Linear Combinations

The Definition of a Linear Combination

Linear combination

The vector v⃗\vec{v} can be written as a linear combination of the set of vectors {a⃗1,a⃗2,…,a⃗n}\{\vec{a}_1, \vec{a}_2, \ldots, \vec{a}_n\}:

v⃗=k1a⃗1+k2a⃗2+⋯+kna⃗n\vec{v} = k_1\vec{a}_1 + k_2\vec{a}_2 + \cdots + k_n\vec{a}_n

where k1,k2,…,knk_1, k_2, \ldots, k_n are real numbers called the coefficients of the linear combination.

Geometric interpretation

A linear combination is the operation of “mixing” a set of vectors in given proportions to obtain a new vector.

The Zero Linear Combination

When all the coefficients are zero, you get the zero vector:

0⃗=0⋅a⃗1+0⋅a⃗2+⋯+0⋅a⃗n\vec{0} = 0\cdot\vec{a}_1 + 0\cdot\vec{a}_2 + \cdots + 0\cdot\vec{a}_n

Linear Dependence and Linear Independence

The Definition of Linear Dependence

Linear dependence

Let a⃗1,a⃗2,…,a⃗m\vec{a}_1, \vec{a}_2, \dots, \vec{a}_m be a set of nn-dimensional vectors. If there exist real numbers k1,k2,…,kmk_1, k_2, \dots, k_m, not all zero, such that:

k1a⃗1+k2a⃗2+⋯+kma⃗m=0⃗k_1\vec{a}_1 + k_2\vec{a}_2 + \dots + k_m\vec{a}_m = \vec{0}

then the set of vectors {a⃗1,a⃗2,…,a⃗m}\{\vec{a}_1, \vec{a}_2, \dots, \vec{a}_m\} is called linearly dependent.

The Definition of Linear Independence

Linear independence

Let a⃗1,a⃗2,…,a⃗m\vec{a}_1, \vec{a}_2, \dots, \vec{a}_m be a set of nn-dimensional vectors. If only when k1=k2=⋯=km=0k_1 = k_2 = \dots = k_m = 0 do we have:

k1a⃗1+k2a⃗2+⋯+kma⃗m=0⃗k_1\vec{a}_1 + k_2\vec{a}_2 + \dots + k_m\vec{a}_m = \vec{0}

then the set of vectors {a⃗1,a⃗2,…,a⃗m}\{\vec{a}_1, \vec{a}_2, \dots, \vec{a}_m\} is called linearly independent.

An Equivalent Condition for Linear Dependence

Theorem1

The set of vectors {a⃗1,a⃗2,…,a⃗m}\{\vec{a}_1, \vec{a}_2, \dots, \vec{a}_m\} is linearly dependent if and only if at least one of its vectors can be written as a linear combination of the others.

Proof:

  • Sufficiency: if a⃗1=k2a⃗2+⋯+kma⃗m\vec{a}_1 = k_2\vec{a}_2 + \dots + k_m\vec{a}_m, then: (−1)a⃗1+k2a⃗2+⋯+kma⃗m=0⃗(-1)\vec{a}_1 + k_2\vec{a}_2 + \dots + k_m\vec{a}_m = \vec{0} in which the coefficients are not all zero, so the set is linearly dependent.

  • Necessity: if the set is linearly dependent, there exist k1,k2,…,kmk_1, k_2, \dots, k_m, not all zero, such that: k1a⃗1+k2a⃗2+⋯+kma⃗m=0⃗k_1\vec{a}_1 + k_2\vec{a}_2 + \dots + k_m\vec{a}_m = \vec{0} Assuming k1≠0k_1 \neq 0, we get: a⃗1=−k2k1a⃗2−⋯−kmk1a⃗m\vec{a}_1 = -\frac{k_2}{k_1}\vec{a}_2 - \dots - \frac{k_m}{k_1}\vec{a}_m

Methods for Testing Linear Dependence

Method 1: The determinant method

Theorem2

nn nn-dimensional vectors are linearly dependent if and only if the determinant they form is zero.

Example: Judge the linear dependence of the set of vectors {(1,2),(3,4)}\{(1,2), (3,4)\}.

Solution: ∣1234∣=4−6=−2≠0\begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = 4 - 6 = -2 \neq 0

so the set of vectors is linearly independent.

Method 2: The matrix rank method

Theorem3

The set of vectors is linearly dependent if and only if the rank of the matrix they form is less than the number of vectors.

Steps:

  1. Take the vectors as the rows (or columns) of a matrix
  2. Compute the rank of the matrix
  3. Compare the rank with the number of vectors

Example: Judge the linear dependence of the set of vectors {(1,2,3),(2,4,6),(1,0,1)}\{(1,2,3), (2,4,6), (1,0,1)\}.

Solution: Construct the matrix: (123246101)\begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 0 & 1 \end{pmatrix}

Reduce it to row echelon form by elementary operations: (1230000−2−2)\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & -2 & -2 \end{pmatrix}

The rank is 2, which is less than the number of vectors 3, so the set is linearly dependent.

The Maximal Linearly Independent Subset of a Set of Vectors

The Definition of a Maximal Linearly Independent Subset

Maximal linearly independent subset

A subset of a set of vectors is called a maximal linearly independent subset of that set if:

  1. the subset is linearly independent
  2. every vector of the set can be written as a linear combination of this subset

Properties of a Maximal Linearly Independent Subset

Theorem4
  1. Any two maximal linearly independent subsets of a set of vectors contain the same number of vectors
  2. Any two maximal linearly independent subsets of a set of vectors are equivalent

How to Find a Maximal Linearly Independent Subset

Steps:

  1. Arrange the vectors as the columns of a matrix
  2. Reduce the matrix to row echelon form by elementary row operations
  3. Take the original vectors corresponding to the non-zero rows of the echelon form

Example: Find a maximal linearly independent subset of {(1,2,3),(2,4,6),(1,0,1)}\{(1,2,3), (2,4,6), (1,0,1)\}.

Solution: Construct the matrix: (123246101)\begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 0 & 1 \end{pmatrix}

Reduce it to row echelon form: (1230000−2−2)\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & -2 & -2 \end{pmatrix}

Take the original vectors corresponding to rows 1 and 3: {(1,2,3),(1,0,1)}\{(1,2,3), (1,0,1)\}

The Rank of a Set of Vectors

The Definition of the Rank of a Set of Vectors

The rank of a set of vectors

The rank of a set of vectors equals the number of vectors in its maximal linearly independent subset.

Properties of the Rank of a Set of Vectors

  1. The inequality for the rank: r({a⃗1,a⃗2,…,a⃗m})≤min⁡(m,n)r(\{\vec{a}_1, \vec{a}_2, \dots, \vec{a}_m\}) \leq \min(m, n)
  2. The transitivity of the rank: if the set of vectors A can be written as a linear combination of the set of vectors B, then r(A)≤r(B)r(A) \leq r(B)
  3. The additivity of the rank: r(A∪B)≤r(A)+r(B)r(A \cup B) \leq r(A) + r(B)

Computing the Rank of a Set of Vectors

Method:

  1. Arrange the vectors as the columns of a matrix
  2. Compute the rank of the matrix
  3. The rank of the matrix is the rank of the set of vectors

Worked Examples

Applications in Physics

In physics, many physical quantities can be expressed as linear combinations of other physical quantities:

  • the resultant force: the vector sum of several forces
  • the resultant displacement: the vector sum of several displacements
  • the resultant velocity: the vector sum of several velocities

Applications in Computer Graphics

In 3D graphics, any vector can be written as a linear combination of the basis vectors:

v⃗=xi⃗+yj⃗+zk⃗\vec{v} = x\vec{i} + y\vec{j} + z\vec{k}


Exercises

Exercise 1

Decide whether the vectors a⃗=(1,2,3)\vec{a} = (1, 2, 3) and b⃗=(2,4,6)\vec{b} = (2, 4, 6) are linearly dependent.

linear dependencea set of vectorsthe coefficient test

Approach:

Check whether there are coefficients, not all zero, making the linear combination the zero vector.

Steps:

  1. Put k1a⃗+k2b⃗=0⃗k_1\vec{a} + k_2\vec{b} = \vec{0}

  2. k1(1,2,3)+k2(2,4,6)=(0,0,0)k_1(1, 2, 3) + k_2(2, 4, 6) = (0, 0, 0)

  3. (k1+2k2,2k1+4k2,3k1+6k2)=(0,0,0)(k_1 + 2k_2, 2k_1 + 4k_2, 3k_1 + 6k_2) = (0, 0, 0)

  4. Solving gives k1=−2k2k_1 = -2k_2, so a nonzero solution exists

Answer: the vectors a⃗\vec{a} and b⃗\vec{b} are linearly dependent.

Exercise 2

Express the vector c⃗=(1,2,3)\vec{c} = (1, 2, 3) as a linear combination of the vectors a⃗=(1,0,1)\vec{a} = (1, 0, 1) and b⃗=(0,1,1)\vec{b} = (0, 1, 1).

linear combinationsolving for the coefficientsa system of equations

Approach:

Solve the system of equations for the coefficients.

Steps:

  1. Put c⃗=xa⃗+yb⃗\vec{c} = x\vec{a} + y\vec{b}

  2. (1,2,3)=x(1,0,1)+y(0,1,1)=(x,y,x+y)(1, 2, 3) = x(1, 0, 1) + y(0, 1, 1) = (x, y, x + y)

  3. Solve the system: {x=1y=2x+y=3\begin{cases} x = 1 \\ y = 2 \\ x + y = 3 \end{cases}

  4. Check: 1+2=31 + 2 = 3, which holds

Answer: c⃗=1a⃗+2b⃗\vec{c} = 1\vec{a} + 2\vec{b}

Exercise 3

Judge the linear dependence of the set of vectors {(1,2,3),(2,4,6),(1,0,1)}\{(1,2,3), (2,4,6), (1,0,1)\}.

linear dependencethe matrix rank methodelementary operations

Approach:

Use the matrix rank method to judge the linear dependence.

Steps:

  1. Construct the matrix: (123246101)\begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 0 & 1 \end{pmatrix}

  2. Reduce it to row echelon form: R2−2R1R_2 - 2R_1: (123000101)\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \end{pmatrix}

    R3−R1R_3 - R_1: (1230000−2−2)\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & -2 & -2 \end{pmatrix}

  3. The rank of the matrix is 2, which is less than the number of vectors 3, so the set is linearly dependent.

Answer: the set of vectors is linearly dependent

Exercise 4

Judge the linear dependence of the set of vectors {(1,2),(3,4)}\{(1,2), (3,4)\}.

linear dependencethe determinant methodthe 2×2 determinant

Approach:

Use the determinant method to judge the linear dependence.

Steps:

∣1234∣=1×4−2×3=4−6=−2≠0\begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = 1 \times 4 - 2 \times 3 = 4 - 6 = -2 \neq 0

The determinant is not zero, so the set of vectors is linearly independent.

Answer: the set of vectors is linearly independent

Exercise 5

Find a maximal linearly independent subset of {(1,2,3),(2,4,6),(1,0,1)}\{(1,2,3), (2,4,6), (1,0,1)\}.

maximal linearly independent subsetelementary operationsrow echelon form

Approach:

Use the elementary-operations method to find a maximal linearly independent subset.

Steps:

  1. Construct the matrix: (123246101)\begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 0 & 1 \end{pmatrix}

  2. Reduce it to row echelon form: R2−2R1R_2 - 2R_1: (123000101)\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \end{pmatrix}

    R3−R1R_3 - R_1: (1230000−2−2)\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & -2 & -2 \end{pmatrix}

  3. Take the original vectors corresponding to the non-zero rows: {(1,2,3),(1,0,1)}\{(1,2,3), (1,0,1)\}

Answer: the maximal linearly independent subset is {(1,2,3),(1,0,1)}\{(1,2,3), (1,0,1)\}

Exercise 6

Find the rank of the set of vectors {(1,2,3),(2,4,6),(3,6,9)}\{(1,2,3), (2,4,6), (3,6,9)\}.

the rank of a set of vectorsthe rank of a matrixlinear dependence

Approach:

Compute the rank of the matrix.

Steps:

  1. Construct the matrix: (123246369)\begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{pmatrix}

  2. Reduce it to row echelon form: R2−2R1R_2 - 2R_1: (123000369)\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 3 & 6 & 9 \end{pmatrix}

    R3−3R1R_3 - 3R_1: (123000000)\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}

  3. There is only 1 non-zero row, so the rank is 1.

Answer: the rank of the set of vectors is 1


Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
a⃗i\vec{a}_iMathematical symbolVector a_iA vector in the set of vectors
0⃗\vec{0}Mathematical symbolZero vectorThe zero vector
kik_iVariablek_iA coefficient of the linear combination
r(A)r(A)Function symbolRank of AThe rank of the matrix AA
k1a⃗1+⋯k_1\vec{a}_1 + \cdotsLinear combinationlinear combinationA linear combination of the set of vectors

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