Basis & Dimension

Goldilocks: not too many, not too few

Two nodes have been circling this. Spanning (from Span) means your set reaches every vector — but a spanning set can be bloated with redundant vectors. Independence (from Linear Independence) means no redundancy — but an independent set might not reach everything. A basis is the set that's just right: both at once.

Definition. A set $B = \{v_1, \dots, v_n\}$ is a basis of a vector space $V$ if:

1. $B$ spans $V$ ($\operatorname{span}(B) = V$), and
2. $B$ is linearly independent.

Three ways to say the same thing, each worth holding in your head — they're all the same beast from different angles:

• A basis is a minimal spanning set — strip any vector and it stops spanning.
• A basis is a maximal independent set — add any vector and it becomes dependent.
• A basis is a coordinate skeleton — the exact frame you measure every other vector against (see below).

THE theorem: all bases have the same size

This is the load-bearing wall of the entire subject. Without it, "dimension" is gibberish — just a word people say to sound smart. We prove it properly or we don't use it.

Theorem (invariance of basis size). If $V$ has a basis with $n$ vectors, then every basis of $V$ has exactly $n$ vectors.

Proof sketch (the exchange / replacement argument). Suppose $B = \{v_1, \dots, v_n\}$ is a basis and $C = \{w_1, \dots, w_m\}$ is another, with $m > n$. Because $B$ spans, each $w_j$ is a combination of the $v_i$. The replacement lemma says: you can swap the $w$'s into $B$ one at a time, kicking out a $v$ each time, and keep a spanning set. After $n$ swaps you've used up all of $B$ and the first $n$ of the $w$'s span $V$ — so the remaining $w$'s (there are $m - n > 0$ of them) are combinations of the first $n$. That makes $C$ dependent, contradicting that $C$ is a basis. So $m \le n$. Run the same argument with the roles flipped ($C$ spans, $B$ independent) to get $n \le m$. Therefore $m = n$. $\blacksquare$

The engine underneath is the pigeonhole-via-pivots fact from the Linear Independence node: you can't have more independent vectors than the size of a spanning set. Once sizes are pinned, the word dimension is born:

$\dim(V) = \text{the number of vectors in any (hence every) basis of } V.$

Standard bases and their dimensions

Concrete frames you should know cold — these will come up throughout the rest of the course and in every exam I can imagine:

• $\mathbb{R}^n$: the standard basis $e_1, \dots, e_n$ (columns of the identity). $\dim(\mathbb{R}^n) = n$.
• $P_n$ (polynomials of degree $\le n$): the basis $\{1, x, x^2, \dots, x^n\}$. That's $n+1$ vectors, so $\dim(P_n) = n + 1$. (Watch the off-by-one — $P_2$ has basis $1, x, x^2$, dimension $3$. Mess this up and you'll be wrong in ways that compound.)
• $M_{m \times n}$ (matrices): the basis of matrices with a single $1$ and zeros elsewhere — there are $mn$ of them, so $\dim(M_{m\times n}) = mn$. So $\dim(M_{2\times 2}) = 4$.
• The zero space $\{\mathbf{0}\}$: its basis is the empty set, and $\dim(\{\mathbf{0}\}) = 0$.

Coordinates: a basis turns vectors into number-tuples

Here's the deepest payoff, and it ties straight back to What Is a Number — the basis is the place-value system for a vector space.

Theorem (unique coordinates). If $B = \{v_1, \dots, v_n\}$ is a basis of $V$, then every $v \in V$ can be written as a linear combination of the $v_i$ in exactly one way.

Proof. Existence is just spanning: since $B$ spans, $v = c_1 v_1 + \cdots + c_n v_n$ for some coefficients. Uniqueness is the two-line independence argument — watch. Suppose two representations: $v = c_1 v_1 + \cdots + c_n v_n = d_1 v_1 + \cdots + d_n v_n.$ Subtract: $(c_1 - d_1)v_1 + (c_2 - d_2)v_2 + \cdots + (c_n - d_n)v_n = \mathbf{0}.$ By independence of $B$, every coefficient must be zero: $c_i - d_i = 0$ for all $i$, i.e. $c_i = d_i$. The two representations were identical all along. $\blacksquare$

Those unique coefficients $(c_1, \dots, c_n)$ are the coordinates of $v$ relative to $B$. This is exactly positional notation: the basis is the "place values," the coordinates are the "digits," and uniqueness is what makes the digits a well-defined name. A basis is the device that turns an abstract vector — a polynomial, a matrix, a function — into an honest tuple of numbers you can compute with. That is why we drag everything back to $\mathbb{R}^n$. Same fucking move as base-ten notation, just in a different costume, exactly as advertised in lesson one.

Dimension as degrees of freedom

The cleanest intuition: $\dim(V)$ is the number of independent choices — degrees of freedom — you make to specify a vector. In $\mathbb{R}^3$ you choose three coordinates freely, so $\dim = 3$. Revisit the Subspaces census of $\mathbb{R}^3$ with this new word:

• $\{\mathbf{0}\}$: zero free choices, $\dim = 0$.
• A line through $\mathbf{0}$: one free choice (how far along), $\dim = 1$.
• A plane through $\mathbf{0}$: two free choices, $\dim = 2$.
• All of $\mathbb{R}^3$: three free choices, $\dim = 3$.

The subspaces of $\mathbb{R}^3$ are exactly the subspaces of dimension $0, 1, 2, 3$ — the geometry and the counting are the same fact, finally named. A subspace's dimension is its "number of independent directions," and it can never exceed the dimension of the space it lives in.