Subspaces

Section 2.3 Subspaces

Subsection Definition of subspaces

The outputs a matrix can “reach” form a subspace. The directions a matrix “forgets” also form a subspace. The definitions below make this precise. In particular, the closure rules are not arbitrary. If two output vectors are reachable, then any linear combination of them is reachable. If two input changes are forgotten by a matrix, then any linear combination of those changes is also forgotten.

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We are already familiar with geometric spaces living inside bigger ones, like lines lying in the plane, or planes lying in 3-space. We are now going to formalize this notion algebraically. Recall the definition of the set \(\R^n\) in Definition 1.1.1.

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Definition 2.3.1. Subspaces.

A set \(U\) of vectors in \(\R^n\) is called a subspace of \(\R^n\) if it satisfies the following properties:

The zero vector (written as \(\mathbf{0}\)) is in \(U\text{.}\)

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If \(\mathbf{x}\) is in \(U\) and \(y\) is in \(U\text{,}\) then \(\mathbf{x} + \mathbf{y}\) is in \(U\text{.}\) This means \(U\) is closed under addition.

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If \(\mathbf{x}\) is in \(U\text{,}\) then \(a \mathbf{x}\) is in \(U\) for every scalar \(a\text{.}\) This means \(U\) is closed under scalar multiplication.

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Example 2.3.2.

The whole set \(\R^n\) is a subspace of itself.

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Example 2.3.3.

The set containing only the zero vector, i.e., the set \(\{ \mathbf{0} \}\text{,}\) is a subspace of \(\R^n\) (called the trivial subspace or zero subspace).

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Activity 2.3.1.

Verify that \(\{ \mathbf{0} \}\) is a subspace of \(\R^n\text{.}\)

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Solution.

To verify that \(\{ \mathbf{0} \}\) is a subspace, we must verify three properties:

Contains the zero vector: 🔗: The zero \(\mathbf{0}\) is evidently an element of \(\{ \mathbf{0} \}\text{.}\)
Closure under addition: 🔗: If \(\mathbf{x}, \mathbf{y} \in \{ \mathbf{0} \}\text{,}\) then \(\mathbf{x} = 0\) and \(\mathbf{y} = 0\text{,}\) so \(\mathbf{x} + \mathbf{y} = 0\text{,}\) which is an element of \(\{ \mathbf{0} \}\text{.}\)
Closure under scalar multiplication: 🔗: If \(\mathbf{x} \in \{ \mathbf{0} \}\) and \(a\) is a scalar, then \(\mathbf{x} = 0\text{,}\) and \(a\mathbf{x} = a \mathbf{0} = 0\) is in \(\{ \mathbf{0} \}\text{.}\)

Since \(\{ \mathbf{0} \}\) satisfies all three of the necessary conditions, it is a subspace.

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Definition 2.3.4. Proper Subspaces.

Any subspace of \(\R^n\) other than \(\R^n\) or \(\{ 0 \}\) is a proper subspace of \(\R^n\text{.}\)

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Example 2.3.5.

Planes and lines through the origin in \(\R^3\) are all subspaces of \(\R^3\text{.}\)

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Activity 2.3.2.

(a)

Verify that planes through the origin in \(\R^3\) are all proper subspaces of \(\R^3\text{.}\)

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Solution.

Let \(R\) be a plane through the origin. Then there exist scalars \(a,b,c\text{,}\) not all zero, such that \(R\) is the set of points \((x,y,z)\) satisfying the equation \(ax + by + cz = 0\text{.}\) The vector

\begin{equation*} \mathbf{n} = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \end{equation*}

is a normal vector for the plane, and we can write \(R\) as the set of vectors \(\mathbf{v} \in \R^3\) such that \(\mathbf{n} \cdot \mathbf{v} = 0\text{.}\)

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Let us verify that \(R\) satisfies the conditions to be a subspace:

Contains the zero vector 🔗: \(\mathbf{0} \in R\) because \(\mathbf{n} \cdot \mathbf{0} = 0\text{.}\)
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Closure under addition 🔗: If \(\mathbf{v} \in R\) and \(\mathbf{w} \in R\) then to verify \(\mathbf{v} + \mathbf{w} \in R\) we must argue that \(\mathbf{n} \cdot ( \mathbf{v} + \mathbf{w}) = 0\text{.}\) Indeed,

\begin{equation*} \mathbf{n} \cdot (\mathbf{v} + \mathbf{w}) = \mathbf{n} \cdot \mathbf{v} + \mathbf{n} \cdot \mathbf{w} = 0\text{.} \end{equation*}

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Closure under scalar multiplication 🔗: If \(\mathbf{v} \in R\text{,}\) and \(a\) is a scalar, to verify \(a \mathbf{v} \in R\) we must argue that \(\mathbf{n} \cdot (a \mathbf{v}) = 0\text{.}\) Indeed,

\begin{equation*} \mathbf{n} \cdot (a \mathbf{v}) = a ( \mathbf{n} \cdot \mathbf{v}) = a (0) = 0\text{.} \end{equation*}

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We have verified the three properties required for \(R\) to be a subspace.

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(b)

Verify that lines through the origin in \(\R^3\) are subspaces of \(\R^3\text{.}\)

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Solution.

Let \(L\) be a line through the origin in \(\R^3\text{.}\) If \(L\) has direction vector \(\mathbf{d}\text{,}\) then \(L = \{ t \mathbf{d}: t \in \R \}\text{,}\) i.e., the line \(L\) consists of all scalar multiples of \(\mathbf{d}\text{.}\)

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Let us verify that \(L\) satisfies the conditions to be a subspace:

Contains the zero vector 🔗: \(\mathbf{0} \in L\) because \(\mathbf{0} = 0 \mathbf{d}\text{.}\)
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Closure under addition 🔗: Suppose \(\mathbf{v} \in L\) and \(\mathbf{w} \in L\text{.}\) Then \(\mathbf{v}\) and \(\mathbf{w}\) are both scalar multiples of \(\mathbf{d}\text{,}\) i.e., \(\mathbf{v} = t_1 \mathbf{d}\) and \(\mathbf{w} = t_2 \mathbf{d}\text{.}\) Then

\begin{equation*} \mathbf{v} + \mathbf{w} = t_1 \mathbf{d} + t_2 \mathbf{d} = (t_1 + t_2) \mathbf{d}\text{.} \end{equation*}

Thus \(\mathbf{v} + \mathbf{w}\) is a scalar multiple of \(\mathbf{d}\text{,}\) and so \(\mathbf{v} + \mathbf{w} \in L\text{.}\)

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Closure under scalar multiplication 🔗: If \(\mathbf{v} \in L\text{,}\) and \(a\) is a scalar, then we may write \(\mathbf{v} = t \mathbf{d}\text{,}\) and then

\begin{equation*} a \mathbf{v} = a (t\mathbf{d}) = (at) \mathbf{d}\text{.} \end{equation*}

So \(a \mathbf{v}\) is a scalar multiple of \(\mathbf{d}\text{,}\) and thus \(a \mathbf{v} \in L\text{.}\)

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We have verified the three properties required for \(L\) to be a subspace.

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Note 2.3.6.

Lines and planes which do not pass through the origin in \(\R^3\) are not subspaces of \(\R^3\text{.}\) Nonetheless, if \(\Pi\) is a plane, and we consider the set of displacement vectors

\begin{equation*} \Delta\Pi = \{ \mathbf{v}_1 - \mathbf{v}_0 : \mathbf{v}_0,\mathbf{v}_1 \in \Pi \}\text{,} \end{equation*}

then \(\Delta\Pi\) forms a plane passing through the origin, and so is a subspace of \(\R^3\text{.}\) A similar result holds for sets of displacement vectors given by a line not passing through the origin.

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Activity 2.3.3.

Show that the set

\begin{equation*} U = \left\{ \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} : x_1 x_2 = 0 \right\} \end{equation*}

is not a subspace of \(\R^2\text{.}\)

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Solution.

It suffices to show one of the conditions required to be a subspace does not hold. \(U\) does contain the zero vector, and \(is\) closed under scalar multiplication. But \(U\) is not closed under addition. Define

\begin{equation*} \mathbf{x} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad\text{and}\quad \mathbf{y} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \text{.} \end{equation*}

Then \(\mathbf{x} \in U\) and \(\mathbf{y} \in U\text{.}\) But

\begin{equation*} \mathbf{x} + \mathbf{y} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \end{equation*}

is not in \(U\text{,}\) since \((1)(1) \neq 0\text{.}\) Since \(U\) fails to be closed under addition, it is not a subspace of \(\R^2\text{.}\)

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Activity 2.3.4.

Let \(W\) be the set of all vectors in \(\R^3\) of the form

\begin{equation*} \begin{bmatrix} r \\ s \\ r + 2s \end{bmatrix}\text{,} \end{equation*}

where \(r,s \in \R\text{.}\) Is \(W\) a proper subspace of \(\R^3\text{?}\)

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Solution.

We check whether the required properties hold in order to be a subspace:

Contains the zero vector: 🔗: The zero vector is in \(W\text{,}\) since
\begin{equation*} \mathbf{0} = \begin{bmatrix} r \\ s \\ r + 2s \end{bmatrix} \end{equation*}
when \(r = 0\) and \(s = 0\text{.}\)
Closed under addition: 🔗: If
\begin{equation*} \mathbf{v} = \begin{bmatrix} r_1 \\ s_1 \\ r_1 + 2s_1 \end{bmatrix} \quad\text{and}\quad \mathbf{w} = \begin{bmatrix} r_2 \\ s_2 \\ r_2 + 2s_2 \end{bmatrix} \end{equation*}
Then
\begin{align*} \mathbf{v} + \mathbf{w} \amp = \begin{bmatrix} r_1 + r_2 \\ s_1 + s_2 \\ r_1 + 2s_1 + r_2 + 2s_2 \end{bmatrix}\\ \amp = \begin{bmatrix} r_1 + r_2 \\ s_1 + s_2 \\ (r_1 + r_2) + 2(s_1 + s_2) \end{bmatrix}\\ \amp = \begin{bmatrix} r \\ s \\ r + 2s \end{bmatrix}\text{,} \end{align*}
where \(r = r_1 + r_2\) and \(s = s_1 + s_2\text{.}\) Thus \(\mathbf{v} + \mathbf{w} \in W\text{.}\)
Closed under scalar multiplication: 🔗: If
\begin{equation*} \mathbf{v} = \begin{bmatrix} r_1 \\ s_1 \\ r_1 + 2s_1 \end{bmatrix}\text{,} \end{equation*}
and \(c\) is a scalar, then
\begin{equation*} c \mathbf{v} = \begin{bmatrix} cr_1 \\ cs_1 \\ c(r_1 + 2s_1) \end{bmatrix} = \begin{bmatrix} r \\ s \\ r + 2s \end{bmatrix}\text{,} \end{equation*}
with \(r = cr_1\) and \(s = cs_1\text{.}\) Thus \(c \mathbf{v} \in W\text{.}\)

Thus \(W\) is a subspace of \(\R^3\text{.}\)

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To check that \(W\) is a proper subspace of \(\R^3\text{,}\) we must show that \(W \neq \R^3\) and \(W \neq \{ \mathbf{0} \}\text{.}\) The vector

\begin{equation*} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \end{equation*}

is not in \(W\text{,}\) since if it were of the form

\begin{equation*} \begin{bmatrix} r \\ s \\ r + 2s \end{bmatrix}\text{,} \end{equation*}

then \(r = 1\) and \(s = 1\text{,}\) and the third entry would have to be \(3\text{.}\) Thus \(W\) does not contain every vector, and thus cannot be equal to \(\R^3\text{.}\) But it also contains some non-zero vectors, such as

\begin{equation*} \begin{bmatrix} 1\\ 1 \\ 3 \end{bmatrix}\text{,} \end{equation*}

and so \(W \neq \{ \mathbf{0} \}\text{.}\)

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Thus \(W\) is a proper subspace of \(\R^3\text{.}\)

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Note 2.3.7.

To show a set is a subspace, you tend to need to parameterize all elements of the set in order to show they all satisfy some property (closure under addition or scalar multiplication), and so your argument should use variables. To show a set is not a subspace, you need to find one or more counterexamples that show that some property does not hold for all elements; a single explicit counterexample usually suffices.

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Subsection Null space and image space

The following important examples are representative of two canonical ways of describing a subspace.

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From here on, homogeneous equations play a special role. To solve \(A\mathbf{x}=\mathbf{0}\) by row reduction, we could write the augmented matrix \([A\mid \mathbf{0}]\text{.}\) Since the right-hand side stays zero throughout the row operations, the coefficient matrix \(A\) carries the essential information.

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Definition 2.3.8. Null Space and Image Space.

Let \(A\) be an \(m \times n\) matrix. The null space of \(A\text{,}\) denoted \(\nulls(A)\text{,}\) is the set of vectors \(\mathbf{x}\) in \(\R^n\) which are solutions to the equation \(A \mathbf{x} = \mathbf{0}\text{,}\) i.e.,

\begin{equation*} \nulls(A) = \{ \mathbf{x} \in \R^n : A \mathbf{x} = \mathbf{0} \}\text{.} \end{equation*}

The image space of \(A\text{,}\) denoted \(\im(A)\text{,}\) is the set of vectors \(\mathbf{y}\) in \(\R^m\) such that \(A\mathbf{x} = \mathbf{y}\) has a solution, i.e.,

\begin{equation*} \im(A) = \{ A\mathbf{x} : \mathbf{x} \in \R^n \}\text{.} \end{equation*}

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Note 2.3.9.

The null space lives in the input space. It records input directions that \(A\) sends to \(\mathbf{0}\text{.}\)

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The image space lives in the output space. It records outputs that \(A\) can reach.

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Table 2.3.10.

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Object	Lives in	How to read it
\(\operatorname{null}(A)\)	\(\mathbb R^n\)	input directions \(A\) forgets
\(\operatorname{im}(A)\)	\(\mathbb R^m\)	outputs \(A\) can reach

Activity 2.3.5. Revisiting the height map with subspace language.

Return to Activity 2.1.1. Let

\begin{equation*} P= \begin{bmatrix} 1&0&0\\ 0&1&0 \end{bmatrix}. \end{equation*}

Describe the output subspace and the subspace of input directions that \(P\) forgets.

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Solution.

For

\begin{equation*} \mathbf{x} = \begin{bmatrix} x\\ y\\ z \end{bmatrix}, \qquad P\mathbf{x} = \begin{bmatrix} x\\ y \end{bmatrix}. \end{equation*}

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As \(x\) and \(y\) vary freely, every vector in \(\mathbb R^2\) occurs as an output. Therefore

\begin{equation*} \operatorname{im}(P)=\mathbb R^2. \end{equation*}

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The directions that \(P\) forgets are the solutions to \(P\mathbf{x}=\mathbf{0}\text{.}\) This requires

\begin{equation*} x=0,\qquad y=0, \end{equation*}

while \(z\) is free. Thus

\begin{equation*} \operatorname{null}(P) = \left\{ \begin{bmatrix} 0\\ 0\\ t \end{bmatrix} : t\in\mathbb R \right\}. \end{equation*}

This subspace records the height direction.

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Theorem 2.3.11.

For any \(m \times n\) matrix, the null space of \(A\) is a subspace of \(\R^n\text{,}\) and the image space of \(A\) is a subspace of \(\R^m\text{.}\)

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Why is this true?.

We start by showing the null space of \(A\) is a subspace. Recall the properties discussed in Theorem 1.3.2:

Contains the zero vector 🔗: \(\mathbf{0} \in \nulls(A)\) because \(\mathbf{A} \mathbf{0} = \mathbf{0}\text{.}\)
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Closure under addition 🔗: Suppose \(\mathbf{v} \in \nulls(A)\) and \(\mathbf{w} \in \nulls(A)\text{.}\) Then \(A \mathbf{v} = \mathbf{0}\) and \(A \mathbf{w} = \mathbf{0}\text{.}\) So

\begin{equation*} A (\mathbf{v} + \mathbf{w}) = A \mathbf{v} + A \mathbf{w} = \mathbf{0} + \mathbf{0} = \mathbf{0}\text{.} \end{equation*}

Thus \(\mathbf{v} + \mathbf{w} \in \nulls(A)\text{.}\)

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Closure under scalar multiplication 🔗: If \(\mathbf{v} \in \nulls(A)\text{,}\) and \(a\) is a scalar, then

\begin{equation*} A(a \mathbf{v}) = a (A \mathbf{v}) = a \mathbf{0} = \mathbf{0} \text{.} \end{equation*}

Thus \(a \mathbf{v} \in \nulls(A)\text{.}\)

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So \(\nulls(A)\) is a subspace of \(\R^n\text{.}\)

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Next, we show that \(\im(A)\) is a subspace.

Contains the zero vector 🔗: \(\mathbf{0} \in \im(A)\) because \(\mathbf{A} \mathbf{0} = \mathbf{0}\text{.}\)
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Closure under addition 🔗: Suppose \(\mathbf{v} \in \im(A)\) and \(\mathbf{w} \in \im(A)\text{.}\) Then the equations \(A\mathbf{x} = \mathbf{v}\) and \(A \mathbf{x} = \mathbf{w}\) both have solutions, i.e., there exists \(\mathbf{x}_1\) and \(\mathbf{x}_2\) such that \(A\mathbf{x}_1 = \mathbf{v}\) and \(A\mathbf{x}_2 = \mathbf{w}\text{.}\) Then

\begin{equation*} A(\mathbf{x}_1 + \mathbf{x}_2) = A\mathbf{x}_1 + A\mathbf{x}_2 = \mathbf{v} + \mathbf{w}\text{,} \end{equation*}

so the equation \(A \mathbf{x} = \mathbf{v} + \mathbf{w}\) has a solution, and so \(\mathbf{v} + \mathbf{w} \in \im(A)\text{.}\)

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Closure under scalar multiplication 🔗: If \(\mathbf{v} \in \im(A)\text{,}\) and \(a\) is a scalar, then we may write \(\mathbf{v} = A\mathbf{x}_1\) for some \(\mathbf{x}_1\text{.}\) But then

\begin{equation*} A(a\mathbf{x}_1) = a (A\mathbf{x}_1) = a \mathbf{v}\text{,} \end{equation*}

so the equation \(A\mathbf{x} = a \mathbf{v}\) has a solution, so that \(a \mathbf{v} \in \im(A)\text{.}\)

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Thus \(\im(A)\) is a subspace of \(\R^m\text{.}\)

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Warning 2.3.12.

The solution set of \(A\mathbf{x}=\mathbf{0}\) is a subspace. The solution set of \(A\mathbf{x}=\mathbf{b}\) for \(\mathbf{b}\neq\mathbf{0}\) is usually not a subspace; when it is nonempty, it is a shifted copy of \(\nulls(A)\text{.}\)

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Subsection Spanning Sets

Even though subspaces of \(\R^n\) can have infinitely many vectors, it is possible to represent them with a finite amount of data. Recall that a linear combination is a sum of scalar multiples of vectors.

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Activity 2.3.6.

Show that there are two vectors (describe them explicitly), so that every vector in the set \(W\) discussed in Activity 2.3.4 can be written as the linear combination of those two vectors.

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Solution.

We may write

\begin{equation*} \begin{bmatrix} r \\ s \\ r + 2s \end{bmatrix} = r \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + s \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}\text{.} \end{equation*}

Thus every vector in \(W\) can be written as a linear combination of

\begin{equation*} \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}\quad\text{and}\quad \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}\text{.} \end{equation*}

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Definition 2.3.13. The Span of a Set of Vectors.

The set of all linear combinations of a set of vectors \(\mathbf{x}_1,\dots,\mathbf{x}_k\) is called the span of the vectors \(\mathbf{x}_1,\dots,\mathbf{x}_k\text{,}\) and is denoted

\begin{equation*} \spans \{ \mathbf{x}_1, \dots, \mathbf{x}_k \} = \{ t_1 \mathbf{x}_1 + \cdots + t_k \mathbf{x}_k : t_1,\dots,t_k \in \R \}\text{.} \end{equation*}

If \(V = \spans \{ \mathbf{x}_1, \dots, \mathbf{x}_k \}\text{,}\) we say that \(V\) is spanned by the vectors \(\mathbf{x}_1,\dots,\mathbf{x}_k\text{,}\) and that the vectors \(\mathbf{x}_1,\dots,\mathbf{x}_k\) span the space \(V\text{.}\)

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For an image space, spanning vectors come from the columns of the matrix. If

\begin{equation*} A=[\mathbf{a}_1\ \mathbf{a}_2\ \cdots\ \mathbf{a}_n] \end{equation*}

and

\begin{equation*} \mathbf{x} = \begin{bmatrix} x_1\\ x_2\\ \vdots\\ x_n \end{bmatrix}, \end{equation*}

then

\begin{equation*} A\mathbf{x} = x_1\mathbf{a}_1+x_2\mathbf{a}_2+\cdots+x_n\mathbf{a}_n. \end{equation*}

Thus every output \(A\mathbf{x}\) is a linear combination of the columns of \(A\text{.}\)

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Theorem 2.3.14.

Let \(U = \spans \{ \mathbf{x}_1,\dots,\mathbf{x}_k \}\) be a set in \(\R^n\text{.}\) Then

\(U\) is a subspace of \(\R^n\) containing each of the vectors \(\mathbf{x}_1,\dots,\mathbf{x}_k\text{.}\)

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If \(W\) is a subspace of \(\R^n\) and each of the vectors \(\mathbf{x}_1,\dots,\mathbf{x}_k\) is in \(W\text{,}\) then \(U\) is a subset of \(W\text{,}\) i.e., \(U \subseteq W\text{.}\)

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Why is this true?.

Define an \(n \times k\) matrix \(A\text{,}\) with the vectors \(\mathbf{x}_1,\dots,\mathbf{x}_k\) as columns. Then \(U = \im(A)\text{,}\) and is thus a subspace of \(\R^n\text{.}\) Since \(\mathbf{x}_i = A \mathbf{e}_i\) for \(1 \leq i \leq k\text{,}\) the set \(U\) contains each of the vectors \(\mathbf{x}_i\text{.}\)

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Conversely, suppose \(W\) is a subspace of \(\R^n\) and \(\mathbf{x}_1,\dots,\mathbf{x}_k \in W\text{.}\) Since \(W\) is closed under addition and scalar multiplication, all of the linear combinations of \(\mathbf{x}_1,\dots,\mathbf{x}_k\) are elements of \(W\text{,}\) and so \(U \subseteq W\text{.}\)

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Example 2.3.15.

Recall the standard basis \(\{ \mathbf{e}_1, \dots, \mathbf{e}_n \}\) of \(\R^n\) from Definition 1.3.8. Note that if

\begin{equation*} \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ \cdots \\ x_n \end{bmatrix}\text{,} \end{equation*}

then

\begin{equation*} \mathbf{x} = x_1 \mathbf{e}_1 + x_2 \mathbf{e}_2 + \dots + x_n \mathbf{e}_n\text{.} \end{equation*}

This means that \(\R^n = \spans \{ \mathbf{e}_1,\dots,\mathbf{e}_n \}\text{.}\)

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Activity 2.3.7.

In \(\R^3\text{,}\) consider the two vectors

\begin{equation*} \mathbf{v}_1 = \begin{bmatrix} 2 \\1 \\ 1 \end{bmatrix} \quad\text{and}\quad \mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \\ 3 \end{bmatrix}\text{.} \end{equation*}

Determine if the vector

\begin{equation*} \mathbf{w} = \begin{bmatrix} 1 \\ 5 \\ -7 \end{bmatrix} \end{equation*}

belongs to \(\spans \{ \mathbf{v}_1, \mathbf{v}_2 \}\text{.}\)

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Solution.

We need to determine if \(\mathbf{w}\) is a linear combination of the vectors \(\mathbf{v}_1\) and \(\mathbf{v}_2\text{,}\) i.e., if it is possible to find scalars \(a\) and \(b\) such that \(\mathbf{w} = a \mathbf{v}_1 + b \mathbf{v}_2\text{.}\) Comparing entries of the vectors in each side of the equation, we obtain a system of equations in \(a\) and \(b\text{,}\) i.e.,

\begin{align*} 2a + b \amp = 1\\ a - b \amp = 5\\ a + 3b \amp = -7 \end{align*}

It now suffices to determine if this system has a solution. We could use the methods we developed in Unit 2 to solve this system, but since there are only two variables we can get by more elementary methods.

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The only values of \(a\) and \(b\) that satisfy the first pair of equations are \(a = 2\) and \(b = -3\text{.}\) And these values also satisfy the third equation. Thus \(\mathbf{w} = 2 \mathbf{v}_1 - 3 \mathbf{v}_2\text{,}\) and so \(\mathbf{w}\) does belong to the span of \(\mathbf{v}_1\) and \(\mathbf{v}_2\text{.}\)

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Activity 2.3.8.

Find a spanning set for the null space of the matrix

\begin{equation*} A = \begin{bmatrix} 1 \amp -2 \amp -4 \amp 3 \\ 0 \amp -1 \amp -3 \amp 2 \\ 2 \amp 1 \amp 7 \amp -4 \end{bmatrix}\text{.} \end{equation*}

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Solution.

To solve this problem, we begin by finding the basic solutions to the linear system \(A \mathbf{x} = 0\text{,}\) where

\begin{equation*} \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}\text{.} \end{equation*}

Using the Gaussian elimination algorithm, we row reduce:

\begin{align*} \amp \begin{bmatrix} 1 \amp -2 \amp -4 \amp 3 \\ 0 \amp -1 \amp -3 \amp 2 \\ 2 \amp 1 \amp 7 \amp -4 \end{bmatrix}\\ \amp \rightarrow \begin{bmatrix} 1 \amp -2 \amp -4 \amp 3 \\ 0 \amp -1 \amp -3 \amp 2 \\ 0 \amp 5 \amp 15 \amp -10 \end{bmatrix}\\ \amp \rightarrow \begin{bmatrix} 1 \amp -2 \amp -4 \amp 3 \\ 0 \amp 1 \amp 3 \amp -2 \\ 0 \amp 0 \amp 0 \amp 0 \end{bmatrix}\\ \amp \rightarrow \begin{bmatrix} 1 \amp 0 \amp 2 \amp -1 \\ 0 \amp 1 \amp 3 \amp -2 \\ 0 \amp 0 \amp 0 \amp 0 \end{bmatrix} \end{align*}

The free variables are \(x_3\) and \(x_4\text{.}\) Setting \(x_3 = s\) and \(x_4 = t\text{,}\) we obtain that \(A\mathbf{x} = 0\) when \(x_1 = -2s + t\) and \(x_2 = -3s + 2t\text{.}\) Thus a general solution to this equation is given by

\begin{equation*} \begin{bmatrix} -2s + t \\ -3s + 2t \\ s \\ t \end{bmatrix} = s \begin{bmatrix} -2 \\ -3 \\ 1 \\ 0 \end{bmatrix} + t \begin{bmatrix} 1 \\ 2 \\ 0 \\ 1 \end{bmatrix}\text{.} \end{equation*}

Thus the null space is spanned by

\begin{equation*} \mathbf{v}_1 = \begin{bmatrix} -2 \\ -3 \\ 1 \\ 0 \end{bmatrix} \quad\text{and}\quad \mathbf{v}_2 = \begin{bmatrix} 1 \\ 2 \\ 0 \\ 1 \end{bmatrix} \text{.} \end{equation*}

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Activity 2.3.9. A spanning set for an image space.

Find a spanning set for the image space of the matrix

\begin{equation*} A = \begin{bmatrix} 1 \amp -2 \amp -4 \amp 3 \\ 0 \amp -1 \amp -3 \amp 2 \\ 2 \amp 1 \amp 7 \amp -4 \end{bmatrix} \text{.} \end{equation*}

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Solution.

Let

\begin{equation*} \mathbf{x} = \begin{bmatrix} t_1\\ t_2\\ t_3\\ t_4 \end{bmatrix}. \end{equation*}

Then

\begin{equation*} A\mathbf{x} = t_1 \begin{bmatrix} 1\\ 0\\ 2 \end{bmatrix} + t_2 \begin{bmatrix} -2\\ -1\\ 1 \end{bmatrix} + t_3 \begin{bmatrix} -4\\ -3\\ 7 \end{bmatrix} + t_4 \begin{bmatrix} 3\\ 2\\ -4 \end{bmatrix}. \end{equation*}

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Therefore the image space is spanned by the columns of \(A\text{:}\)

\begin{equation*} \operatorname{im}(A) = \operatorname{span} \left\{ \begin{bmatrix} 1\\ 0\\ 2 \end{bmatrix}, \begin{bmatrix} -2\\ -1\\ 1 \end{bmatrix}, \begin{bmatrix} -4\\ -3\\ 7 \end{bmatrix}, \begin{bmatrix} 3\\ 2\\ -4 \end{bmatrix} \right\}. \end{equation*}

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This is a spanning set. It is not claiming to be minimal.

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To summarize: basic solutions of \(A\mathbf{x}=\mathbf{0}\) form a spanning set for \(\operatorname{null}(A)\text{,}\) and the columns of \(A\) form a spanning set for \(\operatorname{im}(A)\text{.}\)

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The next section studies when a spanning set has redundant vectors and how to remove them.

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