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Linear Algebra Gathering and Sharpening Tools

Ryan Higginbottom

Section 2.4 Subspaces

Whenever we begin to deal with abstraction in mathematics, we also consider the relationship between these abstract concepts. We have already made it clear how the notions of fields and vector spaces are related—a vector space requires that we know what a field is! But mathematicians also consider how these abstract objects might fit inside or contain each other.
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The most basic example of this is something with which all readers are likely familiar. When we learn about sets, we soon also learn about subsets. For two sets \(A\) and \(B\text{,}\) we say that \(A\) is a subset of \(B\) when all of the elements of \(A\) are contained in \(B\text{.}\) We denote this by \(A \subseteq B\text{.}\)
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A vector space is a set with a lot more structure. So a “sub-vector space” must be a subset with the properties of a vector space. Here is the formal definition.
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Definition 2.4.1.

Suppose that \(V\) is a vector space over the field \(\ff\text{.}\) Then a subset \(W \subseteq V\) is a subspace of \(V\) if \(W\) is also a vector space over \(\ff\) with the same operations of vector addition and scalar multiplication that are used for \(V\text{.}\)
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To put this less formally: A subspace is a subset which is a vector space over the same field with the operations inherited from the larger space.
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Since the operations of a subspace are inherited, we need not check the full list of properties (from Definition 2.3.1) to prove that a subset is a subspace. In fact, when we consult that definition, we see that axioms 3, 4, and 7–10 will automatically be satisfied—these are properties of the operations, not of the set on which the operation is taking place. This leads to the following fact.
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From this fact, there appear to be four conditions to check in order to prove that a subset is a subspace. However, we will refer the reader back to Theorem 2.3.12. By part 6 of that theorem, we know that if a subset of a vector space is closed under scalar multiplication it must also contain all additive inverses. This means that there are only three conditions to check to prove that a subset is a subspace. We summarize this as a theorem.
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As always, some examples are in order!
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Example 2.4.4.

Every vector space \(V\) has a subspace consisting of a single vector—the zero subspace: \(\{\mathbf{0}\}\text{.}\) In a trivial way, this set meets all of the conditions listed in Theorem 2.4.3.
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Example 2.4.5.

Let \(V\) be the vector space of all functions \(\rr \to \rr\) as defined in Example 2.3.5. Let \(W\) be the subset of all polynomials. (This means we include polynomials of any degree.) Then \(W\) is a subspace of \(V\text{.}\)
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Since the sum of two polynomials is another polynomial and the scalar multiple of a polynomial is a polynomial, the first two conditions in Theorem 2.4.3 hold. Additionally, the zero function in \(V\) is the same as the zero polynomial in \(W\text{.}\) This proves that \(W\) is a subspace of \(V\text{.}\)
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Example 2.4.6.

The vector space \(\rr^2\) is not a subspace of the vector space \(\rr^3\text{.}\) This may seem surprising, as the operations for these spaces are clearly compatible, and we often think of \(\rr^2\) as “living inside” of \(\rr^3\text{.}\)
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However, this common way of thinking is wrong because the space \(\rr^2\) is not even a subset of \(\rr^3\text{;}\) thus, it is impossible for \(\rr^2\) to be a subspace of \(\rr^3\text{.}\) (The set \(\rr^2\) consists of ordered pairs of real numbers while \(\rr^3\) consists of ordered triples of real numbers.)
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We can make a slight adjustment to match the way that many people think of \(\rr^2\) as “living inside” of \(\rr^3\text{.}\) We define the set \(A\) in the following way:
\begin{equation*} A = \{(x,y,0) \mid x,y \in \rr \}\text{.} \end{equation*}
Then \(A\) is what we usually call the “\(xy\)-plane” in \(\rr^3\text{.}\) This subset \(A\) is a subspace of \(\rr^3\text{.}\)
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Example 2.4.7.

Here is a collection of examples that generalize Example 2.4.6. Every line through the origin in \(\rr^2\) is a subspace of \(\rr^2\text{,}\) and every line or plane through the origin in \(\rr^3\) is a subspace of \(\rr^3\text{.}\) On the other hand, lines in \(\rr^2\) which do not pass through the origin are not subspaces of \(\rr^2\text{,}\) and lines and planes which do not pass through the origin in \(\rr^3\) are not subspaces of \(\rr^3\text{.}\) The details are left for the exercises.
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The notion of the span of a set of vectors gives us another angle through which we can identify subspaces. We begin with a result.
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Proof.

We let \(W = \spn\{\bfv_1,\ldots,\bfv_n\}\text{.}\) By the definition of the span of a set of vectors, every element \(\bfw\) of \(W\) can be written in the following form:
\begin{equation*} \bfw = c_1\bfv_1 + \cdots + c_n\bfv_n\text{,} \end{equation*}
where \(c_1,\ldots,c_n \in \ff\text{.}\) We first observe that \(\mathbf{0} \in W\) by taking \(c_1= \cdots = c_n = 0\text{.}\) Next, if \(\bfu\) and \(\bfw\) are elements of \(W\text{,}\) we can write these vectors as
\begin{align*} \bfu \amp = c_1\bfv_1 + \cdots + c_n\bfv_n\\ \bfw \amp = d_1\bfv_1 + \cdots + d_n\bfv_n\text{,} \end{align*}
for some scalars \(c_i\) and \(d_i\text{.}\) The sum of these elements is
\begin{align*} \bfu + \bfw \amp = (c_1\bfv_1 + \cdots + c_n\bfv_n) + (d_1\bfv_1 + \cdots + d_n\bfv_n)\\ \amp = (c_1+d_1)\bfv_1 + \cdots + (c_n+d_n)\bfv_n\text{.} \end{align*}
(We are using some of the properties of a vector space from Definition 2.3.1 in order to carry out this algebraic manipulation.) Since \(c_i+d_i \in \ff\) for each \(i\text{,}\) this proves that \(\bfu+\bfw \in W\text{.}\)
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Finally, we let \(c \in \ff\) and \(\bfw \in W\text{.}\) We want to show that \(c\bfw \in W\text{.}\) We can assume that \(\bfw\) has the form
\begin{equation*} \bfw = d_1\bfv_1 + \cdots + d_n\bfv_n\text{,} \end{equation*}
where \(d_1,\ldots,d_n \in \ff\text{.}\) Then we have
\begin{align*} c\bfw \amp = c(d_1\bfv_1 + \cdots + d_n\bfv_n)\\ \amp = c(d_1\bfv_1) + \cdots + (cd_n\bfv_n)\\ \amp = (cd_1)\bfv_1 + \cdots + (cd_n)\bfv_n\text{.} \end{align*}
(Again, we are using properties of the vector space \(V\) here.) Since \(cd_i \in \ff\) for each \(i\text{,}\) this proves that \(W\) is closed under scalar multiplication.
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Since \(W\) has all of the properties from Theorem 2.4.3, we have shown that \(W\) is a subspace of \(V\text{.}\)
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In the final example of this section, we will use Theorem 2.4.8 to prove that a set is a subspace by realizing it as the span of a set of vectors.
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Example 2.4.9.

Consider the following subset of \(\rr^3\text{:}\)
\begin{equation*} A = \left\{ \left[\begin{array}{@{}c@{}} 2a+b \\ 5b-\tfrac{1}{2}a \\ 6a \end{array}\right] \;\middle\vert\; a,b \in \rr \right \}\text{.} \end{equation*}
This notation means that every given pair of real numbers \(a\) and \(b\) specifies an element of \(A\text{.}\) For example, when \(a=1\) and \(b=-1\text{,}\) we have \((1, -\frac{11}{2}, 6) \in A\text{.}\)
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We will use Theorem 2.4.8 to prove that \(A\) is a subspace of \(\rr^3\text{.}\) Any element \(\bfv \in A\) can be written in the following way:
\begin{equation*} \bfv = \left[\begin{array}{@{}c@{}} 2a+b \\ 5b-\tfrac{1}{2}a \\ 6a \end{array}\right] = a \left[\begin{array}{@{}c@{}} 2 \\ -\tfrac{1}{2} \\ 6 \end{array}\right] + b \left[\begin{array}{@{}c@{}} 1 \\ 5 \\ 0 \end{array}\right]\text{.} \end{equation*}
This proves that if
\begin{equation*} \bfu_1 = \left[\begin{array}{@{}c@{}} 2 \\ -\tfrac{1}{2} \\ 6 \end{array}\right] \hspace{6pt} \text{and} \hspace{6pt} \bfu_2 = \left[\begin{array}{@{}c@{}} 1 \\ 5 \\ 0 \end{array}\right]\text{,} \end{equation*}
then \(A = \spn\{\bfu_1,\bfu_2\}\text{.}\) We conclude that \(A\) is a subspace of \(\rr^3\) by Theorem 2.4.8.
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Reading Questions Reading Questions

1.

Let \(V\) be the vector space \(\rr^2\) with the usual addition and scalar multiplication. Let \(A\) be the following set:
\begin{equation*} A = \{(x,y) \in \rr^2 \mid x \ge 0 \text{ and } y \ge 0 \}\text{.} \end{equation*}
(We can identify \(A\) with the first quadrant in \(\rr^2\text{.}\)) Determine whether or not \(A\) is a subspace of \(V\text{.}\) Explain your answer.
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2.

Consider the vector space \(P_4\) and let \(A\) be the following set:
\begin{equation*} A = \{ \text{polynomials in } P_4 \text{ with even degree} \}\text{.} \end{equation*}
Determine whether or not \(A\) is a subspace of \(P_4\text{.}\) Explain your answer.
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3.

Let \(V\) be the vector space of all functions \(\rr \to \rr\text{.}\) Let \(f\) and \(g\) be the following functions:
\begin{equation*} f(t) = 2t^2 \hspace{.4in} g(t) = \cos(3t)\text{.} \end{equation*}
Let \(A = \spn\{f, g\}\text{.}\) We know that \(A\) is a subspace of \(V\) by Theorem 2.4.8. Write down four distinct vectors in \(A\text{.}\)
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Exercises Exercises

1.

For each of the following, determine whether or not the subset \(W\) of \(\rr^3\) is a subspace of \(\rr^3\text{.}\) Explain your answers. (We are writing vectors horizontally for convenience.)
  1. \(\displaystyle W = \{(a,0,0) \mid a \in \rr\}\)
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  2. \(\displaystyle W = \{(a,b,c) \mid b = a+c\}\)
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  3. \(\displaystyle W = \{(a,1,c) \mid a,c \in \rr \}\)
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  4. \(\displaystyle W = \{(a,0,c) \mid a,c \in \rr \}\)
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2.

For each of the following, determine whether or not the subset \(W\) of \(P_3\) is a subspace of \(P_3\text{.}\) Explain your answers.
  1. \(\displaystyle W = \{ at^3 + at \mid a \in \rr \}\)
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  2. \(\displaystyle W = \{ at^2 + bt + c \mid a,b,c \in \rr \}\)
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  3. \(\displaystyle W = \{ p \in P_3 \mid p \text{ has rational coefficients} \}\)
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  4. \(\displaystyle W = \{ p \in P_3 \mid p(0)=0 \}\)
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  5. \(\displaystyle W = \{ a_3t^3 + a_2t^2 + a_1t + a_0 \mid a_3+a_2+a_1+a_0 = 0 \}\)
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Answer.
  1. Yes, \(W\) is a subspace.
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  2. Yes, \(W\) is a subspace.
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  3. No, \(W\) is not a subspace. It is not closed under scalar multiplication.
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  4. Yes, \(W\) is a subspace.
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  5. Yes, \(W\) is a subspace.
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3.

Let \(C[0,10]\) be the vector space of continuous functions defined \([0,10] \to \rr\text{,}\) and let \(D[0,10]\) be the set of differentiable functions defined \([0,10] \to \rr\text{.}\) (The fact that \(C[0,10]\) is a vector space is established in Exercise 2.4.5.)
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For each of the following, determine whether or not the subset \(W\) of \(C[0,10]\) is a subspace of \(C[0,10]\text{.}\) Explain your answers.
  1. \(\displaystyle W = \{ f \in C[0,10] \mid f(1)=0 \}\)
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  2. \(\displaystyle W = \{ f \in C[0,10] \mid f(1)=1 \}\)
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  3. \(\displaystyle W = \{ f \in D[0,10] \mid f'(1)=0 \}\)
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  4. \(\displaystyle W = \{ f \in D[0,10] \mid f'(1)=1 \}\)
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  5. \(\displaystyle W = \{ f \in D[0,10] \mid f'(t) \text{ is constant} \}\)
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4.

For each of the following, determine whether or not the subset \(W\) of \(\rr^3\) is a subspace of \(\rr^3\text{.}\) Explain your answers. (We are writing vectors horizontally for convenience.)
  1. \(\displaystyle W = \{ (2a-b, a+5b, -a) \mid a,b \in \rr \}\)
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  2. \(\displaystyle W = \{ (a+b, 2b, ab) \mid a,b \in \rr \}\)
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  3. \(\displaystyle W = \{ (a+b, b-c, a+2d) \mid a,b,c,d \in \rr \}\)
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  4. \(\displaystyle W = \{ (2a-4b, 5, 3c+b) \mid a,b,c \in \rr \}\)
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Answer.
  1. Yes, \(W\) is a subspace as it can be written as the span of two vectors.
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  2. No, \(W\) is not a subspace. It is not closed under scalar multiplication, for example.
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  3. Yes, \(W\) is a subspace as it can be written as the span of four vectors.
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  4. No, \(W\) is not a subspace as it does not contain the zero vector.
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Writing Exercises

5.
Let \(V\) denote the vector space of all functions \(\rr \to \rr\text{.}\) (See Example 2.3.5 for the relevant definitions and operations.) Let \(a\) and \(b\) be real numbers with \(a < b\text{,}\) and let \(C[a,b]\) denote the set of all continuous functions \([a,b] \to \rr\text{.}\) Prove that \(C[a,b]\) is a subspace of \(V\text{.}\)
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6.
Let \(a\) and \(b\) be real numbers with \(a < b\text{.}\) Show that the set of all functions \(f \in C[a,b]\) such that
\begin{equation*} \int_a^b f(x)\; dx = 0 \end{equation*}
is a subspace of \(C[a,b]\text{.}\) (See Exercise 2.4.5 for the definition of \(C[a,b]\text{.}\))
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Answer.
Let \(V\) denote the set of functions described in the statement of the problem. We first observe that the zero function is in \(V\) since it integrates to zero over any interval \([a,b]\text{.}\) If \(f, g \in V\text{,}\) then
\begin{align*} \int_a^b (f+g)(x)\;dx \amp = \int_a^b (f(x) + g(x))\;dx\\ \amp = \int_a^b f(x)\;dx + \int_a^b g(x)\;dx = 0 + 0 = 0\text{,} \end{align*}
which proves that \(f + g \in V\text{.}\) Finally, if \(c \in \rr\) and \(f \in V\text{,}\) we have
\begin{equation*} \int_a^b (cf)(x)\;dx = \int_a^b cf(x)\;dx = c\int_a^b f(x)\;dx = c \cdot 0 = 0\text{,} \end{equation*}
which shows that \(cf \in V\text{.}\) (These last two calculations rely on the linear properties of the definite integral.) This completes the proof that \(V\) is a subspace.
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7.
Prove that a line in \(\rr^2\) is a subspace of \(\rr^2\) if and only if the line passes through the origin. (Note that all lines in \(\rr^2\) can be written in the form \(ax+by+c=0\text{.}\))
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8.
Prove that a plane in \(\rr^3\) is a subspace of \(\rr^3\) if and only if the plane passes through the origin. (Note that all planes in \(\rr^3\) can be written in the form \(ax+by+cz+d=0\text{.}\))
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9.
Let \(V\) be a vector space and let \(W_1\) and \(W_2\) be subspaces of \(V\text{.}\) Must \(W_1 \cup W_2\) be a subspace of \(V\text{?}\) Justify your answer.
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Answer.
No. Let \(W_1\) and \(W_2\) be the following two subspaces of \(\rr^2\text{:}\)
\begin{equation*} W_1 = \{(x,2x) \mid x \in \rr\}, \hspace{12pt} W_2 = \{(x,3x) \mid x \in \rr\}\text{.} \end{equation*}
Then \((1,2) \in W_1\) and \((1,3) \in W_2\text{,}\) so both of these elements are in \(W_1 \cup W_2\text{.}\) However, \((1,2) + (1,3) = (2,5)\text{,}\) and \((2,5) \not \in W_1 \cup W_2\text{.}\) This shows that \(W_1 \cup W_2\) is not closed under addition, so it is not a subspace.
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10.
Let \(V\) be a vector space and let \(W_1\) and \(W_2\) be subspaces of \(V\text{.}\) Must \(W_1 \cap W_2\) be a subspace of \(V\text{?}\) Justify your answer.
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11.
Let \(V\) be a vector space and let \(W_1\) and \(W_2\) be subspaces of \(V\text{.}\) Define the sum of \(W_1\) and \(W_2\) like this:
\begin{equation*} W_1 + W_2 = \{ \bfv \mid \bfv = \bfw_1 + \bfw_2 \text{ for some } \bfw_1 \in W_1 \text{ and some } \bfw_2 \in W_2 \}\text{.} \end{equation*}
Must \(W_1 + W_2\) be a subspace of \(V\text{?}\) Justify your answer.
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