In Section 4.4, it states that a vector space \(V\) with a basis \(\mathcal{B}\) containing \(n\) vectors is isomorphic to \(\mathbb{R}^n\). The discussion of dimension will give additional insight into properties of bases.
(I) Introduction: Coordinate Systems
Definition
Suppose \(\mathcal{B} = \{\vec{b}_1, \dots, \vec{b}_n\}\) is a basis for a vector space \(V\) and \(\vec{x}\) is a vector in \(V\). The coordinate vector of \(\vec{x}\) relative to \(\mathcal{B}\), or the \(\mathcal{B}\)-coordinate vector, is:
where \(c_1, \dots, c_n\) are weights such that \(\vec{x} = c_1\vec{b}_1 + \dots + c_n\vec{b}_n\).
Coordinate Mapping: This creates a one-to-one linear transformation from \(V\) onto \(\mathbb{R}^n\):
We can write \(\vec{x} = P_{\mathcal{B}} \cdot [\vec{x}]_{\mathcal{B}}\), where \(P_{\mathcal{B}} = \begin{bmatrix} \vec{b}_1 & \vec{b}_2 & \dots & \vec{b}_n \end{bmatrix}\) is the change-of-coordinate matrix from \(\mathcal{B}\) to the standard basis in \(\mathbb{R}^n\).
Let \(\mathcal{B} = \{1, 2t, -2 + 4t^2, -12t + 8t^3\}\) be a basis of \(\mathbb{P}_3\).
Find the coordinate vector of \(p(t) = 7 - 12t - 8t^2 + 12t^3\) relative to \(\mathcal{B}\).
Solution: Set the polynomial equal to a linear combination of the basis vectors:
\[ 7 - 12t - 8t^2 + 12t^3 = c_1(1) + c_2(2t) + c_3(-2 + 4t^2) + c_4(-12t + 8t^3) \]Equating the coefficients of like powers of \(t\):
- \(t^3\): \(8c_4 = 12 \implies c_4 = \frac{3}{2}\)
- \(t^2\): \(4c_3 = -8 \implies c_3 = -2\)
- \(t\): \(2c_2 - 12c_4 = -12 \implies 2c_2 - 12\left(\frac{3}{2}\right) = -12 \implies 2c_2 - 18 = -12 \implies c_2 = 3\)
- Constant: \(c_1 - 2c_3 = 7 \implies c_1 - 2(-2) = 7 \implies c_1 + 4 = 7 \implies c_1 = 3\)
Therefore, the coordinate vector is:
\[ [p]_{\mathcal{B}} = \begin{bmatrix} 3 \\ 3 \\ -2 \\ 3/2 \end{bmatrix} \](II) Dimension
Theorem 10
If a vector space \(V\) has a basis \(\mathcal{B} = \{\vec{b}_1, \dots, \vec{b}_n\}\), then any set in \(V\) containing more than \(n\) vectors must be linearly dependent.
It implies that each linearly independent set in \(V\) has no more than \(n\) vectors.
Theorem 11
If a vector space \(V\) has a basis of \(n\) vectors, then every basis of \(V\) must consist of exactly \(n\) vectors.
Proof: Let \(\mathcal{B}_1\) be a basis of \(n\) vectors and \(\mathcal{B}_2\) be another basis of \(V\).
- \(\mathcal{B}_1\) is a basis and \(|\mathcal{B}_1| = n\). Since \(\mathcal{B}_2\) is linearly independent, Theorem 10 implies \(|\mathcal{B}_2| \leq n\).
- \(\mathcal{B}_2\) is a basis and \(\mathcal{B}_1\) is linearly independent. By the same logic, \(|\mathcal{B}_1| \leq |\mathcal{B}_2| \implies n \leq |\mathcal{B}_2|\).
Therefore, \(|\mathcal{B}_2| = n\).
Definitions
\(V\) is finite-dimensional if \(V\) is spanned by a finite set. The dimension of \(V\), denoted \(\dim V\), is the number of vectors in a basis of \(V\).
The dimension of the zero space \(\{\vec{0}\}\) is defined to be \(0\).
\(V\) is infinite-dimensional if \(V\) is not spanned by a finite set.
- \(\dim \mathbb{R}^3 = 3\) (Standard basis: \(\{\vec{e}_1, \vec{e}_2, \vec{e}_3\}\))
- \(\dim \mathbb{P}_3 = 4\) (Standard basis: \(\{1, t, t^2, t^3\}\))
- \(\mathbb{P}\) (the set of all polynomials) is infinite-dimensional.
(1) Find \(\dim H\) for \(H = \left\{ \begin{bmatrix} a - 3b + 6c \\ 5a + 4d \\ b - 2c - d \\ 5d \end{bmatrix} : a, b, c, d \in \mathbb{R} \right\}\).
Solution: Decompose the general vector into a linear combination:
\[ \begin{bmatrix} a - 3b + 6c \\ 5a + 4d \\ b - 2c - d \\ 5d \end{bmatrix} = a \begin{bmatrix} 1 \\ 5 \\ 0 \\ 0 \end{bmatrix} + b \begin{bmatrix} -3 \\ 0 \\ 1 \\ 0 \end{bmatrix} + c \begin{bmatrix} 6 \\ 0 \\ -2 \\ 0 \end{bmatrix} + d \begin{bmatrix} 0 \\ 4 \\ -1 \\ 5 \end{bmatrix} = a\vec{v}_1 + b\vec{v}_2 + c\vec{v}_3 + d\vec{v}_4 \]So, \(H = \text{Span}\{\vec{v}_1, \vec{v}_2, \vec{v}_3, \vec{v}_4\}\), which is the column space of the matrix \(A = \begin{bmatrix} \vec{v}_1 & \vec{v}_2 & \vec{v}_3 & \vec{v}_4 \end{bmatrix}\).
The pivot columns are 1, 2, and 4. Therefore, \(\{\vec{v}_1, \vec{v}_2, \vec{v}_4\}\) forms a basis for \(H\), and \(\dim H = 3\).
(2) Find \(\dim H'\) for \(H' = \left\{ \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} : a - 3b + c = 0 \right\}\).
Solution: Solve the equation for \(a\): \(a = 3b - c\), with \(b, c, d\) as free variables.
\[ \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} = \begin{bmatrix} 3b - c \\ b \\ c \\ d \end{bmatrix} = b \begin{bmatrix} 3 \\ 1 \\ 0 \\ 0 \end{bmatrix} + c \begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \end{bmatrix} + d \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} = b\vec{u}_1 + c\vec{u}_2 + d\vec{u}_3 \]\(\{\vec{u}_1, \vec{u}_2, \vec{u}_3\}\) forms a basis for \(H'\), and \(\dim H' = 3\).
Classify the subspaces of \(\mathbb{R}^3\) (Revisit):
- 0-dim: Only the zero subspace \(\{\vec{0}\}\).
- 1-dim: Any subspace spanned by a single nonzero vector is a line through the origin.
- 2-dim: Any subspace spanned by 2 linearly independent vectors is a plane through the origin.
- 3-dim: Only \(\mathbb{R}^3\) itself. Any 3 linearly independent vectors in \(\mathbb{R}^3\) span all of \(\mathbb{R}^3\) by the Invertible Matrix Theorem.
Theorem 12 (Counterpart to the Spanning Set Theorem)
Let \(H\) be a subspace of a finite-dimensional vector space \(V\). Any linearly independent set in \(H\) can be expanded, if necessary, to a basis for \(H\). Also, \(H\) is finite-dimensional and:
When the dimension of a vector space is known, the search for a basis is simplified by the next theorem. It says that if a set has the right number of vectors, one only has to show either that the set is linearly independent or that it spans the space.
Theorem 13 (The Basis Theorem)
Let \(V\) be a \(p\)-dimensional vector space with \(p \geq 1\).
- Any linearly independent set of exactly \(p\) elements in \(V\) is automatically a basis for \(V\).
- Any set of exactly \(p\) elements that spans \(V\) is automatically a basis for \(V\).
Show that \(\{1+t, 1-t, t+t^2\}\) is a basis for \(\mathbb{P}_2\).
Solution: We know that \(\dim \mathbb{P}_2 = 3\), and this set contains exactly 3 vectors. So, by the Basis Theorem, we just need to show that the set is linearly independent.
Set a linear combination equal to the zero polynomial:
\[ c_1(1+t) + c_2(1-t) + c_3(t+t^2) = 0 \] \[ (c_1 + c_2) + (c_1 - c_2 + c_3)t + c_3t^2 = 0 \]This yields the system:
\[ \begin{cases} c_3 = 0 \\ c_1 - c_2 + c_3 = 0 \\ c_1 + c_2 = 0 \end{cases} \]Since \(c_3 = 0\), the second equation becomes \(c_1 - c_2 = 0 \implies c_1 = c_2\). Substituting this into the first gives \(2c_1 = 0 \implies c_1 = c_2 = c_3 = 0\). The set is linearly independent, and therefore it is a basis for \(\mathbb{P}_2\).
(III) The Dimension of Nul A, Col A and Row A (\(A\) is an \(m \times n\) matrix)
- \( \text{Rank } A = \dim \text{Col } A = \text{number of pivot columns of } A \)
- \( \text{Nullity } A = \dim \text{Nul } A = \text{number of free variables in } A\vec{x} = \vec{0} \)
-
\( \dim \text{Row } A = \dim \text{Col } A^T = \text{Rank } A^T = \text{Rank } A \)
(Since the number of pivot rows in \(A\) equals the number of pivot columns of \(A\))
Theorem 14 (The Rank Theorem)
If \(A\) is an \(m \times n\) matrix, then:
(Number of pivot columns + Number of non-pivot columns = Total number of columns in \(A\)).
Written Homework Note: Show \(\dim \text{Row } A + \text{Nullity } A^T = \text{number of rows in } A\).
(a) If \(A\) is a \(7 \times 9\) matrix with a 2-dimensional null space, what is \(\text{Rank } A\)?
Solution: By the Rank Theorem, \(\text{Rank } A = n - \dim \text{Nul } A = 9 - 2 = 7\).
Is \(\text{Col } A = \mathbb{R}^7\)?
Yes. Since \(\text{Col } A\) is a subspace of \(\mathbb{R}^7\) and \(\dim \text{Col } A = 7\), then \(\text{Col } A = \mathbb{R}^7\).
Is \(\text{Nul } A = \mathbb{R}^2\)?
No. Although \(\dim \text{Nul } A = 9 - 7 = 2\), \(\text{Nul } A \neq \mathbb{R}^2\) since \(\text{Nul } A\) is a subspace of \(\mathbb{R}^9\) (the domain), not \(\mathbb{R}^2\).
(b) Could a \(6 \times 9\) matrix have a 2-dimensional null space?
Solution: No. Assume \(B\) is a \(6 \times 9\) matrix with \(\dim \text{Nul } B = 2\). By the Rank Theorem, \(\text{Rank } B = 9 - 2 = 7\). But \(\text{Col } B\) is a subspace of \(\mathbb{R}^6\), meaning its dimension (\(\text{Rank } B\)) cannot exceed 6. Thus, a contradiction exists.
(c) What is the largest possible rank of a \(6 \times 9\) matrix? How about a \(9 \times 6\) matrix?
Solution: The rank is the same as the number of pivots in the matrix. Since there are only 6 columns in a \(9 \times 6\) matrix, and only 6 rows in a \(6 \times 9\) matrix, there can be at most 6 pivots for either matrix. Therefore, the largest possible rank for either matrix is 6.
Let \(\mathbb{P}_3\) be the subspace of all polynomials of degree at most 3. Let \(H\) be the set of polynomials \(p(t)\) in \(\mathbb{P}_3\) such that \(p(0) = p(1)\). Find a basis for \(H\).
Solution: Let the general polynomial be \(p(t) = at^3 + bt^2 + ct + d\).
Evaluate at \(t=0\) and \(t=1\):
\[ p(0) = d \] \[ p(1) = a + b + c + d \]Since \(p(0) = p(1)\):
\[ d = a + b + c + d \implies a + b + c = 0 \]Solve for \(a\): \(a = -b - c\). The variables \(b, c, d\) are free variables.
Substitute back into the polynomial expression:
\[ p(t) = (-b-c)t^3 + bt^2 + ct + d = b(t^2 - t^3) + c(t - t^3) + d(1) \]Thus, \(H = \text{Span}\{1, t - t^3, t^2 - t^3\}\).
To confirm it forms a basis, we check for linear independence:
\[ c_1(1) + c_2(t - t^3) + c_3(t^2 - t^3) = 0 \] \[ c_1 + c_2t + c_3t^2 + (-c_2 - c_3)t^3 = 0 \]Equating coefficients to zero gives \(c_1 = 0\), \(c_2 = 0\), and \(c_3 = 0\). The set is linearly independent.
Therefore, \(\{1, t - t^3, t^2 - t^3\}\) forms a basis for \(H\).