Math Core

A spanning set lets you reach every vector in a subspace, but it may contain redundant vectors that add nothing. A basis is a spanning set with all the redundancy squeezed out: big enough to reach everything, small enough that nothing is wasted. Bases are what let you attach coordinates to vectors in any vector space.

What a basis is

Definition

Basis

Let $H$ be a subspace of a vector space $V$ . An indexed set $\mathcal{B} = \lbrace \mathbf{b}_1, \dots, \mathbf{b}_p \rbrace$ in $V$ is a basis for $H$ if

$\mathcal{B}$ is linearly independent, and
$\mathcal{B}$ spans $H$ , that is, $H = \operatorname{Span}\lbrace \mathbf{b}_1, \dots, \mathbf{b}_p \rbrace$ .

The two conditions pull in opposite directions. Adding vectors makes spanning easier but risks dependence; removing vectors keeps independence but risks losing the span. A basis sits exactly at the balance point.

Some standard bases:

The columns $\mathbf{e}_1, \dots, \mathbf{e}_n$ of the $n \times n$ identity matrix form the standard basis of $\mathbb{R}^n$ .
$\lbrace 1, t, t^2, \dots, t^n \rbrace$ is the standard basis of $\mathbb{P}_n$ . It spans because every polynomial of degree at most $n$ is a combination of powers of $t$ , and it is independent because a polynomial that is zero for all $t$ must have all coefficients zero.

For $n$ vectors in $\mathbb{R}^n$ there is a quick test. Put them into an $n \times n$ matrix $A$ . By the Invertible Matrix Theorem, the columns are independent exactly when they span $\mathbb{R}^n$ , and both happen exactly when $A$ is invertible, that is, when $\det A \ne 0$ .

Worked example: Testing a basis of R^3

Is $\lbrace (1, 0, 1), (2, 1, 0), (0, 1, -2) \rbrace$ a basis for $\mathbb{R}^3$ ?

Form the matrix with these columns and expand along the first row:

\det \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & -2 \end{bmatrix} = 1(1 \cdot (-2) - 1 \cdot 0) - 2(0 \cdot (-2) - 1 \cdot 1) + 0 = -2 + 2 = 0.

The determinant is $0$ , so the columns are dependent and the set is not a basis. Indeed, $2(1, 0, 1) - (2, 1, 0) = (0, -1, 2)$ , which is $-1$ times the third vector.

Trimming a spanning set

Suppose $\mathbf{v}_3$ is a combination of $\mathbf{v}_1$ and $\mathbf{v}_2$ . Then any combination of all three vectors can be rewritten using only $\mathbf{v}_1$ and $\mathbf{v}_2$ , so throwing $\mathbf{v}_3$ away does not shrink the span.

The spanning set theorem

Let $S = \lbrace \mathbf{v}_1, \dots, \mathbf{v}_p \rbrace$ span a subspace $H \ne \lbrace \mathbf{0} \rbrace$ .

If some $\mathbf{v}_k$ is a linear combination of the other vectors in $S$ , then removing $\mathbf{v}_k$ leaves a set that still spans $H$ .
Repeating this, some subset of $S$ is a basis for $H$ .

So a basis is a smallest possible spanning set. Viewed from the other side, it is also a largest possible independent set in $H$ : adding any further vector of $H$ creates a dependence, because that vector is already a combination of the basis.

Bases for Nul A and Col A

Null space. The method from the previous lesson (solve $A\mathbf{x} = \mathbf{0}$ and write the parametric vector form) already produces a linearly independent spanning set. So those vectors form a basis for $\operatorname{Nul} A$ , with one vector per free variable.

Column space. Row reduction does not change the linear relations among the columns. If $B$ is an echelon form of $A$ , then $A\mathbf{x} = \mathbf{0}$ and $B\mathbf{x} = \mathbf{0}$ have the same solutions, so any dependence among the columns of $B$ holds with the same weights among the columns of $A$ . In $B$ the pivot columns are clearly independent and every non-pivot column is a combination of pivot columns to its left. The same is therefore true in $A$ .

Basis for a column space

The pivot columns of $A$ form a basis for $\operatorname{Col} A$ . Use row reduction to find which columns are pivot columns, then take those columns from the original matrix $A$ .

Worked example: A basis for a column space

Find a basis for $\operatorname{Col} A$ , where

A = \begin{bmatrix} 1 & 2 & 0 & 3 \\ 2 & 4 & 1 & 4 \\ 1 & 2 & 1 & 1 \end{bmatrix}.

$R_2 - 2R_1$ gives $(0, 0, 1, -2)$ and $R_3 - R_1$ gives $(0, 0, 1, -2)$ . Then $R_3 - R_2$ :

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

The pivots are in columns $1$ and $3$ . Taking those columns from $A$ :

\operatorname{Col} A \text{ has basis } \left\lbrace \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} \right\rbrace.

The echelon form also shows the relations: $\mathbf{a}_2 = 2\mathbf{a}_1$ and $\mathbf{a}_4 = 3\mathbf{a}_1 - 2\mathbf{a}_3$ . You can check the second one directly in $A$ : $3(1, 2, 1) - 2(0, 1, 1) = (3, 4, 1)$ .

Common mistake

Do not use the pivot columns of the echelon form as the basis. Row operations change the column space. Here every column of the echelon form has third entry $0$ , but $\operatorname{Col} A$ contains $(1, 2, 1)$ , whose third entry is not $0$ . The echelon form tells you which columns; the original matrix supplies the vectors.

Worked example: A basis in a polynomial space

Show that $\lbrace 1,\; 1 + t,\; 1 + t + t^2 \rbrace$ is a basis for $\mathbb{P}_2$ .

Independent: suppose $c_1 + c_2(1 + t) + c_3(1 + t + t^2) = 0$ for all $t$ . Collecting powers of $t$ :

(c_1 + c_2 + c_3) + (c_2 + c_3)t + c_3 t^2 = 0.

Every coefficient must be $0$ , so $c_3 = 0$ , then $c_2 = 0$ , then $c_1 = 0$ .

Spans: given $a_0 + a_1 t + a_2 t^2$ , set $c_3 = a_2$ , $c_2 = a_1 - a_2$ and $c_1 = a_0 - a_1$ . Then the combination above has coefficients $a_0$ , $a_1$ and $a_2$ .

Both conditions hold, so the set is a basis.

Tip

When a set has more vectors than entries (for example, $4$ vectors in $\mathbb{R}^3$ ), it is automatically dependent and can't be a basis. When it has fewer, it can't span. In $\mathbb{R}^n$ , only sets of exactly $n$ vectors are candidates.

Practice

Practice 1

Which set is a basis for $\mathbb{R}^2$ ?

Practice 2

Which set is a basis for $\mathbb{R}^3$ ?

Practice 3

For what value of $h$ is $\lbrace (1, 0, 2), (0, 1, h), (1, 1, 5) \rbrace$ not a basis for $\mathbb{R}^3$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Let $A = \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 3 \end{bmatrix}$ . A basis for $\operatorname{Nul} A$ consists of one vector of the form $(a, b, 1)$ . Find it.

Enter a point like (2, -3)

Practice 5

Let $A = \begin{bmatrix} 1 & 3 & 1 \\ 2 & 6 & 0 \\ -1 & -3 & 2 \end{bmatrix}$ . An echelon form of $A$ is $\begin{bmatrix} 1 & 3 & 1 \\ 0 & 0 & -2 \\ 0 & 0 & 0 \end{bmatrix}$ . Which set is a basis for $\operatorname{Col} A$ ?

Practice 6

Which set is a basis for $\mathbb{P}_2$ ?