5.2. Subspaces#

We have seen that a vector space is a set, and hence we can consider studying its subsets. Think of the Euclidean space \(\mathbb{R}^3\) (which we think of with the usual \(x,y,z\)-coordinates). This is a vector space, and consider the plane \(p = \{(x,y,z) : x + y + z = 0\} \subseteq \mathbb{R}^3\). In fact, we can view \(p\) itself as a vector space, meaning it satisfies the vector space axioms A1 – A4 and M1 – M4, where we define addition and scalar multiplication as coming from \(\mathbb{R}^3\). We saw in the chapter on co-ordinate geometry that this vector space \(p\) actually closely resembles \(\mathbb{R}^2\).

Definition 5.4 (Subspace)

A non-empty subset \(W\) of a vector space \(V\) is called a subspace (or a vector subspace) if it satisfies all the vector space axioms A1 – A4 and M1 – M4.

Most of the axioms A1 – A4 and M1 – M4 are automatically satisfied, simply because they’re satisfied in \(V\). The only things we really need to check are that addition and scalar multiplication are well-defined. In other words we need to check that when we ‘add’ or ‘multiply’ in \(W\) we are not taken outside of \(W\) (Fig. 5.1). This idea is formalised in the subspace condition.

../_images/5_subspaces.svg

Fig. 5.1 Illustration of a subspace \(W\) of a vector space \(V\).#

Theorem 5.2 (Subspace condition)

Let \(V\) be a vector space over \(F\) and \(W\) be a non-empty subset of \(V\) then \(W\) is a subspace if and only if the following condition is satisfied

(5.1)#\[ u + \alpha w \in W,\]

where \(u, w \in W\) and \(\alpha \in F\).

To demonstrate the use of the subspace condition let’s check whether \(W = \{(x, y, z)^\mathsf{T} : x + 2y - 3z = 0\}\) is a subspace of \(\mathbb{R}^3\). First we need to verify that \(W\) is non-empty. To do this we simply need to find an element of \(W\), e.g., \(\vec{0} \in W\) so \(W\) is non-empty.

Next we need to take two arbitrary vectors in \(W\), \(\vec{u} = (u_1, u_2, u_3)^\mathsf{T}, \vec{v} = (v_1, v_2, v_3)^\mathsf{T} \in W\) say and a scalar \(\alpha \in \mathbb{R}\) then

\[\begin{split} \vec{u} + \alpha \vec{v} = \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix} + \alpha \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} = \begin{pmatrix} u_1 + \alpha v_1 \\ u_2 + \alpha v_2 \\ u_3 + \alpha v_3 \end{pmatrix}. \end{split}\]

We now need to show that \((u_1 + \alpha v_1, u_2 + \alpha v_2, u_3 + \alpha v_3)^\mathsf{T} \in W\), i.e., that \(x + 2y - 3z = 0\), so since \(x = u_1 + \alpha v_1\), \(y = u_2 + \alpha v_2\) and \(z = u_3 + \alpha v_3\) then

\[\begin{split} \begin{align*} x + 2y - 3z &= (u_1 + \alpha v_1) + 2(u_2 + \alpha v_2) - 3(u_3 + \alpha v_3) \\ &= (u_1 + 2u_2 - 3u_3) + \alpha (v_1 + 2v_2 - 3v_3) \\ &= 0 + \alpha \cdot 0 = 0, \end{align*} \end{split}\]

therefore \(\vec{u} + \alpha \vec{v} \in W\) and \(W\) is a subspace of \(\mathbb{R}^3\)

Example 5.3

(i)   Let \(W\) be the subset \(M_{2\times 2}\) given by

\[\begin{split} \begin{align*} W = \left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} : a + b + c + d = 0 \right\}. \end{align*} \end{split}\]

Show that \(W\) is a subspace of \(M_{2\times 2}\).

Solution

Clearly \(\vec{0}_{2\times 2} \in W\) so \(W\) is non-empty. Let \(U, V \in W\) and \(\alpha \in \mathbb{R}\). Then

\[\begin{split} \begin{align*} U + \alpha V = \begin{pmatrix} u_{11} & u_{12} \\ u_{21} & u_{22} \end{pmatrix} + \alpha \begin{pmatrix} v_{11} & v_{12} \\ v_{21} & v_{22} \end{pmatrix} = \begin{pmatrix} u_{11} + \alpha v_{11} & u_{12} + \alpha v_{12} \\ u_{21} + \alpha v_{21} & u_{22} + \alpha v_{22} \end{pmatrix}. \end{align*} \end{split}\]

Checking that \(U + \alpha V \in W\)

\[\begin{split} \begin{align*} a + b + c + d &= (u_{11} + \alpha v_{11}) + (u_{12} + \alpha v_{12}) + (u_{21} + \alpha v_{21}) + (u_{22} + \alpha v_{22}) \\ & = (u_{11} + u_{12} + u_{21} + u_{22}) + \alpha (v_{11} + v_{12} + v_{21} + v_{22}) \\ & = 0 + \alpha \cdot 0 = 0, \end{align*} \end{split}\]

so \(U + \alpha V \in W\) therefore \(W\) is a subspace of \(M_{2\times 2}\).

(ii)   Let \(W = \{ p(x) \in P_n(\mathbb{R}) : p(x) = p(-x)\} \subseteq P(\mathbb{R}_n)\) be the set of all even functions. Show that \(W\) is a subspace of \(P_n(\mathbb{R})\).

Solution

The zero polynomial \(0 \in W\) so \(W\) is non-empty. Let \(u, v \in W\) and \(\alpha \in \mathbb{R}\), we need to show that \(u + \alpha v \in W\). Now we know that \(u(-x) = u(x)\) and \(v(-x) = v(x)\) so

\[ \begin{align*} (u + \alpha v)(x) = u(x) + \alpha v(x) = u(-x) + \alpha v(-x) = (u + \alpha v)(-x), \end{align*} \]

so \((u + \alpha v)(x) \in W\) is also an even function therefore \(W\) is a subspace.

5.2.1. Non-examples of Subspaces#

Not all subsets of a vector space will be a subspace. For example, define \(p\) as the polynomial

\[ p=a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n = \sum_{i = 0}^n a_ix^i, \]

with \(a_n\neq 0\), we say \(p\) has degree \(n\) and write \(\operatorname{deg}(p)=n\). Let

\[ W = \{p \in P_n(\mathbb{R}) : \operatorname{deg}(p) \in \{0,2,4\}\} \]

i.e, the set of all polynomials of even degree at most 4 together with the zero polynomial. Is \(W\) a subspace of \(P_4(\mathbb{R})\)?

Let \(u = x^2\) and \(v = -x ^ 2 + x\) then \(u, v \in W\) but \(u + v = x^2 - x^2 + x = x \notin W\) since \(\operatorname{deg}(x) = 1\) is odd. Therefore \(W\) is not a subspace of \(P_4(\mathbb{R})\). Here we came up with a single counter example to show that the linear combination \(u + v \notin W\).

This counterexample is great but how does one come up with such a counterexample if you can’t immediately think of one? If you do not see an obvious counterexample, try and construct one: we are looking for two even-degree polynomials, whose sum is not an even polynomial. Let’s start at the lowest degree: \(0\). The sum of two polynomials of degree \(0\) is just the sum of two real numbers, so there is no counterexample here. Let’s try degree \(2\). Now we need to find: \(u = ax^2 + bx + c\) and \(v = dx^2 + ex + f\), such that \(u + v\) has an odd degree. Now

\[ u + v = (a + d)x^2 + (b + e)x + (c + f) \]

has degree one when \((a + d) = 0\) and \((b + e) \neq 0\). So let’s take \(a = 1\), \(d = -1\) and \(b = 0\), \(e = 1\) (we don’t care about \(c\) or \(f\), so let’s just set both to be 0). This yields \(u = x^2\) and \(v = -x^2 + x\), which was our counterexample.

If we try constructing a counterexample and but fail, then it might mean that either a counterexample is not easy to find or the statement is actually true. With a lot of pure maths research, mathematicians are often in this situation, trying to find a counterexample or a proof, not knowing which one it should be.

Example 5.4

Show that the following are not subspaces:

(i)   \(W = \{z = (a+bi) : a + b = 5\} \subseteq \mathbb{C}\);

Solution

Here \(0 \notin W\) so \(W\) is not a subspace.

(ii)   \(W = \{z = (a + bi) : a^2 + b^2 < 0\} \subseteq \mathbb{C}\);

Solution

\(W\) is the empty set since \(a^2 + b^2 \geq 0\) for all \(a, b \in \mathbb{R}\). Hence \(W\) is not a subspace.

(iii)   \(W = \{(x,y,z)^\mathsf{T} : x + y + z = 5\} \subseteq{\mathbb{R}^3}\).

Solution

Let \(\vec{u} = (1, 2, 2)^\mathsf{T} \in W\) and \(\vec{v} = (2, 1, 2)^\mathsf{T} \in W\), then \(\vec{u} + \vec{v} = (3, 3, 4) \notin W\). We could’ve also just noted that \(\vec{0} \notin W\).