Vector Spaces & Bases

Study Sheet

Vector Spaces & Bases

Span, independence, basis, dimension

The Language of Vector Spaces

Tip
Span, independence, basis

The SPAN of a set is everything reachable by linear combinations. A set is INDEPENDENT when no member is a combination of the others. A BASIS is both: independent AND spanning — every vector then has exactly ONE coordinate representation, and every basis of a space has the same size: its DIMENSION.

Concept
Subspaces cut dimension by conditions

Each independent linear condition on Rn\mathbb R^n removes one dimension: the hyperplane x1+x2+x3+x4=0x_1 + x_2 + x_3 + x_4 = 0 in R4\mathbb R^4 has dimension 33. Solution sets of homogeneous systems ARE subspaces — this is where geometry and equations meet.

Example
Coordinates in a new basis

To write (7,3)(7,3) in the basis {(1,1),(1,−1)}\{(1,1), (1,-1)\}: solve a+b=7a + b = 7, a−b=3a - b = 3, giving a=5a = 5, b=2b = 2. Changing basis is solving a linear system — and orthogonal bases make it a pair of one-line dot products.

Side note
Significance: the same axioms, everywhere

Polynomials, audio signals, quantum states, and solutions of linear ODEs all form vector spaces — the span/basis/dimension language transfers wholesale. That is why "the solution space of y′′+y=0y'' + y = 0 is 2-dimensional with basis {cos⁡,sin⁡}\{\cos, \sin\}" is a sentence, not an analogy: the differential equations course runs on this chapter.

Try it
Try it: dimensions by conditions

What is the dimension of the space of 2×22\times2 matrices with trace 00? Work: the space of all 2×22\times2 matrices has dimension 44; trace zero is ONE linear condition; dimension 4−1=34 - 1 = 3. A basis: (100−1),(0100),(0010)\begin{pmatrix}1&0\\0&-1\end{pmatrix}, \begin{pmatrix}0&1\\0&0\end{pmatrix}, \begin{pmatrix}0&0\\1&0\end{pmatrix}.

Proofs & Why It Matters

Tip
Proof: coordinates in a basis are unique

Suppose v=∑aibi=∑cibi\mathbf v = \sum a_i\mathbf b_i = \sum c_i\mathbf b_i in a basis {bi}\{\mathbf b_i\}. Subtract: ∑(ai−ci)bi=0\sum (a_i - c_i)\mathbf b_i = \mathbf 0. Independence forces every coefficient to vanish: ai=cia_i = c_i. Existence comes from spanning, uniqueness from independence — a basis delivers both at once. ■\blacksquare

Tip
Proof sketch: every basis has the same size

The exchange argument: if {u1,…,um}\{\mathbf u_1, \ldots, \mathbf u_m\} is independent and {b1,…,bn}\{\mathbf b_1, \ldots, \mathbf b_n\} spans, insert the u\mathbf us one at a time into the spanning set, each time ejecting some b\mathbf b (writing the new u\mathbf u in the current spanning set must use a surviving b\mathbf b, which can then be solved for). The process never runs out of b\mathbf bs to eject, so m≤nm \le n. Applying this both ways to two bases gives m≤nm \le n and n≤mn \le m: dimension is well-defined. ■\blacksquare

Going Deeper: Explanations & Worked Problems

Concept
Independence testing is elimination in disguise

To decide whether vectors v1,…,vk\mathbf v_1, \ldots, \mathbf v_k are independent, ask whether c1v1+⋯+ckvk=0c_1\mathbf v_1 + \cdots + c_k\mathbf v_k = \mathbf 0 forces all ci=0c_i = 0 — but that is a homogeneous linear SYSTEM in the cic_i, with the vectors as columns.

Row reduce: if every column earns a pivot, only the zero combination works — independent; a pivotless column means a free variable, hence a nonzero combination summing to zero — dependent, and back-substitution hands you the actual dependence. The same computation delivers dimension: the pivot columns form a basis of the span, so dim⁡(span)=rank\dim(\text{span}) = \text{rank}. This is the great economy of the subject — span, independence, basis, dimension, and solvability are all read off ONE echelon form.

Example
Worked: dependence exposed, coefficients and all

Test (1,0,1),(0,1,1),(1,1,2)(1,0,1), (0,1,1), (1,1,2).

Step 1 — put them as columns and reduce: (101011112)→R3−R1:(101011011)→R3−R2:(101011000)\begin{pmatrix}1&0&1\\0&1&1\\1&1&2\end{pmatrix} \to R_3 - R_1: \begin{pmatrix}1&0&1\\0&1&1\\0&1&1\end{pmatrix} \to R_3 - R_2: \begin{pmatrix}1&0&1\\0&1&1\\0&0&0\end{pmatrix}.

Step 2 — read: pivots in columns 1, 2 only — DEPENDENT, dimension of the span is 22.

Step 3 — recover the dependence: the free column says c3c_3 is free; rows give c1=−c3c_1 = -c_3 and c2=−c3c_2 = -c_3; take c3=1c_3 = 1: −(1,0,1)−(0,1,1)+(1,1,2)=(0,0,0)-(1,0,1) - (0,1,1) + (1,1,2) = (0,0,0) ✓ — i.e. v3=v1+v2\mathbf v_3 = \mathbf v_1 + \mathbf v_2, the hidden relation, now explicit.

Step 4 — basis: keep the pivot columns {(1,0,1),(0,1,1)}\{(1,0,1), (0,1,1)\}; they span the same plane with nothing wasted.

Example
Worked: change of basis with an orthogonal shortcut

Write (7,3)(7, 3) in the basis b1=(1,1)\mathbf b_1 = (1,1), b2=(1,−1)\mathbf b_2 = (1,-1). THE SYSTEM ROUTE: a(1,1)+b(1,−1)=(7,3)a(1,1) + b(1,-1) = (7,3) means a+b=7a + b = 7 and a−b=3a - b = 3; adding, 2a=102a = 10, a=5a = 5; subtracting, b=2b = 2.

THE ORTHOGONAL SHORTCUT: because b1⋅b2=1−1=0\mathbf b_1\cdot\mathbf b_2 = 1 - 1 = 0, each coefficient is an independent projection: a=(7,3)⋅(1,1)(1,1)⋅(1,1)=102=5a = \dfrac{(7,3)\cdot(1,1)}{(1,1)\cdot(1,1)} = \dfrac{10}{2} = 5 and b=(7,3)⋅(1,−1)2=42=2b = \dfrac{(7,3)\cdot(1,-1)}{2} = \dfrac{4}{2} = 2 — no system solved, no interaction between coordinates. Check: 5(1,1)+2(1,−1)=(7,3)5(1,1) + 2(1,-1) = (7, 3) ✓. This is precisely why orthogonal bases (and eventually Fourier series) are prized: coordinates decouple.