One exercise on subspaces

Exercise 1 Prove that set of differentiable functions on {[a, b]} and satisfying {f' + 4f = 0} that is a subspace of {C_{[a, b]}}. Find a basis of this subspace and compute it’s dimension.

Hint: Let {W} be above set. Suppose that {f, g \in W}, we have {\begin{cases} f' + 4f&=0\\ g' + 4g&=0 \end{cases}}. Hence, {h = \alpha f + \beta g} also satisfies differential equation {h' + 4h = 0}. This implies {W} is a subspace of {C_{[a, b]}}.

{f' + 4f = 0 \Leftrightarrow \dfrac{df}{f} = -4dx \Leftrightarrow f(x) = C.e^{-4x}}. Each of vectors {f} is represented by the function {e^{-4x}, x \in [a, b]}, that implies the basis and the dimension of this subspace. \Box

Some properties of span of a vector space

In this post, we consider “span” of a vector space V. As we know, definition of span is:

Definition 1 Let {\{v_1, v_2, \dots, v_n\}} be a finite set of vectors in a vector space {V}. The subset of {V} spanned by {\{v_1, v_2, \dots, v_n\}} is the set of all linear combinations of {v_1, v_2, \dots, v_n}. This set is called the span of {\{v_1, v_2, \dots, v_n\}} and is denoted

\displaystyle \text{span}\{v_1, v_2, \dots, v_n\}.

For example, you can see

Example 1 Let {\{v_1=(1,0), v_2=(1,1)\}}, we have {span\{v_1, v_2\}=\{x(1,0) + y(1,1)| x, y \in \mathbb{R}\}}. Of course, {span\{v_1, v_2\} \subset \mathbb{R}^2}. It is easy to see that {\mathbb{R}^2 \subset span\{v_1, v_2\}} because {v = (x',y') \in \mathbb{R}^2 \Rightarrow (x',y') = x(1,0)+y(1,1) \Leftrightarrow \begin{cases} x'&= x+ y\\ y'& = y \end{cases} }. That system has always solution {x, y \in \mathbb{R}}. This implies {span\{v_1, v_2\} = \mathbb{R}^2}.

We have the following result

Theorem 2 The span of any finite set {\{v_1, v_2, \dots, v_n\}} of vectors in a vector space {V} is a subspace of {V}.

Some properties:

  1. Suppose {W} is a subspace of a vector space {V}. Prove that if {v_1, v_2, \dots, v_n \in W}, then {span\{v_1, \dots, v_n\} \subseteq W}.
  2. Suppose {v_1, \dots, v_m} and {w_1, \dots, w_n} are vectors in a vector space satisfy:
    if {w_1, \dots, w_n \in span\{v_1, \dots, v_m\}} and {v_1, \dots, v_m \in span\{w_1, \dots, w_n\}} then {span\{v_1, \dots, v_m\} = span\{w_1, \dots, w_n\}}.

References:

R. Messer, Linear algebra gateway to mathematics, Harper Collins College Publishers

D. Lay, Linear algebra and its applications, Addison-Wesley, 2012.

(cont.)

Some examples of subspaces – 1

As we knew, a subset {W} of a vector space {V} is called a subspace if it is nonempty and closed under addition and scalar multiplication.

  • The vector spaces {\mathbb{R}, \mathbb{R}^2, \mathbb{R}^3} are the usual Euclidean spaces of analytic geometry. There are three types of subspaces of {\mathbb{R}^2}: {\{\theta\},} a line through the origin, and {\mathbb{R}^2} itself. There are four types of subspaces of {\mathbb{R}^3}: {\{\theta\}}, a line through the origin, and {\mathbb{R}^3} itself.
  • Let {P[x]} be the set of all polynomials in the single variable {x}, with coefficients from {\mathbb{R}} and {P_n[x]} be the subset of {P[x]} consisting of all polynomials of degree {n} or less. We have a result: {P_n[x]} is a subspace of {P[x]}.
  • When {n > 0}, the set of all polynomials of degree exactly {n} is not a subspace of {P[x]}.
  • Let {\mathcal{F}(X, F)} be the set of all functions {f: X \rightarrow F} ({F} is field {\mathbb{R}} or {\mathbb{C}}). {\mathcal{F}(X, F)} is a vector space. Why?
  • {\mathcal{F}(X, F)} have many subspaces, for example, we consider the case of {X \subset \mathbb{R}^n} and {F = \mathbb{R}}. The set {C(X)} of all continuous functions {f: X \rightarrow \mathbb{R}} is a subspace of {\mathcal{F}(X, \mathbb{R})}. The set {\mathcal{D}(X)} of all differentiable functions {f: X \rightarrow \mathbb{R}} is a subspace {C(X)} and also of {\mathcal{F}(X, \mathbb{R})}. Prove that?

References

D. Lay, Linear algebra and its applications, Addison-Wesley, 2012.

Nguyen Huu Viet Hung, Linear algebra (Vietnamese), VNU Publisher .