Bases and dimension

Overview §

Every vector space has a basis, which combines notions from spans and linear independence. We can study vector spaces by studying their bases.

One special basis is the standard basis for ${\mathbb{F}}^n$ is the set $\{e_1,\cdots ,e_n\}$, where the only non-zero coordinate in each vector $e_i$ is the $i$th coordinate.

Dimensions tell us exactly how many coordinates are needed to describe an arbitrary element of a vector space.

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Bases §

Definition: Basis

If $V$ is a finite-dimensional vector space, a basis for $V$ is a set of vectors $\{v_1,\cdots ,v_n\}\subset V$ which:

• Spans the entire space, so every vector in $V$ is some linear combination of these vectors;
• Is linearly independent, so only the trivial linear combination produces $\vec{0}$.

• (Lemma 35) If $V$ is a vector space, a set $S=\{v_1,\cdots ,v_n\}$ is a basis for $V$ if and only if there is a unique way to express each $v\in V$ as a linear combination of the elements of $S$.
• (Proposition 37) Every finite-dimensional vector space has a finite basis.
  • (Lemma 36) Basis reduction lemma. If $S=\{v_1,\cdots ,v_n\}$ is any spanning set for $V$, there is a subset $S^{\prime }\subset S$ which is a basis for $V$.
    • Proposition 37 can be proven by choosing a spanning set and applying the basis reduction lemma to find a basis contained in the set.
  • (Lemma 38) Basis extension lemma. If $S=\{v_1,\cdots ,v_n\}$ is any linearly independent set for a finite-dimensional vector space $V$, there is a larger set $S^{\prime }=\{v_1,\cdots ,v_n,w_1,\cdots ,w_m\}$ which is a basis for $V$ (with $k=0$ if $\text{span}(S)=V$).
• (Proposition 39) If $W\subset V$ is a subspace of a finite-dimensional vector space, then there exists a basis $(w_1,\cdots ,w_n,v_1,\cdots ,v_k)$ for $V$ which begins with the basis for $W$.
  • (Corollary 40) Dimension is well-defined. Any two bases for a finite-dimensional vector space $V$ have the same number of elements.

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Dimensions §

Definition: Dimension

If $V$ is a finite-dimensional vector space, then its dimension $\text{dim}(V)\in \mathbb{N}$ is the number of elements in a basis for $V$.

By the “dimension is well-defined” corollary, we know the choice of basis doesn’t matter since this number is unambiguous.

• (Theorem) A subspace of a finite-dimensional vector space has dimension less than or equal to the latter’s
• If $(v_1,\cdots ,v_n)$ is a basis for $V$, then there is a bijection $A:{\mathbb{F}}^n\to V$ defined by $A(a_1\cdots a_n)=a_1{v}_1+\cdots +a_n{v}_n.$ The transformation is injective because $(v_1,\cdots ,v_n)$ is linearly independent and surjective because $(v_1,\cdots ,v_n)$ spans $V$.
• If two subspaces $U,W\subset V$ intersect trivially so that $U\cap W=\{\vec{0}\}$, then $\text{dim}V=\text{dim}U+\text{dim}W=\text{dim}(U+W).$
  • Proof. The union of two bases for $U,W$ forms a basis for $V$.

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Examples §

• The standard basis for ${\mathbb{F}}^n$ is the set $\{e_1,\cdots ,e_n\}$ where the only non-zero coordinate in each $e_i$ is the $i$th coordinate.

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Review §

Honors Math A §

• Prove Lemma 35.
• Prove Corollary 40.