# 2.7: Basis and Dimension

- Page ID
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- Understand the basis theorem.
*Recipes:*basis for a column space, basis for a null space, basis of a span.*Picture:*basis of a subspace of \(\mathbb{R}^2 \) or \(\mathbb{R}^3 \).*Theorem:*basis theorem.*Essential vocabulary words:***basis**,**dimension**.

## Basis of a Subspace

As we discussed in Section 2.6, a subspace is the same as a span, except we do not have a set of spanning vectors in mind. There are infinitely many choices of spanning sets for a nonzero subspace; to avoid redundancy, usually it is most convenient to choose a spanning set with the *minimal* number of vectors in it. This is the idea behind the notion of a basis.

Let \(V\) be a subspace of \(\mathbb{R}^n \). A **basis **of \(V\) is a set of vectors \(\{v_1,v_2,\ldots,v_m\}\) in \(V\) such that:

- \(V = \text{Span}\{v_1,v_2,\ldots,v_m\}\text{,}\) and
- the set \(\{v_1,v_2,\ldots,v_m\}\) is linearly independent.

Recall that a set of vectors is *linearly independent* if and only if, when you remove any vector from the set, the span shrinks (Theorem 2.5.1 in Section 2.5). In other words, if \(\{v_1,v_2,\ldots,v_m\}\) is a basis of a subspace \(V\text{,}\) then no proper subset of \(\{v_1,v_2,\ldots,v_m\}\) will span \(V\text{:}\) it is a *minimal* spanning set. Any subspace admits a basis by Theorem 2.6.1 in Section 2.6.

A nonzero subspace has *infinitely many* different bases, but they all contain the same number of vectors.

We leave it as an exercise to prove that any two bases have the same number of vectors; one might want to wait until after learning the invertible matrix theorem in Section 3.5.

Let \(V\) be a subspace of \(\mathbb{R}^n \). The number of vectors in any basis of \(V\) is called the **dimension **of \(V\text{,}\) and is written \(\dim V\).

Find a basis of \(\mathbb{R}^2 \).

###### Solution

We need to find two vectors in \(\mathbb{R}^2 \) that span \(\mathbb{R}^2 \) and are linearly independent. One such basis is \(\bigl\{{1\choose 0},{0\choose 1}\bigr\}\text{:}\)

- They span because any vector \(a\choose b\) can be written as a linear combination of \({1\choose 0},{0\choose 1}\text{:}\)

\[\left(\begin{array}{c}a\\b\end{array}\right)=a\left(\begin{array}{c}1\\0\end{array}\right)+b\left(\begin{array}{c}0\\1\end{array}\right).\nonumber\] - They are linearly independent: if

\[x\left(\begin{array}{c}1\\0\end{array}\right)+y\left(\begin{array}{c}0\\1\end{array}\right)=\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}0\\0\end{array}\right)\nonumber\]

then \(x=y=0\).

This shows that the plane \(\mathbb{R}^2 \) has dimension 2.

Figure \(\PageIndex{1}\)

Find all bases of \(\mathbb{R}^2 \).

###### Solution

We know from the previous Example \(\PageIndex{1}\) that \(\mathbb{R}^2 \) has dimension 2, so any basis of \(\mathbb{R}^2 \) has two vectors in it. Let \(v_1,v_2\) be vectors in \(\mathbb{R}^2 \text{,}\) and let \(A\) be the matrix with columns \(v_1,v_2\).

- To say that \(\{v_1,v_2\}\) spans \(\mathbb{R}^2 \) means that \(A\) has a pivot, Definition 1.2.5 in Section 1.2, in every
*row*: see Theorem 2.3.1 in Section 2.3. - To say that \(\{v_1,v_2\}\) is linearly independent means that \(A\) has a pivot in every
*column*: see Recipe: Checking linear independence in Section 2.5.

Since \(A\) is a \(2\times 2\) matrix, it has a pivot in every row exactly when it has a pivot in every column. Hence any two noncollinear vectors form a basis of \(\mathbb{R}^2 \). For example,

\[\left\{\left(\begin{array}{c}1\\0\end{array}\right),\:\left(\begin{array}{c}1\\1\end{array}\right)\right\}\nonumber\]

is a basis.

Figure \(\PageIndex{2}\)

One shows exactly as in the above Example \(\PageIndex{1}\) that the standard coordinate vectors

\[e_1=\left(\begin{array}{c}1\\0\\ \vdots \\ 0\\0\end{array}\right),\quad e_2=\left(\begin{array}{c}0\\1\\ \vdots \\ 0\\0\end{array}\right),\quad\cdots ,\quad e_{n-1}=\left(\begin{array}{c}0\\0\\ \vdots \\1\\0\end{array}\right),\quad e_n=\left(\begin{array}{c}0\\0\\ \vdots \\0\\1\end{array}\right)\nonumber\]

form a basis for \(\mathbb{R}^n \). This is sometimes known as the **standard basis**.

In particular, \(\mathbb{R}^n \) has dimension \(n\).

The previous Example \(\PageIndex{3}\) implies that any basis for \(\mathbb{R}^n \) has \(n\) vectors in it. Let \(v_1,v_2,\ldots,v_n\) be vectors in \(\mathbb{R}^n \text{,}\) and let \(A\) be the \(n\times n\) matrix with columns \(v_1,v_2,\ldots,v_n\).

- To say that \(\{v_1,v_2,\ldots,v_n\}\) spans \(\mathbb{R}^n \) means that \(A\) has a pivot position, Definition 1.2.5 in Section 1.2, in every
*row*: see this Theorem 2.3.1 in Section 2.3. - To say that \(\{v_1,v_2,\ldots,v_n\}\) is linearly independent means that \(A\) has a pivot position in every
*column*: see Recipe: Checking linear independence in Section 2.5.

Since \(A\) is a square matrix, it has a pivot in every row if and only if it has a pivot in every column. We will see in Section 3.5 that the above two conditions are equivalent to the *invertibility* of the matrix \(A\).

Let

\[V=\left\{\left(\begin{array}{c}x\\y\\z\end{array}\right)\text{ in }\mathbb{R}^{3}|x+3y+z=0\right\}\quad\mathcal{B}=\left\{\left(\begin{array}{c}-3\\1\\0\end{array}\right),\:\left(\begin{array}{c}0\\1\\-3\end{array}\right)\right\}.\nonumber\]

Verify that \(V\) is a subspace, and show directly that \(\mathcal{B}\) is a basis for \(V\).

###### Solution

First we observe that \(V\) is the solution set of the homogeneous equation \(x + 3y + z = 0\text{,}\) so it is a subspace: see this note in Section 2.6, Note 2.6.3. To show that \(\mathcal{B}\) is a basis, we really need to verify three things:

- Both vectors are in \(V\) because

\[\begin{array}{rrrrrrl} (-3) &+& 3(1) &+& (0) &=& 0\\ (0) &+& 3(1) &+& (-3) &=& 0.\end{array}\nonumber\] *Span:*suppose that \(\left(\begin{array}{c}x\\y\\z\end{array}\right)\) is in \(V\). Since \(x + 3y + z = 0\) we have \(y = -\frac 13(x+z)\text{,}\) so

\[\left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}x\\ {-\frac{1}{3}(x+z)} \\ z\end{array}\right)=-\frac{x}{3}\left(\begin{array}{c}-3\\1\\0\end{array}\right)-\frac{z}{3}\left(\begin{array}{c}0\\1\\-3\end{array}\right).\nonumber\]

Hence \(\mathcal{B}\) spans \(V\).*Linearly independent:*

\[c_1\left(\begin{array}{c}-3\\1\\0\end{array}\right)+c_2\left(\begin{array}{c}0\\1\\-3\end{array}\right)=0\implies\left(\begin{array}{c}-3c_1 \\ c_1+c_2 \\ -3c_2\end{array}\right)=\left(\begin{array}{c}0\\0\\0\end{array}\right)\implies c_1=c_2=0.\nonumber\]

Alternatively, one can observe that the two vectors are not collinear.

Since \(V\) has a basis with two vectors, it has dimension two: it is a *plane*.

This example is somewhat contrived, in that we will learn systematic methods for verifying that a subset is a basis. The intention is to illustrate the defining properties of a basis.

## Computing a Basis for a Subspace

Now we show how to find bases for the column space of a matrix and the null space of a matrix. In order to find a basis for a given subspace, it is usually best to rewrite the subspace as a column space or a null space first: see this note in Section 2.6, Note 2.6.3

### A Basis for the Column Space

First we show how to compute a basis for the column space of a matrix.

The pivot columns of a matrix \(A\) form a basis for \(\text{Col}(A)\).

**Proof**-
This is a restatement of Theorem 2.5.3 in Section 2.5.

The above theorem is referring to the pivot columns in the *original* matrix, not its reduced row echelon form. Indeed, a matrix and its reduced row echelon form generally have different column spaces. For example, in the matrix \(A\) below:

Figure \(\PageIndex{4}\)

the pivot columns are the first two columns, so a basis for \(\text{Col}(A)\) is

\[\left\{\left(\begin{array}{c}1\\-2\\2\end{array}\right),\:\left(\begin{array}{c}2\\-3\\4\end{array}\right)\right\}.\nonumber\]

The first two columns of the reduced row echelon form certainly span a different subspace, as

\[\text{Span}\left\{\left(\begin{array}{c}1\\0\\0\end{array}\right),\:\left(\begin{array}{c}0\\1\\0\end{array}\right)\right\}=\left\{\left(\begin{array}{c}a\\b\\0\end{array}\right)|a,b\text{ in }\mathbb{R}\right\}=(x,y\text{-plane}),\nonumber\]

but \(\text{Col}(A)\) contains vectors whose last coordinate is nonzero.

The dimension of \(\text{Col}(A)\) is the number of pivots of \(A\).

### A Basis of a Span

Computing a basis for a span is the same as computing a basis for a column space. Indeed, the span of finitely many vectors \(v_1,v_2,\ldots,v_m\) *is* the column space of a matrix, namely, the matrix \(A\) whose columns are \(v_1,v_2,\ldots,v_m\text{:}\)

\[A=\left(\begin{array}{cccc}|&|&\quad &| \\ v_1 &v_2 &\cdots &v_m \\ |&|&\quad &|\end{array}\right).\nonumber\]

Find a basis of the subspace

\[V=\text{Span}\left\{\left(\begin{array}{c}1\\-2\\2\end{array}\right),\:\left(\begin{array}{c}2\\-3\\4\end{array}\right),\:\left(\begin{array}{c}0\\4\\0\end{array}\right),\:\left(\begin{array}{c}-1\\5\\-2\end{array}\right)\right\}.\nonumber\]

###### Solution

The subspace \(V\) is the column space of the matrix

\[A=\left(\begin{array}{cccc}1&2&0&-1 \\ -2&-3&4&5 \\ 2&4&0&-2\end{array}\right).\nonumber\]

The reduced row echelon form of this matrix is

\[\left(\begin{array}{cccc}1&0&-8&-7 \\ 0&1&4&3 \\ 0&0&0&0\end{array}\right).\nonumber\]

The first two columns are pivot columns, so a basis for \(V\) is

\[\left\{\left(\begin{array}{c}1\\-2\\2\end{array}\right),\:\left(\begin{array}{c}2\\-3\\4\end{array}\right)\right\}.\nonumber\]

Find a basis of the subspace

\[V=\text{Span}\left\{\left(\begin{array}{c}1\\-2\\2\end{array}\right),\:\left(\begin{array}{c}2\\-3\\4\end{array}\right),\:\left(\begin{array}{c}0\\4\\0\end{array}\right),\:\left(\begin{array}{c}-1\\5\\-2\end{array}\right)\right\}\nonumber\]

which does not consist of the first two vectors, as in the previous Example \(\PageIndex{6}\).

###### Solution

The point of this example is that the above Theorem \(\PageIndex{1}\) gives *one* basis for \(V\text{;}\) as always, there are infinitely more.

Reordering the vectors, we can express \(V\) as the column space of

\[A'=\left(\begin{array}{cccc}0&-1&1&2 \\ 4&5&-2&-3 \\ 0&-2&2&4\end{array}\right).\nonumber\]

The reduced row echelon form of this matrix is

\[\left(\begin{array}{cccc}1&0&3/4 &7/4 \\ 0&1&-1&-2 \\ 0&0&0&0\end{array}\right).\nonumber\]

The first two columns are pivot columns, so a basis for \(V\) is

\[\left\{\left(\begin{array}{c}0\\4\\0\end{array}\right),\:\left(\begin{array}{c}-1\\5\\-2\end{array}\right)\right\}.\nonumber\]

These are the *last* two vectors in the given spanning set.

### A Basis for the Null Space

In order to compute a basis for the null space of a matrix, one has to find the parametric vector form of the solutions of the homogeneous equation \(Ax=0\).

The vectors attached to the free variables in the parametric vector form of the solution set of \(Ax=0\) form a basis of \(\text{Nul}(A)\).

The proof of the theorem has two parts. The first part is that every solution lies in the span of the given vectors. This is automatic: the vectors are exactly chosen so that every solution is a linear combination of those vectors. The second part is that the vectors are linearly independent. This part was discussed in Example 2.5.3 in Section 2.5.

### A Basis for a General Subspace

As mentioned at the beginning of this subsection, when given a subspace written in a different form, in order to compute a basis it is usually best to rewrite it as a column space or null space of a matrix.

Let \(V\) be the subspace defined by

\[V=\left\{\left(\begin{array}{c}x\\y\\z\end{array}\right)|x+2y=z\right\}.\nonumber\]

Find a basis for \(V\). What is \(\dim(V)\text{?}\)

###### Solution

First we notice that \(V\) is exactly the solution set of the homogeneous linear equation \(x + 2y - z = 0\). Hence \(V = \text{Nul}\left(\begin{array}{ccc}1&2&-1\end{array}\right).\) This matrix is in reduced row echelon form; the parametric form of the general solution is \(x = -2y + z\text{,}\) so the parametric vector form is

\[\left(\begin{array}{c}x\\y\\z\end{array}\right)=y\left(\begin{array}{c}-2\\1\\0\end{array}\right)=z\left(\begin{array}{c}1\\0\\1\end{array}\right).\nonumber\]

It follows that a basis is

\[\left\{\left(\begin{array}{c}-2\\1\\0\end{array}\right),\:\left(\begin{array}{c}1\\0\\1\end{array}\right)\right\}.\nonumber\]

Since \(V\) has a basis with two vectors, its dimension is \(2\text{:}\) it is a plane.

## The Basis Theorem

Recall that \(\{v_1,v_2,\ldots,v_n\}\) forms a basis for \(\mathbb{R}^n \) if and only if the matrix \(A\) with columns \(v_1,v_2,\ldots,v_n\) has a pivot in every row and column (see this Example \(\PageIndex{4}\)). Since \(A\) is an \(n\times n\) matrix, these two conditions are equivalent: the vectors span if and only if they are linearly independent. The basis theorem is an abstract version of the preceding statement, that applies to any subspace.

Let \(V\) be a subspace of dimension \(m\). Then:

- Any \(m\) linearly independent vectors in \(V\) form a basis for \(V\).
- Any \(m\) vectors that span \(V\) form a basis for \(V\).

**Proof**-
Suppose that \(\mathcal{B} = \{v_1,v_2,\ldots,v_m\}\) is a set of linearly independent vectors in \(V\). In order to show that \(\mathcal{B}\) is a basis for \(V\text{,}\) we must prove that \(V = \text{Span}\{v_1,v_2,\ldots,v_m\}.\) If not, then there exists some vector \(v_{m+1}\) in \(V\) that is not contained in \(\text{Span}\{v_1,v_2,\ldots,v_m\}.\) By the increasing span criterion Theorem 2.5.2 in Section 2.5, the set \(\{v_1,v_2,\ldots,v_m,v_{m+1}\}\) is also linearly independent. Continuing in this way, we keep choosing vectors until we eventually do have a linearly independent spanning set: say \(V = \text{Span}\{v_1,v_2,\ldots,v_m,\ldots,v_{m+k}\}\). Then \(\{v_1,v_2,\ldots,v_{m+k}\}\) is a basis for \(V\text{,}\) which implies that \(\dim(V) = m+k > m\). But we were assuming that \(V\) has dimension \(m\text{,}\) so \(\mathcal{B}\) must have already been a basis.

Now suppose that \(\mathcal{B} = \{v_1,v_2,\ldots,v_m\}\) spans \(V\). If \(\mathcal{B}\) is not linearly independent, then by this Theorem 2.5.1 in Section 2.5, we can remove some number of vectors from \(\mathcal{B}\) without shrinking its span. After reordering, we can assume that we removed the last \(k\) vectors without shrinking the span, and that we cannot remove any more. Now \(V = \text{Span}\{v_1,v_2,\ldots,v_{m-k}\}\text{,}\) and \(\{v_1,v_2,\ldots,v_{m-k}\}\) is a basis for \(V\) because it is linearly independent. This implies that \(\dim V=m-k < m\). But we were assuming that \(\dim V = m\text{,}\) so \(\mathcal{B}\) must have already been a basis.

In other words, if you *already* know that \(\dim V = m\text{,}\) and if you have a set of \(m\) vectors \(\mathcal{B} = \{v_1,v_2,\ldots,v_m\}\) in \(V\text{,}\) then you only have to check *one* of:

- \(\mathcal{B}\) is linearly independent,
*or* - \(\mathcal{B}\) spans \(V\text{,}\)

in order for \(\mathcal{B}\) to be a basis of \(V\). If you did not already know that \(\dim V = m\text{,}\) then you would have to check *both* properties.

To put it yet another way, suppose we have a set of vectors \(\mathcal{B} = \{v_1,v_2,\ldots,v_m\}\) in a subspace \(V\). Then if any two of the following statements is true, the third must also be true:

- \(\mathcal{B}\) is linearly independent,
- \(\mathcal{B}\) spans \(V\text{,}\) and
- \(\dim V = m.\)

For example, if \(V\) is a plane, then any two noncollinear vectors in \(V\) form a basis.

Find a basis of the subspace

\[V=\text{Span}\left\{\left(\begin{array}{c}1\\-2\\2\end{array}\right),\:\left(\begin{array}{c}2\\-3\\4\end{array}\right),\:\left(\begin{array}{c}0\\4\\0\end{array}\right),\:\left(\begin{array}{c}-1\\5\\-2\end{array}\right)\right\}\nonumber\]

which is different from the bases in this Example \(\PageIndex{6}\) and this Example \(\PageIndex{7}\).

###### Solution

We know from the previous examples that \(\dim V = 2\). By the Theorem \(\PageIndex{3}\), it suffices to find any two noncollinear vectors in \(V\). We write two linear combinations of the four given spanning vectors, chosen at random:

\[w_1=\left(\begin{array}{c}1\\-2\\2\end{array}\right)+\left(\begin{array}{c}2\\-3\\4\end{array}\right)=\left(\begin{array}{c}3\\-5\\6\end{array}\right)\quad w_2=-\left(\begin{array}{c}2\\-3\\4\end{array}\right)+\frac{1}{2}\left(\begin{array}{c}0\\4\\0\end{array}\right)=\left(\begin{array}{c}-2\\5\\-4\end{array}\right).\nonumber\]

Since \(w_1,w_2\) are not collinear, \(\mathcal{B} = \{w_1,w_2\}\) is a basis for \(V\).

Find a basis for the plane

\[V=\left\{\left(\begin{array}{c}x_1\\x_2\\x_3\end{array}\right)|x_1 +x_2=x_3\right\}\nonumber\]

by inspection. (This plane is expressed in set builder notation, Note 2.2.3 in Section 2.2.)

###### Solution

First note that \(V\) is the null space of the matrix \(\left(\begin{array}{ccc}1&1&-1\end{array}\right)\) this matrix is in reduced row echelon form and has two free variables, so \(V\) is indeed a plane. We write down two vectors satisfying \(x_1 + x_2 = x_3\text{:}\)

\[v_1=\left(\begin{array}{c}1\\0\\1\end{array}\right)\quad v_2=\left(\begin{array}{c}0\\1\\1\end{array}\right).\nonumber\]

Since \(v_1\) and \(v_2\) are not collinear, they are linearly independent; since \(\dim(V) = 2\text{,}\) the basis theorem implies that \(\{v_1,v_2\}\) is a basis for \(V\).