8.2 Basis, Dimension, and Coordinate Representations

8.2 Basis, Dimension, and Coordinate Representations

The concepts of basis and dimension formalize the degree of freedom in a vector space. By choosing an ordered basis, any abstract finite-dimensional vector space over $\mathbb{F}$ becomes isomorphic to the concrete coordinate space $\mathbb{F}^n$. The GRE Mathematics Subject Test evaluates both the theoretical properties of bases (extension, reduction, and dimension invariance) and computational proficiency with coordinate vectors and change-of-basis transition matrices.

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Bases: Minimal Spanning and Maximal Independent Sets

Let $V$ be a vector space over $\mathbb{F}$. An ordered subset $\mathcal{B} = (v_1, v_2, \dots, v_n)$ of $V$ is a basis for $V$ if:

1. $\mathcal{B}$ is linearly independent: $\sum_{i=1}^n c_i v_i = 0_V \implies c_1 = \dots = c_n = 0$.
2. $\mathcal{B}$ spans $V$: $\operatorname{span}(\mathcal{B}) = V$.

Equivalent Characterizations of a Basis

For a subset $\mathcal{B} \subset V$, the following statements are equivalent:

• $\mathcal{B}$ is a basis of $V$.
• $\mathcal{B}$ is a minimal spanning set: removing any vector from $\mathcal{B}$ yields a set that does not span $V$.
• $\mathcal{B}$ is a maximal linearly independent set: adding any vector from $V$ to $\mathcal{B}$ yields a linearly dependent set.
• Unique Representation Property: Every vector $v \in V$ can be expressed uniquely as a linear combination of vectors in $\mathcal{B}$: $v = \sum_{i=1}^n c_i v_i, \quad c_i \in \mathbb{F}.$

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Dimension and Fundamental Basis Theorems

Steinitz Exchange Lemma & Dimension Invariance

• Steinitz Exchange Lemma: If $S = {u_1, \dots, u_k}$ is linearly independent in $V$ and $G = {w_1, \dots, w_m}$ spans $V$, then $k \le m$. Furthermore, there exists a subset of $m - k$ vectors from $G$ which, together with $S$, spans $V$.
• Invariance of Dimension: If $V$ has a finite basis of $n$ elements, every basis of $V$ contains exactly $n$ elements. This invariant integer is the dimension of $V$, written $\dim(V) = n$. By convention, $\dim({0_V}) = 0$.

Basis Reduction and Extension Theorems

1. Basis Reduction Theorem: Every finite spanning set $S$ for $V$ contains a subset that is a basis of $V$.
2. Basis Extension Theorem: Every linearly independent subset $S$ in a finite-dimensional space $V$ can be extended to a basis of $V$.
3. The Two-Out-of-Three Rule: If $\dim V = n$ and $S \subset V$ contains exactly $n$ vectors: $S \text{ is linearly independent} \iff S \text{ spans } V \iff S \text{ is a basis of } V.$ On the GRE, this eliminates half the work: to verify that $n$ vectors form a basis in an $n$-dimensional space, check either independence or span, not both.

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Standard Vector Spaces and Dimensions

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Coordinate Vectors and the Coordinate Isomorphism

Let $\mathcal{B} = (v_1, \dots, v_n)$ be an ordered basis of $V$. By the unique representation property, for each $v \in V$ there exist unique scalars $c_1, \dots, c_n \in \mathbb{F}$ such that $v = \sum_{i=1}^n c_i v_i$. The coordinate vector of $v$ relative to $\mathcal{B}$ is the column vector: $[v]_\mathcal{B} = \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix} \in \mathbb{F}^n.$

The Coordinate Isomorphism Theorem

The mapping $\Phi_\mathcal{B}: V \to \mathbb{F}^n$ defined by $\Phi_\mathcal{B}(v) = [v]_\mathcal{B}$ is a linear isomorphism:

• $\Phi_\mathcal{B}$ is linear: $[c u + v]\mathcal{B} = c [u]\mathcal{B} + [v]_\mathcal{B}$.
• $\Phi_\mathcal{B}$ is bijective (one-to-one and onto).
• Preservation of Structure: Vectors ${u_1, \dots, u_k}$ are linearly independent in $V$ if and only if their coordinate vectors ${[u_1]\mathcal{B}, \dots, [u_k]\mathcal{B}}$ are linearly independent in $\mathbb{F}^n$.

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Change-of-Basis and Transition Matrices

Let $\mathcal{B} = (u_1, \dots, u_n)$ and $\mathcal{C} = (v_1, \dots, v_n)$ be two ordered bases of $V$. The change-of-basis matrix from $\mathcal{B}$ to $\mathcal{C}$, denoted $P_{\mathcal{C} \leftarrow \mathcal{B}}$, converts $\mathcal{B}$-coordinates into $\mathcal{C}$-coordinates: $[v]_\mathcal{C} = P_{\mathcal{C} \leftarrow \mathcal{B}} [v]_\mathcal{B}.$

Constructing the Transition Matrix

The columns of $P_{\mathcal{C} \leftarrow \mathcal{B}}$ are the coordinate vectors of the original basis vectors $\mathcal{B}$ relative to the target basis $\mathcal{C}$: $P_{\mathcal{C} \leftarrow \mathcal{B}} = \begin{bmatrix} [u_1]_\mathcal{C} & [u_2]_\mathcal{C} & \dots & [u_n]_\mathcal{C} \end{bmatrix}.$

Key Algebraic Properties

1. Invertibility: $P_{\mathcal{C} \leftarrow \mathcal{B}}$ is always invertible, with $(P_{\mathcal{C} \leftarrow \mathcal{B}})^{-1} = P_{\mathcal{B} \leftarrow \mathcal{C}}$.
2. Chain Rule: For three bases $\mathcal{B}, \mathcal{C}, \mathcal{D}$: $P_{\mathcal{D} \leftarrow \mathcal{B}} = P_{\mathcal{D} \leftarrow \mathcal{C}} P_{\mathcal{C} \leftarrow \mathcal{B}}.$
3. Transition in Coordinate Space $\mathbb{R}^n$: If $\mathcal{E}$ is the standard basis, let $M_\mathcal{B} = [u_1 \dots u_n]$ and $M_\mathcal{C} = [v_1 \dots v_n]$ be the matrices containing basis vectors as columns. Then $P_{\mathcal{E} \leftarrow \mathcal{B}} = M_\mathcal{B}$, yielding: $P_{\mathcal{C} \leftarrow \mathcal{B}} = M_\mathcal{C}^{-1} M_\mathcal{B}.$

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Step-by-Step Worked Problems

Problem 1: Coordinate Representation in Polynomial Spaces

Let $\mathcal{B} = (1, 1 + x, 1 + x + x^2)$ be an ordered basis of $P_2(\mathbb{R})$. Find the coordinate vector $[p(x)]_\mathcal{B}$ for $p(x) = 2 - 3x + 5x^2$.

Solution:

1. Express $p(x)$ as a linear combination: $2 - 3x + 5x^2 = c_1(1) + c_2(1 + x) + c_3(1 + x + x^2).$
2. Collect powers of $x$: $(c_1 + c_2 + c_3) + (c_2 + c_3)x + c_3 x^2 = 2 - 3x + 5x^2.$
3. Solve by back-substitution:
   • $x^2$ coefficient: $c_3 = 5$.
   • $x$ coefficient: $c_2 + 5 = -3 \implies c_2 = -8$.
   • Constant term: $c_1 + (-8) + 5 = 2 \implies c_1 - 3 = 2 \implies c_1 = 5$.
4. Thus $[p(x)]_\mathcal{B} = \begin{pmatrix} 5 \ -8 \ 5 \end{pmatrix}$.

Problem 2: Transition Matrix in $\mathbb{R}^2$

Let $\mathcal{B} = (\begin{pmatrix} 1 \ 2 \end{pmatrix}, \begin{pmatrix} 3 \ 5 \end{pmatrix})$ and $\mathcal{C} = (\begin{pmatrix} 1 \ 0 \end{pmatrix}, \begin{pmatrix} 0 \ 1 \end{pmatrix})$ be bases of $\mathbb{R}^2$. Find $P_{\mathcal{B} \leftarrow \mathcal{C}}$.

Solution:

1. Note that $\mathcal{C}$ is the standard basis $\mathcal{E}$. The matrix whose columns are vectors of $\mathcal{B}$ is $P_{\mathcal{E} \leftarrow \mathcal{B}} = M_\mathcal{B} = \begin{pmatrix} 1 & 3 \ 2 & 5 \end{pmatrix}$.
2. The desired matrix is $P_{\mathcal{B} \leftarrow \mathcal{C}} = (P_{\mathcal{E} \leftarrow \mathcal{B}})^{-1} = M_\mathcal{B}^{-1}$.
3. Determinant: $\det(M_\mathcal{B}) = (1)(5) - (3)(2) = 5 - 6 = -1$.
4. Matrix inverse: $M_\mathcal{B}^{-1} = \frac{1}{-1} \begin{pmatrix} 5 & -3 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} -5 & 3 \\ 2 & -1 \end{pmatrix}.$
5. Thus $P_{\mathcal{B} \leftarrow \mathcal{C}} = \begin{pmatrix} -5 & 3 \ 2 & -1 \end{pmatrix}$.

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GRE Exam Traps & Pitfalls

Trap 1: The Off-By-One Polynomial Dimension Always remember that $\dim(P_n) = n + 1$, not $n$, because of the non-zero degree zero constant term $x^0 = 1$. For example, $P_3$ has basis ${1, x, x^2, x^3}$, giving dimension 4.

Trap 2: Transposed Transition Matrix The columns of $P_{\mathcal{C} \leftarrow \mathcal{B}}$ are the coordinates $[u_j]_\mathcal{C}$. Never write the coordinates as rows; writing coordinates as rows transposes the matrix and invalidates subsequent coordinate multiplications.

Trap 3: Direction of Transition Remember that $[v]\mathcal{C} = P{\mathcal{C} \leftarrow \mathcal{B}} [v]_\mathcal{B}$. The arrow points from the right subscript to the left subscript: target basis on the left, source basis on the right.