6.5 Basis and dimension

A spanning list represents every vector in a space, but a dependent list can give several coefficient lists for the same vector. Requiring independence as well as spanning makes each representation unique. A basis combines these two conditions.

Definition 6.5.1

(Basis)

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

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

Here the scalars are real and the displayed bases are finite. Fixing an order $(u_1,\ldots,u_m)$ gives unique coordinates: spanning provides an expression for $x$, and if $x=\sum a_i u_i=\sum b_i u_i$, subtraction gives $\sum(a_i-b_i)u_i=0$. Independence forces $a_i=b_i$ for every $i$. Thus existence and uniqueness come from different parts of the definition.

Spanning and independence answer different questions

The following sequence separates the two basis conditions: spanning supplies every target, while independence removes redundant representations. Dimension records the common size of a basis; the counting proof comes next.

Compare a pair of scalar-multiple generators with an independent pair, and identify the span in each case.

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 6.5.2

(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 $R^3$.

Once the standard basis is fixed, the coefficients $a$, $b$, and $c$ rebuild the vector uniquely.

Why the number of basis vectors matters

Different bases can use different vectors. To define dimension, we must first show that their lengths cannot differ:

Theorem 6.5.3

(Every basis of the same space has the same size)

If $B_1$ and $B_2$ are finite bases of the same real vector space $V$, then they 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 6.5.4

(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.

Proof

(Count coefficients before defining dimension)

The goal is to produce a nontrivial relation among $n\gt m$ vectors $u_j\in V$, using only the fact that $m$ vectors $v_i$ span $V$. Write

$$u_j=\sum_{i=1}^m a_{ij}v_i,\qquad C=(a_{ij})\in M_{m,n}(\mathbb R).$$

The coefficient matrix has $m$ rows and $n$ columns. Elimination gives at most $m$ pivots, so the homogeneous system $Cc=0$ has a free variable and a nonzero solution $c\in\mathbb R^n$. For that solution,

$$\sum_{j=1}^n c_ju_j =\sum_{i=1}^m\left(\sum_{j=1}^n a_{ij}c_j\right)v_i=0.$$

This is the required nontrivial relation. We imposed the coefficient equations as a sufficient condition for the vector sum to vanish. We did not assume the spanning vectors $v_i$ were independent, so we must not infer conversely that every vanishing sum has zero coefficients relative to that list. When $m=0$, the span is the zero space and any nonempty list in it contains only zero vectors, hence is dependent directly.

Now let the two finite bases have sizes $r,s$. The first spans and the second is independent, so $s\le r$. Interchanging their roles gives $r\le s$, hence $r=s$. This uses the earlier homogeneous-system theorem, not dimension itself; therefore defining dimension by basis size is not circular.

Dimension is a count of independent directions

Definition 6.5.5

(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:

$$\dim(\{0\}) = 0.$$

Its basis is the empty list: its only linear combination is the empty sum, defined as zero, and it has no nontrivial coefficient relation. The singleton $\{0\}$ spans the zero space but is dependent, so it is not a basis.

A subspace has its own dimension

A vector can have three entries while belonging to a space with only two independent directions. The next example separates the ambient coordinate count from the dimension of the subspace.

Worked example 6.5.6

(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:

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

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 6.5.7

(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.

A nontrivial zero relation detects redundancy in the given list.

Extending and reducing lists

Theorem 6.5.8

(Extension and reduction consequences)

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

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

A basis is also an inclusion-minimal spanning set: deleting any member loses spanning, since otherwise that member would be a combination of the others. Conversely, a dependent spanning list can be shortened by the redundancy theorem. A basis is inclusion-maximal among independent lists inside $V$: every further vector already lies in its span. Conversely, if an independent list does not span $V$, appending a vector outside its span preserves independence. These are inclusion statements, not merely guesses from list length.

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

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

An RREF with $r$ pivots has $q-r$ free coordinates. Setting one free coordinate to one and the others to zero gives one null-space vector for each free coordinate. Every solution is their combination, and their free coordinates force the coefficients of any zero combination to vanish. This gives the dimension formula, including the empty basis when $q=r$. Here rank is the pivot count; 6.6 Column space, row space, and rank identifies it with the dimensions of the row and column spaces.

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

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

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

Worked example 6.5.9

(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 $\operatorname{rank}(A)=2$. 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 6.5.10

(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 6.5.11

(Extracting a basis from a generating list)

Let $W=\operatorname{Span}\{u_1,u_2,u_3\}$, 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. Suppose $W$ is finite-dimensional, $V$ is a subspace of $W$, and both use the same real scalars. 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.

To see the equality implication, take a basis of $V$. It is an independent list inside $W$ of size $\dim W$, so the dimension criterion makes it a basis of $W$ too. Its span is both spaces. The finite-dimensional hypothesis is necessary for this counting argument; the earlier infinite polynomial space does not license extending this criterion to arbitrary infinite-dimensional inclusions.

Worked example 6.5.12

(Using dimension to prove equality of subspaces)

Let

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

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

Common mistake

(Dimension is not just the number of coordinates)

A subspace of $R^3$ can have dimension 0, 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

No. The two vectors point in the same direction, so they are linearly dependent. They also fail to span the vertical direction. Therefore they are not a basis for $R^2$.

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

Nothing essential remains. In a 3-dimensional space, any three linearly independent vectors automatically form a basis, so they already span $V$.

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

Set

$$\alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3 = 0.$$

Comparing coordinates gives

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

Subtract the first two equations to get $\alpha_2 - \alpha_3 = 0$, so $\alpha_2 = \alpha_3$. Then the third equation becomes $2\alpha_2 = 0$, so $\alpha_2 = 0$, and therefore $\alpha_1 = \alpha_3 = 0$ too.

So the three vectors are linearly independent. Because $R^3$ has dimension 3, they automatically form a basis for $R^3$.

Read this first

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