4.1. Subspaces of \(\R^n\) — Linear Algebra (2024)

A (linear) subspace of \(\R^n\) is a subset \(S\) of \(\R^n\) with the following three properties:

1. \(S\) contains the zero vector.

2. If two vectors \(\vect{u}\) and \(\vect{v}\) are in \(S\), then their sum is in \(S\) too:

   
   

   \[\vect{u} \in S, \vect{v} \in S \quad \Longrightarrow \quad\vect{u}+ \vect{v} \in S.\]

3. If a vector \(\vect{u}\) is in \(S\), then every scalar multiple of \(\vect{u}\) is in \(S\) too:

   
   

   \[\vect{u} \in S, c \in \R \quad \Longrightarrow \quadc\vect{u} \in S.\]

Property (ii) is also expressed as: a subspace is closed under sums. Likewise property (iii) says that a subspace is closed under taking scalar multiples.

The set in \(\R^n\) that consists of only the zero vector, i.e. \(S = \{\vect{0}\}\), is a subspace.

We will check that it has the three properties mentioned in the definition:

1. \(S\) certainly contains the zero vector.

2. If two vectors \(\vect{u}\) and \(\vect{v}\) are in \(S\), then their sum is in \(S\) too:

   
   

   \[\vect{u} \in S, \vect{v} \in S \,\,\Longrightarrow \,\, \vect{u} = \vect{v} = \vect{0} \,\,\Longrightarrow \,\, \vect{u} + \vect{v} = \vect{0} + \vect{0} =\vect{0} \in S.\]

3. If a vector \(\vect{u}\) is in \(S\), then every scalar multiple of \(\vect{u}\) is in \(S\) too:

   
   

   \[\vect{u} \in S\quad\Longrightarrow \quad \vect{u} = \vect{0} \quad\Longrightarrow \quad c\vect{u} = c\vect{0} = \vect{0} \in S.\]

The trivial subspaces of \(\R^n\) are the set \(\{\vect{0}\}\) and the set \(\R^n\) itself.

In \(\R^2\), a line through the origin is a non-trivial subspace. A line not containing the origin is not.In fact, the latter does not satisfy any of the three properties of a subspace, as may be clear from Figure 4.1.1.In the picture on the right, for two vectors \(\vect{u}\) and \(\vect{v}\) on the line \(\mathcal L\),

\[\vect{u}+\vect{v} \, \text{ and } \, -\tfrac32\vect{u} \, \text{ do not lie on } \,{\mathcal L}\]

Examples of subspaces in \(\R^3\) are lines and planes through the origin.Try to visualize that these sets do satisfy the properties of a subspace. A sketch may help.It is good practice to keep these examples in mind as typical examples of subspaces.

A disk \(D\) specified by the inequality \(x^2 + y^2 \leq a^2\), where \(a\) is some positive number, is not a subspace of \(\R^2\). It has neither of the properties(ii) and (iii). See Figure 4.1.2.

1. Give an example of a subset in \(\R^2\) that has property i and ii, but not property iii.

2. Also give a set with only the properties i and iii.

A non-empty subset \(S\) of \(\R^n\) is a subspace if and only if

(4.1.1)#\[\text{for all } \vect{u}, \vect{v} \in S, c_1, c_2 \in \R\,\,\text{ it holds that }\,\,c_1\vect{u}+ c_2 \vect{v} \in S.\]

By repeatedly applying the last proposition, we find for any subspace \(S\)

\[\vect{u}_1, \ldots , \vect{u}_k \in S, c_1, \ldots , c_k \in \R\quad \Longrightarrow \quadc_1\vect{u}_1+ \ldots + c_k\vect{u}_k \in S.\]

So we can more generally say that a subspace is closed under taking linear combinations.
This also means that if \(\vect{u}_1, \ldots , \vect{u}_k \) are vectors in a subspace \(S\),
then \(\Span{\vect{u}_1, \ldots , \vect{u}_k} \) is contained in \(S\).

If \(\vect{v}_1,\vect{v}_2, \ldots , \vect{v}_r\) are vectors in \(\R^n\), then

\[\Span{\vect{v}_1,\vect{v}_2, \ldots , \vect{v}_r} \quad \text{is a subspace in } \R^n.\]

In this situation the vectors are said to generate the subspace, or to be a set of generators for the subspace.Recall Definition 2.2.2: the span of zero vectors in \(\R^n\) (in other words, the span of the empty set) is defined to be the set \(\{\vect{0}\}\).

Give a proof of property (iii).

In the previous proposition we do not impose any restrictions on the set of vectors\(\{ \vect{v}_1,\vect{v}_2, \ldots , \vect{v}_r \}\). In the sequel we will see that it will be advantageous to have a linear independent set of generators.

Each subspace \(S\) in \(\R^3\) has one of the following forms:

In other words

\[S = \text{Span}\{\vect{v}_i\, |\,\ i = 1,\ldots, r\} \quad \text{where }\, r = 0, 1, 2 \text{ or } 3,\]

and we may assume that the vectors \(\vect{v}_i\) are linearly independent.

Once more we recall the convention that the span of zero vectors (i.e., when \(r = 0\)) is the set only containing the zero vector.

Every subspace of \(\R^n\) is of the form

\[S = \Span{\vect{v}_1, \ldots , \vect{v}_r} \quad \text{for some } \, r \leq n,\]

where

\[\{\vect{v}_1, \ldots , \vect{v}_r\} \,\, \text{is linearly independent.}\]

The column space of an \(m\times n\) matrix \(A= \begin{bmatrix} \vect{a}_1 & \vect{a}_2 & \ldots & \vect{a}_n \end{bmatrix}\) is the span of the columns of \(A\):

\[\Col{A} = \Span{\vect{a}_1,\vect{a}_2,\ldots,\vect{a}_n}.\]

The null space of an \(m\times n\) matrix \(A\) is the solution set of the hom*ogeneous equation \(A\vect{x} = \vect{0}\):

\[\Nul{A} = \{\vect{x} \in \mathbb{R}^n \,|\, A\vect{x} = \vect{0}\}.\]

For an \(m\times n\) matrix \(A\), Col \(A\) is the set of all vectors of the form \(A\vect{x}\), for \(\vect{x}\in\R^n\). The column spaceCol \({A}\) can also be interpreted as the range of the linear transformation \(T:\R^n \to \R^m\) defined via\(T(\vect{x}) = A\vect{x}\). (Cf. Proposition 3.1.1.)

Note that for an \(m\times n \) matrix \(A\) the column space is a subset of \(\R^m\) and the null space lives in \(\R^n\). In short,

\[\Col{A} \subseteq \R^m ,\quad \Nul{A} \subseteq \R^n.\]

Let \(A\) be an \(m\times n\) matrix.

1. The column space of \(A\) is a subspace of \(\R^m\).




2. The null space of \(A\) is a subspace of \(\R^n\).

The above proof, that the null space is a subspace, is as basic as possible. That is, we started from the definitions (of null space and subspace) and used properties of the matrix product to connect the two.

Alternatively we could have used knowledge already acquired earlier. In Section 2.3 we have seen that the solution set of a hom*ogeneoussystem

\[A\vect{x} = \vect{0}\]

can be written in parametric vector form

\[\vect{x} = c_1\vect{u}_1 + c_2\vect{u}_2 + \ldots + c_k\vect{u}_k.\]

Thus: it is the span of a set of vectors, and as suchby Proposition 4.1.2 it is a subspace.

Suppose that \(A\) and \(B\) are matrices for which the product \(AB\) is defined.

1. Show that the column space of \(AB\) is a subset of the column space of \(A\), i.e.

   \[\Col{AB} \subseteq \Col{A}.\]

2. Can you find a similar formula relating the null space of \(AB\) to the null space of either \(A\) or \(B\) (or both)?

For an \(n\times n\) matrix \(A\), the null space and the column space are both subspaces of (the same) \(\R^n\). Prove or disprove the following statement.

For a square matrix \(A\):

\[A^2 = O \quad \iff \quad \Col{(A)} \subseteq \Nul{(A)}.\]