Linear Independence

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Lecture 10

• Linear Independence

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Lecture 10 Objectives

• Decide if a vector is a linear combination
  (belongs to the span) of a set of vectors
• Describe the span of a set of vectors
• geometrically
• algebraically
• Decide if a set of vectors is linearly dependent
  (and if it is, find a dependence relationship)

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Example

• Let v1 1,2,3, v2 2,?1,1, v3 ?3,2,?4.
• Can the vector u ?4, 10, 3 be written as x
  v1 y v2 z v3 , for some scalars x, y, z?
• Note If we conveniently write the vectors as
  column vectors, we can try solving

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Solution

• The vector equation

is equivalent to the matrix equation
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Solution

• Which is also equivalent to the system of
  equations (considered in Lecture 8)
• x 2y ? 3z ?4 2x ? y 2z
  10 3x y ? 4z 3
• which has the unique solution x 3,
  y ?2, z 1.
• Since we found a solution, the answer is yes.

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In general

• The system of linear equations
• a11x1 a12x2 ? a1nxn b1
• a21x1 a22x2 ? a2nxn b2
• ?
• am1x1 am2x2 ? amnxn bm
• which has the form Ax b, can also be expressed
  as

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Linear Combination

• Definition If the vector equation
• u c1v1 c2v2 ? cnvn
• holds, then we say that The vector u is
  a linear combination of the vectors v1, v2, ?,
  vn.
• Note We only use addition and scalar
  multiplication to get u from v1, v2,?, vn.

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Example

• Is the vector u 1, 2, 3 a linear combination
  of the vectors v1 4, 5, 6 and v2
  7, 8, 9?
• Interpret the answer geometrically.

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Solution

• We can solve the equation u xv1 yv2 to get
  x 2 and y ?1, or we can just check
  that 1, 2, 3 24, 5, 6 ? 7, 8, 9.
• Thus, u is indeed a linear combination of the
  vectors v1 and v2.
• Geometrically, this means that u lies in the
  plane through the origin extended by the vectors
  v1,v2.

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Example

• Is the vector u 1, 2, 3, 4 a linear
  combination of the vectors v1 0, 1, 1,
  1, v2 1, 0, 1, 0, v3 0, 1, 0,
  1.
• Interpret the answer geometrically.

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Solution

• We can check to see
• that the equation
• has no solution.
• Thus, u is NOT a linear combination of v1, v2,
  and v3.
• Geometrically, this means that u does not lie in
  the hyperplane through the origin extended by the
  vectors v1,v2, and v3.

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Span

• Definition The set of all linear combinations of
  the set of vectors S v1, v2, ?, vn is called
  span(S), i.e.
• span(S) c1v1 c2v2 ? cnvn c1,?,cn?R
• Notes span(S) is a set of vectors.
• It also represents the subset of the space that
  can be reached from the origin by possible linear
  combinations of the vectors.

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Example

• Let v1 0, 0, 1, v2 1, 0, 1,
  and v3 1, 1, 0.
• Geometrically describe the following
• span(v1) , span(v1,v2) , and span(v1,v2,v3).
• Solution span(v1) The z-axis.
• span(v1,v2) The x-z plane
• span(v1,v2,v3) R3

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Example

• Can the vectors v1 0, 0, 1, 1,
  v2 1, 0, 1, 0, v3 0, 1, 0, 1,
  span the whole 4 dimensional space R4, i.e.
  is any vector u x, y, z, w ? R4 a linear
  combination of the vectors v1,v2,v3?

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Linear Dependence

• Definition The vectors v1, v2, ?, vn are called
  linearly dependent iff at least one of them is a
  linear combination of the others.
  Otherwise, they are called linearly independent.
• Example Show by inspection that the vectors e1
  1, 0, 0, e2 0, 1, 0, e3
  0, 0, 1, are linearly independent.

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An Equivalent Definition

• Theorem The vectors v1, v2, ?, vn are linearly
  dependent iff there is a nonzero solution for the
  equation c1v1 c2v2 ? cnvn 0.
• Note A nonzero solution means values for the
  coefficients c1, c2, ?, cn that are not all
  zeros. Observe that c1 c2 ? cn 0 gives the
  so-called trivial (zero) solution.

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Example

• Are the vectors v1 1, 3, ?2,
  v2 1, 2, 0, v3 1, 0, 4,
• linearly dependent?
• If yes, express one of them as a linear
  combination of the others.

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Solution

• Using the theorem we try solving the equation
  c1v1 c2v2 c3v3 0, by getting the equivalent
  augmented matrices

Since there is a nonzero solution, e.g. c1 2, c2
?3, c3 1, the vectors are dependent. Also, we
get the relation 2v1 ?3v2 v3 0, i.e. v3 3v2
? 2v1.
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• Thank you for listening.
• Wafik