(I) Vector Space
Definition
A vector space is a nonempty set \(V\) of objects, called vectors, on which are defined two operations, called addition and multiplication by scalars (real numbers), subject to the ten axioms/rules listed below.
The rules hold for all \(\vec{u}\), \(\vec{v}\), and \(\vec{w}\) in \(V\) and for all scalars \(c\) and \(d\).
- The sum \(\vec{u} + \vec{v}\) is in \(V\). (Closed under addition)
- \(\vec{u} + \vec{v} = \vec{v} + \vec{u}\)
- \((\vec{u} + \vec{v}) + \vec{w} = \vec{u} + (\vec{v} + \vec{w})\)
- There is a zero vector \(\vec{0}\) in \(V\) such that \(\vec{u} + \vec{0} = \vec{u}\). (The \(\vec{0}\) is unique)
- For each \(\vec{u}\) in \(V\), there is a vector \(-\vec{u}\) in \(V\) such that \(\vec{u} + (-\vec{u}) = \vec{0}\). (\(-\vec{u}\) is unique)
- The scalar multiple of \(\vec{u}\) by \(c\), denoted by \(c\vec{u}\), is in \(V\). (Closed under scalar multiplication)
- \(c(\vec{u} + \vec{v}) = c\vec{u} + c\vec{v}\)
- \((c + d)\vec{u} = c\vec{u} + d\vec{u}\)
- \(c(d\vec{u}) = (cd)\vec{u}\)
- \(1 \cdot \vec{u} = \vec{u}\)
Note: Only Axioms 1, 4, and 6 need to be explicitly checked when determining if a subset is a subspace.
Also, for each \(\vec{u}\) in \(V\) and scalar \(c\), the following properties hold:
- \(0 \cdot \vec{u} = \vec{0}\)
- \(c\vec{0} = \vec{0}\)
- \(-\vec{u} = (-1)\vec{u}\)
The space \(\mathbb{R}^n\), where \(n \geq 1\), are standard examples of vector spaces.
Definition (Polynomials of degree \(\leq n\))
For \(n \geq 0\), the set \(\mathbb{P}_n\) of polynomials of degree at most \(n\) consists of all polynomials of the form:
where \(a_0, a_1, \dots, a_n\) and the variable \(t\) are real numbers.
The degree of \(p(t)\) is the highest power of \(t\) whose coefficient is not \(0\). The zero polynomial is a polynomial where all coefficients are \(0\).
Show that \(\mathbb{P}_n\) is a vector space.
Solution: Suppose \(p(t) = a_0 + a_1t + \dots + a_nt^n\) and \(q(t) = b_0 + b_1t + \dots + b_nt^n\).
- Sum: \(p(t) + q(t) = (a_0 + b_0) + (a_1 + b_1)t + \dots + (a_n + b_n)t^n\), which is in \(\mathbb{P}_n\).
- Scalar Multiplication: \((cp)(t) = ca_0 + ca_1t + \dots + ca_nt^n\), which is in \(\mathbb{P}_n\).
Other axioms can be easily verified too. Thus, \(\mathbb{P}_n\) is a vector space.
(II) Subspaces
In many problems, a vector space consists of an appropriate subset of vectors from some larger vector space. In this case, only Axioms 1, 4, and 6 need to be checked; the rest are automatically true.
Definition
A subspace of a vector space \(V\) is a subset \(H\) of \(V\) that has 3 properties:
- The zero vector of \(V\) is in \(H\).
- \(H\) is closed under vector addition. That is, for each \(\vec{u}\) and \(\vec{v}\) in \(H\), the sum \(\vec{u} + \vec{v}\) is in \(H\).
- \(H\) is closed under multiplication by scalars. That is, for each \(\vec{u}\) in \(H\) and each scalar \(c\), the vector \(c\vec{u}\) is in \(H\).
Notes: Axioms 2, 3, and 7-10 are automatically true in \(H\) since they apply to all elements of \(V\), including those in \(H\). For Axiom 5, for any \(\vec{u}\) in \(H\), \(-\vec{u} = (-1)\vec{u}\) is in \(H\) by Property 3 above.
The zero subspace is the set containing only the zero vector in a vector space \(V\).
Let \(\mathbb{P}\) be the set of all polynomials with real coefficients. \(\mathbb{P}_n\) is a subspace of \(\mathbb{P}\).
Let \(V\) be the first quadrant in the \(xy\)-plane; that is, \( V = \left\{ \begin{bmatrix} x \\ y \end{bmatrix} : x \geq 0 \text{ and } y \geq 0 \right\} \). Is \(V\) a subspace of \(\mathbb{R}^2\)?
Solution:
- The origin \(\begin{bmatrix} 0 \\ 0 \end{bmatrix}\) is in \(V\).
- If \(\vec{u}\) and \(\vec{v}\) are in \(V\), then \(\vec{u} + \vec{v}\) is also in \(V\).
- However, if \(\vec{u}\) is in \(V\) and we multiply by a scalar \(c = -1\), \(-\vec{u} = (-1)\vec{u}\) is not in \(V\) (it maps to the third quadrant).
\(V\) is not closed under scalar multiplication. Therefore, \(V\) is not a subspace of \(\mathbb{R}^2\).
Is \(\mathbb{R}^2\) a subspace of \(\mathbb{R}^3\)?
Solution: The vector space \(\mathbb{R}^2\) is not a subspace of \(\mathbb{R}^3\) because \(\mathbb{R}^2\) is not even a subset of \(\mathbb{R}^3\). A vector in \(\mathbb{R}^2\) has two entries, while a vector in \(\mathbb{R}^3\) has three.
However, the set \( H = \left\{ \begin{bmatrix} s \\ t \\ 0 \end{bmatrix} : s \text{ and } t \text{ are real} \right\} \) is a subspace of \(\mathbb{R}^3\). \(H\) acts like \(\mathbb{R}^2\) but is logically distinct from \(\mathbb{R}^2\).
(III) A Subspace Spanned by a Set
The set \(\text{Span}\{\vec{v}_1, \dots, \vec{v}_p\}\), the subspace spanned/generated by \(\{\vec{v}_1, \dots, \vec{v}_p\}\), is the set of all vectors that can be written as linear combinations of \(\vec{v}_1, \dots, \vec{v}_p\).
\(\text{Span}\{\vec{v}_1, \dots, \vec{v}_p\}\) represents an infinite number of vectors generated by \(\{\vec{v}_1, \dots, \vec{v}_p\}\) (which is a set of \(p\) vectors).
Given any subspace \(H\) of \(V\), a spanning (or generating) set for \(H\) is a set \(\{\vec{v}_1, \dots, \vec{v}_p\}\) in \(H\) such that \(H = \text{Span}\{\vec{v}_1, \dots, \vec{v}_p\}\).
Theorem 1
If \(\vec{v}_1, \dots, \vec{v}_p\) are in a vector space \(V\), then \(\text{Span}\{\vec{v}_1, \dots, \vec{v}_p\}\) is a subspace of \(V\).
(This can be proved directly using the definition of a subspace.)
Let \(\mathbb{P}_3\) be the space of all polynomials of degree at most 3. Which of the following sets are subspaces of \(\mathbb{P}_3\)?
(1) The set of all polynomials \(p(t)\) in \(\mathbb{P}_3\) such that \(p(2) = 0\).
Solution: Let \(p(t) = a_0 + a_1t + a_2t^2 + a_3t^3\). Since \(p(2) = 0\):
\[ p(2) = a_0 + 2a_1 + 4a_2 + 8a_3 = 0 \] \[ a_0 = -2a_1 - 4a_2 - 8a_3 \]Substitute \(a_0\) back into the general form:
\[ p(t) = (-2a_1 - 4a_2 - 8a_3) + a_1t + a_2t^2 + a_3t^3 \] \[ p(t) = a_1(t - 2) + a_2(t^2 - 4) + a_3(t^3 - 8) \]This set is exactly \(\text{Span}\{t - 2, t^2 - 4, t^3 - 8\}\), and by Theorem 1, it is a subspace of \(\mathbb{P}_3\).
(Note: If the condition was changed to \(p(2) = 1\), it would not be a subspace since it wouldn't contain the zero polynomial.)
(2) All polynomials of the form \(p(t) = a + t^2\), where \(a \in \mathbb{R}\).
Solution: It is not a subspace of \(\mathbb{P}_3\) since the zero polynomial is not included (the \(t^2\) term is always present).
(3) All polynomials of the form \(p(t) = a + bt + abt^2\), where \(a,b \in \mathbb{R}\).
Solution: This set is not a subspace since it is not closed under vector addition. For example, let \(a=1, b=0 \implies p_1(t) = 1\). Let \(a=0, b=1 \implies p_2(t) = t\). Both \(1\) and \(t\) are in the set. Their sum is \(1 + t\). But for \(1 + t\) to be in the set (where \(a=1\) and \(b=1\)), its \(t^2\) term must be \(ab t^2 = (1)(1)t^2 = 1t^2\). Since the sum \(1 + t\) lacks the \(t^2\) term, it is not in the set.
Let \(H = \{(a - 3b, b - a, a, b) : a, b \in \mathbb{R}\}\). Show that \(H\) is a subspace of \(\mathbb{R}^4\).
Solution: Write the general vector in \(H\) as a linear combination:
\[ \begin{bmatrix} a - 3b \\ b - a \\ a \\ b \end{bmatrix} = \begin{bmatrix} a \\ -a \\ a \\ 0 \end{bmatrix} + \begin{bmatrix} -3b \\ b \\ 0 \\ b \end{bmatrix} = a \begin{bmatrix} 1 \\ -1 \\ 1 \\ 0 \end{bmatrix} + b \begin{bmatrix} -3 \\ 1 \\ 0 \\ 1 \end{bmatrix} \]Thus, \( H = \text{Span}\left\{ \begin{bmatrix} 1 \\ -1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -3 \\ 1 \\ 0 \\ 1 \end{bmatrix} \right\} \).
By Theorem 1, \(H\) is a subspace of \(\mathbb{R}^4\).