1) Linear Combinations and Span - Building Vectors 🏗️
In Part 1, we introduced the abstract concept of vector spaces. Now, let's explore how to construct vectors within a vector space using a fundamental operation: linear combinations. Linear combinations allow us to generate new vectors from a given set of vectors by scaling them and adding them together. The set of all possible vectors we can create in this way is called the span of the original set.
A Linear Combination of vectors \( \mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_p \) in a vector space \(V\) is a vector of the form: \( c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + ... + c_p\mathbf{v}_p \) where \( c_1, c_2, ..., c_p \) are scalars.
The Span of a set of vectors \( \{ \mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_p \} \) in \(V\), denoted by \( \text{Span}\{ \mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_p \} \), is the set of all possible linear combinations of \( \mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_p \). In set notation:
\( \text{Span}\{ \mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_p \} = \{ c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + ... + c_p\mathbf{v}_p \mid c_1, c_2, ..., c_p \in \mathbb{R} \} \)
The span of a set of vectors is always a subspace of the original vector space. It's the "smallest" subspace that contains all the original vectors.
Example 1: Linear Combination in \( \mathbb{R}^2 \)
Let \( \mathbf{v}_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \) and \( \mathbf{v}_2 = \begin{bmatrix} -1 \\ 1 \end{bmatrix} \) be vectors in \( \mathbb{R}^2 \). Find the linear combination \( 3\mathbf{v}_1 - 2\mathbf{v}_2 \).
\( 3\mathbf{v}_1 - 2\mathbf{v}_2 = 3\begin{bmatrix} 1 \\ 2 \end{bmatrix} - 2\begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 6 \end{bmatrix} + \begin{bmatrix} 2 \\ -2 \end{bmatrix} = \begin{bmatrix} 5 \\ 4 \end{bmatrix} \)
So, \( \begin{bmatrix} 5 \\ 4 \end{bmatrix} \) is a linear combination of \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \).
Example 2: Span in \( \mathbb{R}^2 \)
What is \( \text{Span}\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix} \} \) in \( \mathbb{R}^2 \)?
The span is the set of all linear combinations of \( \begin{bmatrix} 1 \\ 0 \end{bmatrix} \), which are vectors of the form \( c \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} c \\ 0 \end{bmatrix} \). This is the set of all vectors on the x-axis in \( \mathbb{R}^2 \). So, \( \text{Span}\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix} \} \) is the x-axis, which is a subspace of \( \mathbb{R}^2 \).
Example 3: Span in \( \mathbb{R}^3 \)
What is \( \text{Span}\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \} \) in \( \mathbb{R}^3 \)?
The span is the set of all linear combinations of \( \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \) and \( \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \), which are vectors of the form \( c_1 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \\ 0 \end{bmatrix} \). This is the set of all vectors in the xy-plane in \( \mathbb{R}^3 \). So, \( \text{Span}\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \} \) is the xy-plane, a subspace of \( \mathbb{R}^3 \).
2) Linear Independence and Linear Dependence - Efficient Vector Sets 💡
Now, let's think about the "efficiency" of a set of vectors in terms of spanning a space. A set of vectors is linearly independent if none of the vectors in the set can be written as a linear combination of the others. In other words, they are all "essential" for spanning their span, and none are redundant. If at least one vector can be written as a linear combination of the others, the set is linearly dependent.
2.1 Definition of Linear Independence and Dependence
A set of vectors \( \{ \mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_p \} \) in a vector space \(V\) is said to be Linearly Independent if the vector equation: \( c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + ... + c_p\mathbf{v}_p = \mathbf{0} \) has only the trivial solution \( c_1 = c_2 = ... = c_p = 0 \).
A set of vectors \( \{ \mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_p \} \) is Linearly Dependent if there exist scalars \( c_1, c_2, ..., c_p \), not all zero, such that: \( c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + ... + c_p\mathbf{v}_p = \mathbf{0} \)
In essence:
- Linearly Independent: The only way to get the zero vector as a linear combination is by using all zero scalars.
- Linearly Dependent: There is a way to get the zero vector as a linear combination using at least one non-zero scalar.
Example 4: Linear Independence in \( \mathbb{R}^2 \)
Are the vectors \( \mathbf{v}_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \) and \( \mathbf{v}_2 = \begin{bmatrix} -1 \\ 1 \end{bmatrix} \) linearly independent or linearly dependent?
Consider the equation \( c_1\mathbf{v}_1 + c_2\mathbf{v}_2 = \mathbf{0} \), which is \( c_1\begin{bmatrix} 1 \\ 2 \end{bmatrix} + c_2\begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \). This gives the system of equations: \( c_1 - c_2 = 0 \) \( 2c_1 + c_2 = 0 \) Solving this system, we find that the only solution is \( c_1 = 0 \) and \( c_2 = 0 \). Thus, \( \{ \mathbf{v}_1, \mathbf{v}_2 \} \) is linearly independent. Neither vector is a scalar multiple of the other.
Example 5: Linear Dependence in \( \mathbb{R}^2 \)
Are the vectors \( \mathbf{v}_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \), \( \mathbf{v}_2 = \begin{bmatrix} -2 \\ -4 \end{bmatrix} \), and \( \mathbf{v}_3 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \) linearly independent or linearly dependent?
Notice that \( \mathbf{v}_2 = -2\mathbf{v}_1 \), so \( 2\mathbf{v}_1 + \mathbf{v}_2 + 0\mathbf{v}_3 = \mathbf{0} \). We found a non-trivial solution (coefficients 2, 1, 0 are not all zero). Thus, \( \{ \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 \} \) is linearly dependent. Specifically, \( \mathbf{v}_2 \) is a linear combination of \( \mathbf{v}_1 \) (and also \( \mathbf{v}_1 \) is a linear combination of \( \mathbf{v}_2 \)).
2.2 Properties of Linear Dependence
Some key properties of linear dependence:
- A set containing the zero vector is always linearly dependent (e.g., \( \{ \mathbf{v}_1, \mathbf{v}_2, \mathbf{0} \} \) is always linearly dependent because \( 0\mathbf{v}_1 + 0\mathbf{v}_2 + 1\mathbf{0} = \mathbf{0} \)).
- A set of two vectors \( \{ \mathbf{v}_1, \mathbf{v}_2 \} \) is linearly dependent if and only if one is a scalar multiple of the other.
- In \( \mathbb{R}^n \), any set containing more than \(n\) vectors must be linearly dependent.
3) Basis of a Vector Space - Minimal Spanning Sets 🔑
Now, we combine the concepts of span and linear independence to define a basis. A basis of a vector space is a set of vectors that is both linearly independent and spans the entire vector space. A basis is a kind of "coordinate system" for the vector space – it provides a minimal and efficient set of vectors to describe all vectors in the space.
3.1 Definition of a Basis
A Basis for a vector space \(V\) is a set of vectors \( \mathcal{B} = \{ \mathbf{b}_1, \mathbf{b}_2, ..., \mathbf{b}_k \} \) in \(V\) such that:
- Spanning Set: \( \text{Span}\{ \mathbf{b}_1, \mathbf{b}_2, ..., \mathbf{b}_k \} = V \) (The set \( \mathcal{B} \) spans the entire vector space \(V\)).
- Linear Independence: The set \( \{ \mathbf{b}_1, \mathbf{b}_2, ..., \mathbf{b}_k \} \) is linearly independent.
If a set of vectors satisfies both conditions, it is a basis for \(V\). A basis is a minimal spanning set in the sense that if you remove any vector from a basis, it will no longer span the entire space. And it is linearly independent, so it doesn't contain any redundant vectors.
Example 6: Standard Basis of \( \mathbb{R}^2 \)
Consider the set \( \mathcal{B} = \{ \mathbf{e}_1, \mathbf{e}_2 \} = \{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix} \} \) in \( \mathbb{R}^2 \). Is \( \mathcal{B} \) a basis for \( \mathbb{R}^2 \)?
- Spanning Set? Yes. Any vector \( \begin{bmatrix} x \\ y \end{bmatrix} \) in \( \mathbb{R}^2 \) can be written as a linear combination of \( \mathbf{e}_1 \) and \( \mathbf{e}_2 \): \( \begin{bmatrix} x \\ y \end{bmatrix} = x\begin{bmatrix} 1 \\ 0 \end{bmatrix} + y\begin{bmatrix} 0 \\ 1 \end{bmatrix} = x\mathbf{e}_1 + y\mathbf{e}_2 \). So, \( \text{Span}\{ \mathbf{e}_1, \mathbf{e}_2 \} = \mathbb{R}^2 \).
- Linearly Independent? Yes. Consider \( c_1\mathbf{e}_1 + c_2\mathbf{e}_2 = \mathbf{0} \), which is \( c_1\begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2\begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \). This gives \( \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \), so \( c_1 = 0 \) and \( c_2 = 0 \). Thus, \( \{ \mathbf{e}_1, \mathbf{e}_2 \} \) is linearly independent.
Example 7: Standard Basis of \( \mathbb{R}^3 \)
Similarly, the standard basis of \( \mathbb{R}^3 \) is \( \mathcal{B} = \{ \mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3 \} = \{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \} \). You can verify that it is both linearly independent and spans \( \mathbb{R}^3 \).
3.2 Dimension of a Vector Space
A fundamental property of a vector space is its dimension. If a vector space \(V\) has a basis consisting of \(k\) vectors, then every basis for \(V\) must consist of exactly \(k\) vectors. This number \(k\) is called the dimension of \(V\), denoted as \( \dim(V) = k \). If a vector space is spanned by a finite set, it is called finite-dimensional. Otherwise, it is infinite-dimensional.
Examples:
- \( \dim(\mathbb{R}^2) = 2 \) (standard basis \( \{ \mathbf{e}_1, \mathbf{e}_2 \} \) has 2 vectors).
- \( \dim(\mathbb{R}^3) = 3 \) (standard basis \( \{ \mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3 \} \) has 3 vectors).
- In general, \( \dim(\mathbb{R}^n) = n \).
- The dimension of the polynomial space \( P_n \) (polynomials of degree at most \(n\)) is \( n+1 \). A basis for \( P_n \) is \( \{ 1, x, x^2, ..., x^n \} \).
4) Practice Questions 🎯
5.1 Fundamental – Build Skills
1. Let \( \mathbf{v}_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \) and \( \mathbf{v}_2 = \begin{bmatrix} 2 \\ 1 \end{bmatrix} \). Is \( \begin{bmatrix} 4 \\ -1 \end{bmatrix} \) in \( \text{Span}\{ \mathbf{v}_1, \mathbf{v}_2 \} \)? If yes, express it as a linear combination of \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \).
2. Determine if the set of vectors \( \{ \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 2 \\ 4 \end{bmatrix} \} \) in \( \mathbb{R}^3 \) is linearly independent or linearly dependent.
3. Determine if the set of vectors \( \{ \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ -1 \end{bmatrix} \} \) is a basis for \( \mathbb{R}^2 \). Justify your answer.
4. Determine if the set of vectors \( \{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \} \) is a basis for \( \mathbb{R}^3 \). Justify your answer.
5. Find the dimension of the subspace \( \text{Span}\{ \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}, \begin{bmatrix} 2 \\ 4 \\ 6 \end{bmatrix} \} \) in \( \mathbb{R}^3 \).
6. Find the dimension of the subspace \( \text{Span}\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \} \) in \( \mathbb{R}^3 \).
7. Are the polynomials \( p_1(x) = 1 + x \), \( p_2(x) = x - x^2 \), \( p_3(x) = 1 + x^2 \) linearly independent or linearly dependent in \( P_2 \)?
8. Is the set of polynomials \( \{ 1, x^2 \} \) a basis for \( P_2 \)? Why or why not?
9. What is the dimension of the vector space \( M_{2 \times 2} \) (the space of \( 2 \times 2 \) matrices)? Can you find a basis for \( M_{2 \times 2} \)?
10. Let \(V\) be a vector space and \( \{ \mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_n \} \) be a basis for \(V\). Is the set \( \{ 2\mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_n \} \) also a basis for \(V\)? Explain.
5.2 Challenging – Push Limits 💪🚀
1. Let \(S = \{ \mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_p \} \) be a set of vectors in a vector space \(V\). Show that \( \text{Span}(S) \) is indeed a subspace of \(V\). (Verify the three subspace conditions).
2. Let \( \{ \mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_p \} \) be a linearly independent set in \( \mathbb{R}^n \). Can \( p \) be greater than \( n \)? Can \( p \) be equal to \( n \)? Can \( p \) be less than \( n \)? Explain.
3. Let \( \mathcal{B} = \{ \mathbf{b}_1, \mathbf{b}_2 \} \) be a basis for a vector space \(V\). Show that if \( \mathbf{v} \in V \), then there is a unique way to write \( \mathbf{v} \) as a linear combination of \( \mathbf{b}_1 \) and \( \mathbf{b}_2 \), i.e., \( \mathbf{v} = c_1\mathbf{b}_1 + c_2\mathbf{b}_2 \) for a unique pair of scalars \( c_1, c_2 \).
4. (Conceptual) Explain why a basis of a vector space can be considered a "minimal spanning set". What happens if you remove a vector from a basis? What happens if you add a vector to a basis?
5. (Abstract) Let \(H\) be a subspace of a finite-dimensional vector space \(V\). Show that \( \dim(H) \leq \dim(V) \). Furthermore, show that if \( \dim(H) = \dim(V) \), then \( H = V \). (Subspaces cannot be "larger" than the space they are in, and if they have the same dimension, they must be the same space).
6) Summary 🎉
- Linear Combination: A sum of scalar multiples of vectors: \( c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + ... + c_p\mathbf{v}_p \).
- Span: The set of all linear combinations of a set of vectors \( \{ \mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_p \} \), denoted \( \text{Span}\{ \mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_p \} \). It's always a subspace.
- Linear Independence: A set \( \{ \mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_p \} \) is linearly independent if \( c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + ... + c_p\mathbf{v}_p = \mathbf{0} \) only when \( c_1 = c_2 = ... = c_p = 0 \). Otherwise, it's linearly dependent.
- Basis: A set of vectors \( \mathcal{B} \) that is both a spanning set for \(V\) and linearly independent.
- Dimension: The number of vectors in any basis of a vector space \(V\), denoted \( \dim(V) \). It's a measure of the "size" of the vector space.
Fantastic progress in understanding Vector Spaces! You've now grasped the core concepts of span, linear independence, and basis – essential building blocks of linear algebra. These ideas will be crucial as we delve deeper into vector spaces and their applications. Keep up the excellent work, and get ready for more explorations in the exciting world of linear algebra! 🚀💡
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