📊 Section 1 — Interactive Simulation
Controls
Vectors
Display
💡 Section 2 -- The Idea, Step by Step
Imagine giving directions on a flat map using only two moves: "go east" and "go north." By taking some amount of each, you can reach any spot on the map. That whole collection of reachable spots is what mathematicians call a span; the moves are vectors (arrows that say "go this far, this way"); and the entire playground of arrows you are allowed to add and stretch is a vector space.
A vector is just an arrow. You are allowed to do two things with arrows: add them tip-to-tail, and scale them (make them longer, shorter, or flip them around). If your two arrows point in genuinely different directions, mixing them lets you reach every point on the plane. But if they point the same way -- one arrow is just a stretched copy of the other -- then no matter how you combine them you can only ever slide back and forth along a single line.
Call the arrows $\mathbf{v}_1$ and $\mathbf{v}_2$, and call the scaling numbers $c_1$ and $c_2$. A linear combination is $c_1\mathbf{v}_1+c_2\mathbf{v}_2$, and the span is the set of all of them. Try $\mathbf{v}_1=(2,0)$ and $\mathbf{v}_2=(0,3)$: the combination $1\cdot\mathbf{v}_1+1\cdot\mathbf{v}_2=(2,3)$, and because the arrows point in different directions you can land on any point in the plane, so they span $\mathbb{R}^2$. The parallelogram they make has area $|\det|=|2\cdot3-0\cdot0|=6$, and that nonzero area is the signal that the two arrows are genuinely different.
Vectors are linearly independent when $c_1\mathbf{v}_1+c_2\mathbf{v}_2=\mathbf{0}$ forces $c_1=c_2=0$: the only way to combine them and get nowhere is to not move at all. A basis is a set that is both independent and spanning -- the smallest collection of arrows that still reaches everything. The number of arrows in a basis is the dimension ($\dim\mathbb{R}^2=2$). For two arrows in the plane, $|\det[\,\mathbf{v}_1\ \mathbf{v}_2\,]|$ equals the parallelogram's area and is zero exactly when the arrows are dependent. In the sim, the angle sliders set each arrow's direction and the length sliders set $|\mathbf{v}_1|$ and $|\mathbf{v}_2|$; watch $\det$ slide toward $0$ as the two angles close in on each other.
Set both angle sliders equal (or to 0° and 180°): $\det\to0$, the dim(span) readout drops to 1, and the blue fill collapses to a line. Then switch the preset to "Rotated 45 deg" -- the arrows are independent again, the area returns, and the dimension climbs back to 2. Finally, shrink $|\mathbf{v}_2|$ to its minimum: the area gets small but stays nonzero, proving that shortening an arrow never destroys independence -- only matching directions does.
📐 Section 3 -- Vector Spaces, Basis & Dimension
Dimension Theorem: All bases of a finite-dimensional vector space have the same number of elements (the dimension).
Rank-Nullity: For $T:V\to W$: $\dim V=\text{nullity}(T)+\text{rank}(T)$.
| Concept | Definition | Example in $\mathbb{R}^2$ |
|---|---|---|
| Subspace | Closed under $+$ and scalar mult.; contains $\mathbf{0}$ | Lines through origin |
| Span | All linear combinations of a set | $\text{span}\{(1,0),(0,1)\}=\mathbb{R}^2$ |
| Lin. indep. | $c_1\mathbf{v}_1+c_2\mathbf{v}_2=\mathbf{0}\Rightarrow c_i=0$ | $(1,0),(0,1)$ are LI |
| Basis | LI spanning set | $\{(1,0),(0,1)\}$ |
| Dimension | Size of any basis | $\dim\mathbb{R}^2=2$ |
Any two bases $B_1,B_2$ of $V$ have $|B_1|=|B_2|$. Proof: $B_2$ spans, so $B_1$ has no more elements than $B_2$ (by the Steinitz exchange lemma). By symmetry $|B_1|=|B_2|$.
$T:V\to W$ linear, $\dim V=n$. Then $n=\text{nullity}(T)+\text{rank}(T)$. The null space (kernel) and image (column space) together account for all dimensions of $V$.
$\ker A$: solve $x+2y=0\Rightarrow x=-2y$. Basis: $\{(-2,1)\}$, dim 1. Rank $=2-1=1$. Column space: $\text{span}\{(1,2)\}$. Rank-nullity: $2=1+1$ ✓.
The area of the parallelogram spanned by $\mathbf{v}_1,\mathbf{v}_2\in\mathbb{R}^2$ equals $|\det[\mathbf{v}_1|\mathbf{v}_2]|$. Zero area $\iff$ dependent vectors. The simulation shows this live.
Row reduce $[\mathbf{v}_1|\mathbf{v}_2|\cdots]$ and count pivots. Number of pivots $=$ rank $=\dim(\text{span})$. If rank $<$ number of vectors, they are dependent.
$\mathcal{P}_n$ = polynomials of degree $\leq n$ is a vector space of dimension $n+1$. Basis: $\{1,x,x^2,\ldots,x^n\}$. The space of continuous functions $C[a,b]$ is infinite-dimensional (no finite basis).
❓ Section 4 -- FAQ
No. The zero matrix is not invertible, so the set does not contain the zero vector (violating axiom 1). Also, $I+(-I)=0$ is not invertible -- not closed under addition. The invertible matrices form a group under multiplication, not a vector space.
Key takeaway: Invertible matrices do NOT form a subspace; zero matrix and closure under addition both fail.The parallelogram with edges along $\mathbf{v}_1$ and $\mathbf{v}_2$ represents their span. When vectors are independent, it fills a 2D region. When dependent (parallel), it collapses to a line. The |det| equals the parallelogram area -- zero iff dependent.
Key takeaway: LI vectors span a 2D region; dependent vectors span only a line. |det| = area of parallelogram.Signal processing: signals are in $L^2$ (infinite-dimensional); Fourier modes are a basis. ML: word embeddings are vectors in $\mathbb{R}^{300}$; PCA finds a lower-dimensional subspace. Quantum mechanics: state space is a Hilbert space. Computer graphics: homogeneous coordinates for 3D transformations.
Key takeaway: Fourier analysis, word embeddings, quantum states, PCA -- all linear algebra in abstract vector spaces.Yes -- infinitely many. $\{(1,0),(0,1)\}$ and $\{(1,1),(1,-1)\}$ are both bases for $\mathbb{R}^2$. All have the same size (dimension). Orthonormal bases are preferred for numerical stability. Any $n$ linearly independent vectors in an $n$-dimensional space form a basis.
Key takeaway: Infinitely many bases exist for any space; all have the same cardinality (dimension).Row reduce the matrix with these as rows. Last row becomes zero -- rank 2, not 3. The third vector $= v_1+v_2$ (dependent). These three vectors span only a 2D subspace of $\mathbb{R}^3$. NOT a basis.
Key takeaway: Row reduce to count pivots = rank = dim(span). Three vectors with rank 2 are dependent.$\mathbb{R}$ viewed as a $\mathbb{Q}$-vector space has an uncountable Hamel basis -- every real number is a finite $\mathbb{Q}$-linear combination of basis elements. Its existence requires the Axiom of Choice and cannot be explicitly constructed. It enables pathological solutions to the Cauchy functional equation $f(x+y)=f(x)+f(y)$ that are not $f(x)=cx$.
Key takeaway: Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$: uncountable, requires Axiom of Choice, enables pathological functions.