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Linear Algebra

Vectors

This chapter systematically studies the basic concepts of vectors, linear combinations, linear dependence, vector spaces, bases and dimension, inner product, and orthogonalization.

Basic Concepts of Vectors

  • n-dimensional vector: a=(a1,a2,,an)\vec{a} = (a_1, a_2, \dots, a_n)
  • Vector addition, scalar multiplication

Linear Combination and Linear Dependence

  • Linear combination: k1a1+k2a2++kmamk_1\vec{a}_1 + k_2\vec{a}_2 + \dots + k_m\vec{a}_m
  • Linear dependence: there exist coefficients, not all zero, such that the combination is zero
  • Linear independence: only when all coefficients are zero does the combination equal zero

Maximal Linearly Independent Set and Rank

  • Maximal linearly independent set: the largest linearly independent subset of a set of vectors
  • Rank of a set of vectors: the number of vectors in a maximal linearly independent set

Vector Spaces, Bases, and Dimension

  • Vector space: a set closed under addition and scalar multiplication, satisfying certain rules
  • Basis: a linearly independent set that can uniquely represent any vector in the space
  • Dimension: the number of vectors in a basis

Coordinate Transformation and Transition Matrix

  • Definition and computation of coordinate transformation and transition matrix

Inner Product and Orthogonalization

  • Inner product: ab=aibi\vec{a} \cdot \vec{b} = \sum a_i b_i
  • Schmidt orthogonalization method
  • Orthogonal basis, orthogonal matrix

Exercises

  1. Determine the linear dependence of the set {(1,2,3),(2,4,6),(1,0,1)}\{(1,2,3), (2,4,6), (1,0,1)\}.
  2. Find the standard basis of R3\mathbb{R}^3.
  3. Let a=(1,2,3),b=(2,1,0)\vec{a} = (1,2,3), \vec{b} = (2,1,0), compute ab\vec{a} \cdot \vec{b}.
  4. Use the Schmidt orthogonalization method to orthogonalize a1=(1,1,0),a2=(1,0,1)\vec{a}_1 = (1,1,0), \vec{a}_2 = (1,0,1).
  5. Find the rank of the set {(1,2,3),(2,4,6),(3,6,9)}\{(1,2,3), (2,4,6), (3,6,9)\}.
Reference Answers

1. Determine linear dependence

(2,4,6)=2×(1,2,3)(2,4,6) = 2\times(1,2,3), so they are linearly dependent.


2. Standard basis of R3\mathbb{R}^3

(1,0,0),(0,1,0),(0,0,1)(1,0,0), (0,1,0), (0,0,1)


3. ab\vec{a} \cdot \vec{b}

1×2+2×1+3×0=2+2+0=41\times2 + 2\times1 + 3\times0 = 2 + 2 + 0 = 4


4. Schmidt orthogonalization

b1=(1,1,0)\vec{b}_1 = (1,1,0)

b2=(1,0,1)(1,0,1)(1,1,0)(1,1,0)(1,1,0)(1,1,0)=(1,0,1)12(1,1,0)=(0.5,0.5,1)\vec{b}_2 = (1,0,1) - \frac{(1,0,1)\cdot(1,1,0)}{(1,1,0)\cdot(1,1,0)}(1,1,0) = (1,0,1) - \frac{1}{2}(1,1,0) = (0.5,-0.5,1)


5. Rank

The three vectors are linearly dependent, so the rank is 1.