typical dependence patterns before you use that shortcut.

Read and try

Test one set for dependence

The live checker compares small vector sets and explains whether a nontrivial linear relation exists.

Verdict

Independent

The only way to solve c1e1 + c2e2 = 0 is c1 = c2 = 0, so this pair is linearly independent.

Key relation

No nontrivial linear relation appears.

A practical checklist for basis questions

When you are asked whether a set is a basis, work in this order:

Identify the space you are trying to span.
Count how many vectors you were given.
Check linear independence, or check spanning.
Use dimension to finish the argument when the count matches.

For example, in $R^3$ , three independent vectors already form a basis. You do not need a second long spanning calculation after that.

The same idea also gives two important repair operations.

Theorem

Extension and reduction consequences

If $dim(V)=m>0$ , then:

no set of fewer than m vectors can span $V$ ;
every linearly independent set with fewer than m vectors can be extended to a basis of $V$ ;
every spanning set with more than m vectors can be reduced to a basis of $V$ by deleting redundant vectors.

These consequences are not separate tricks. They all say the same thing: dimension is the exact size of a just-right list. A list that is too short cannot cover the whole space; an independent list that is too short can still accept more directions; a spanning list that is too long must contain unnecessary vectors.

Null spaces: a rank-nullity basis test

For a homogeneous system, the dimension shortcut becomes especially concrete. Let $A$ be a $p \times q$ matrix with rank r. Then

dim(N(A)) = q-r.

So vectors $u_1,\ldots,u_n$ form a basis for N(A) exactly when all three checks hold:

each vector is actually in the null space, so $Au_j=0$ ;
the number of proposed vectors is correct, so $n=q-r$ ;
the proposed vectors are linearly independent.

This avoids proving the full set equality $N(A)=Span\{u_1,\ldots,u_n\}$ directly.

Worked example

Testing a proposed basis for a null space

Let

A= \begin{bmatrix} 1&2&0&1\\ 1&1&1&-1\\ 3&1&5&-7 \end{bmatrix}, \qquad W=N(A),

and

u_1=\begin{bmatrix}1\\-1\\1\\1\end{bmatrix}, \qquad u_2=\begin{bmatrix}5\\-3\\-1\\1\end{bmatrix}.

First check membership. Direct multiplication gives $Au_1=0$ and $Au_2=0$ , so both vectors lie in $W$ .

Next compute the dimension of $W$ . A row-reduced form of $A$ is

\begin{bmatrix} 1&0&2&-3\\ 0&1&-1&2\\ 0&0&0&0 \end{bmatrix},

so $rank(A)=2$ . Since $A$ has four columns,

dim(N(A))=4-2=2.

Finally check independence. Put the proposed vectors into a matrix:

U=[u_1\ u_2] = \begin{bmatrix} 1&5\\ -1&-3\\ 1&-1\\ 1&1 \end{bmatrix}.

Row reduction gives a pivot in each column, so $u_1,u_2$ are linearly independent. We now have membership, the correct dimension count, and independence. Therefore $u_1,u_2$ form a basis for N(A).

The failure modes are just as important:

if some $u_j$ does not satisfy $Au_j=0$ , the list cannot be a basis for N(A);
if the list has fewer or more than $q-r$ vectors, the size is wrong;
if the list has the correct size but is dependent, it still fails to be a basis.

Extracting a minimal spanning set

Sometimes the input is not a clean proposed basis. You may be given a long list of vectors and asked for the dimension of its span. The efficient method is to extract a basis from the original list.

Theorem

Pivot columns give a minimal spanning set

Let

W=\operatorname{Span}\{u_1,\ldots,u_q\}\subseteq R^n, \qquad U=[u_1\ u_2\ \cdots\ u_q].

Row-reduce $U$ . If the pivot columns of the reduced matrix occur in positions $d_1,\ldots,d_r$ , then

\{u_{d_1},\ldots,u_{d_r}\}

is a basis for $W$ , and $dim(W)=r$ .

The positions are read from a row-reduced form, but the basis vectors themselves come from the original list. This is the same rule used for column space.

Worked example

Extracting a basis from a generating list

Let $W=Span\{u_1,u_2,u_3\}$ , where

u_1=\begin{bmatrix}0\\-1\\2\end{bmatrix}, \quad u_2=\begin{bmatrix}1\\-2\\7\end{bmatrix}, \quad u_3=\begin{bmatrix}-2\\3\\-12\end{bmatrix}.

Put the vectors into the columns of $U$ :

U= \begin{bmatrix} 0&1&-2\\ -1&-2&3\\ 2&7&-12 \end{bmatrix} \sim \begin{bmatrix} 1&0&1\\ 0&1&-2\\ 0&0&0 \end{bmatrix}.

The pivot columns are columns 1 and 2, so

\{u_1,u_2\}

is a basis for $W$ , and $dim(W)=2$ . The non-pivot column also tells us the redundancy relation:

u_3=u_1-2u_2.

Subspaces with the same dimension

Dimension also controls when one subspace can equal another. If $V$ and $W$ are subspaces of the same ambient space and $V \subseteq W$ , then

dim(V) \le dim(W).

Moreover,

dim(V)=dim(W) \quad\text{if and only if}\quad V=W.

The containment hypothesis matters. Equal dimension alone does not force two subspaces to be equal. For example, two different lines through the origin in $R^2$ both have dimension 1, but neither line contains the other. The theorem applies when one subspace is already known to sit inside the other.

Worked example

Using dimension to prove equality of subspaces

Let

V=Span\left\{ \begin{bmatrix}1\\0\\0\end{bmatrix}, \begin{bmatrix}0\\1\\0\end{bmatrix} \right\}

and let

W=\left\{ \begin{bmatrix}x\\y\\0\end{bmatrix}:x,y\in R \right\}.

Every vector in $V$ has third coordinate zero, so $V \subseteq W$ . We also know from the earlier plane example that $dim(W)=2$ . Since the two displayed generators of $V$ are linearly independent, $dim(V)=2$ as well.

Thus $V$ is a subspace of $W$ and both have dimension 2. The comparable-subspace theorem gives $V=W$ .

Common mistake

Dimension is not just the number of coordinates

A subspace of $R^3$ can have dimension 1, 2, or 3. The ambient space tells you how many entries vectors have; the dimension tells you how many independent directions the subspace itself has.

Another common mistake is to check spanning and stop there. A spanning set can still fail to be a basis if one vector is redundant.

Quick checks

Quick check

Can `\{(1,0), (2,0)\}` be a basis for $R^2$ ?

Think about both basis conditions, not just the number of vectors.

Solution

Answer

Quick check

Suppose $dim(V) = 3$ and you already found three linearly independent vectors in $V$ . What remains to prove?

Use the dimension shortcut from this note.

Solution

Answer

Exercises

Quick check

Do the vectors $u_1 = (1,1,0)$ , $u_2 = (1,0,1)$ , and $u_3 = (0,1,1)$ form a basis for $R^3$ ?

Try to test independence first. If they are independent, dimension does the rest.

Solution

Guided solution

Read this first

This note depends on 6.4 Linear dependence and independence and 6.3 Linear combinations and span.

6.5 Basis and dimension

MATH1030: Linear algebra I

Intuition first: enough directions, but not extra baggage

Basis

Build one vector from a span

The standard basis is the model example

Why the standard basis of R3R^3R3 really is a basis

Why the number of basis vectors matters

Every basis of the same space has the same size

Too many vectors in a spanned space force dependence

Dimension is a count of independent directions

Dimension

A basis does not have to look like the standard one

A basis for a plane inside R3R^3R3

How dimension becomes a shortcut

Using dimension to avoid a second calculation in R3R^3R3

Test one set for dependence

A practical checklist for basis questions

Extension and reduction consequences

Null spaces: a rank-nullity basis test

Testing a proposed basis for a null space

Extracting a minimal spanning set

Pivot columns give a minimal spanning set

Extracting a basis from a generating list

Subspaces with the same dimension

Using dimension to prove equality of subspaces

Common mistake

Dimension is not just the number of coordinates

Quick checks

Can \{(1,0), (2,0)\} be a basis for R2R^2R2?

Answer

Suppose dim(V)=3dim(V) = 3dim(V)=3 and you already found three linearly independent vectors in VVV. What remains to prove?

Answer

Exercises

Do the vectors u1=(1,1,0)u_1 = (1,1,0)u1​=(1,1,0), u2=(1,0,1)u_2 = (1,0,1)u2​=(1,0,1), and u3=(0,1,1)u_3 = (0,1,1)u3​=(0,1,1) form a basis for R3R^3R3?

Guided solution

Read this first

Section mastery checkpoint

Prerequisites

Key terms in this unit

Premium learning add-ons

More notes in this series

Why the standard basis of $R^3$ really is a basis

A basis for a plane inside $R^3$

Using dimension to avoid a second calculation in $R^3$

Can `\{(1,0), (2,0)\}` be a basis for $R^2$ ?

Suppose $dim(V) = 3$ and you already found three linearly independent vectors in $V$ . What remains to prove?

Do the vectors $u_1 = (1,1,0)$ , $u_2 = (1,0,1)$ , and $u_3 = (0,1,1)$ form a basis for $R^3$ ?