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Linear Algebra
· Formula🧮 Formula Reference Sheet
Vector space: closed under + and scalar × Linear combination: Σ cᵢvᵢ Linear independence: only c=0 gives Σ cᵢvᵢ = 0 Basis: smallest spanning independent set Dimension: number of basis vectors Eigenvector: Av = λv
⚡ Example per formula
Vector space: closed under + and scalar ×
↳ℝⁿ is closed under + and scalar ×
Linear combination: Σ cᵢvᵢ
↳2v₁ + 3v₂ in span of v₁, v₂
Linear independence: only c=0 gives Σ cᵢvᵢ = 0
↳(1,0), (0,1) are independent in ℝ²
Basis: smallest spanning independent set
↳standard basis of ℝ³: (1,0,0), (0,1,0), (0,0,1)
Dimension: number of basis vectors
↳dim(ℝ³) = 3
Eigenvector: Av = λv
↳A·v = 5v → v is eigenvector, λ=5
✏️ Worked Example
Are (1, 2) and (2, 4) linearly independent?
👁 Show step-by-step solution
(2, 4) = 2 × (1, 2). They are linearly dependent. ✅ Answer: No (dependent)
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