Linear Combination and spanning Sets

Linear Combinations and Spanning Sets

Understanding the concepts is crucial for future topics.

Two Dimensions

  • Vector V = (3, 4) can be written as a sum of vectors:
    (3, 4) = (3, 0) + (0, 4)
  • Alternatively, it can be expressed as a linear combination:
    (3, 4) = 3(1, 0) + 4(0, 1)
  • The set {(1, 0), (0, 1)} spans \mathbb{R}^2.
  • Any point on the 2D plane can be reached by multiplying (1, 0) and (0, 1) by some constants.

Explanation

Multiplying the vector (0, 1) and (1, 0) by scalars, then adding them allows you to get to any point in the 2D plane. These two vectors span the whole plane.

Three Dimensions

  • Vector OP = (a, b, c) can be written as a linear combination:
    (a, b, c) = a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1)
  • The vectors (1, 0, 0), (0, 1, 0), and (0, 0, 1) (basis vectors) span \mathbb{R}^3.
  • These basis vectors are often referred to as i, j, k.

Explanation

By multiplying each of the basis vectors by a scalar, any vector in 3D space can be created. For example take vector (2, 3, 4), by multiplying i by 2, j by 3, and k by 4 we can create the desired vector.

Important Notes for \mathbb{R}^2

  • Any pair of non-zero, non-collinear vectors will span \mathbb{R}^2.
  • Collinear vectors lie on same line, if vectors are scalar multiples of each other then they are collinear. If the vectors are collinear, they will not span the plane because they will only produce vectors along that line.
  • Every vector in the plane can be expressed as a linear combination involving the pair of non-collinear vectors.

Planes in \mathbb{R}^3

  • There are many planes in \mathbb{R}^3 (e.g., XY-plane, YZ-plane, XZ-plane).
  • Think of a movie scene with laser lines representing vectors in a plane.

Question 9 (Textbook Example)

  • Given the set of vectors {(1, 0, 0), (0, 1, 0)} in \mathbb{R}^3.
  • This set spans the XY-plane because the z-component is always zero.
  • The vector (-2, 4, 0) can be written as a linear combination:
    (-2)(1, 0, 0) + (4)(0, 1, 0) = (-2, 4, 0)
  • The vector (3, 5, 8) cannot be a linear combination because the z-component is always zero.
  • Adding the vector (100, 110) to the set does not change the span because the z-component is still zero.

Graphical Example

  • Express the vector (8, 7) as a linear combination of (1, 2) and (2, 1).
  • (8, 7) = 2(1, 2) + 3(2, 1)
  • Graphically:
    • Multiply (1, 2) by 2 to get (2, 4).
    • Multiply (2, 1) by 3 to get (6, 3).
    • Adding these vectors head to tail results in the vector (8, 7).
  • Any combination of these vectors multiplied by scalar values allows endpoints anywhere on the grid to be reached.

Mathematical Example: Showing Linear Combination

  • Show that (-12, 20) is a linear combination of (-1, 2) and (4, -6).
  • Set up the equation:
    a(-1, 2) + b(4, -6) = (-12, 20)
  • Create two equations with two unknowns:
    • -a + 4b = -12
    • 2a - 6b = 20
  • Solve using elimination. Multiply the first equation by 2:
    • -2a + 8b = -24
  • Add the modified first equation to the second equation:
    2b = -4
    b = -2
  • Substitute b back into the first equation:
    -a + 4(-2) = -12
    -a - 8 = -12
    -a = -4
    a = 4
  • Concluding statement:
    4(-1, 2) - 2(4, -6) = (-12, 20)

Collinear Vectors

  • Two vectors are collinear if one is a scalar multiple of the other.
  • Example: (1, 2, 3) and (-2, -4, -6) are collinear.
  • Example: (2, 1) and (6, 3) are collinear.

Determining if Vectors are Collinear

  • Check if the ratios of corresponding components are equal.
  • Example: Is (6, -10, -21) collinear with (35, 9, -15)? The negatives don't matter in this case since they all have a negative.
    \frac{6}{-10} = \frac{-21}{35} = \frac{9}{-15}
    -\frac{3}{5} = -\frac{3}{5} = -\frac{3}{5}
  • Since the ratios are equal, the vectors are collinear.

Coplanar Vectors

  • Any two non-collinear vectors in \mathbb{R}^3 determine a plane.
  • There are an infinite number of planes in \mathbb{R}^3.
  • If three vectors lie on the same plane, they are coplanar.
  • Test to see if three vectors are coplanar is to see if they can be written as a linear combination of the vectors.

Example: Show Vectors Lie on the Same Plane

  • Given vectors u = (-1, 3, 4), v = (0, -1, 1), and w = (-3, 14, 7), show they lie on the same plane.
  • Write w as a linear combination of u and v:
    s(-1, 3, 4) + t(0, -1, 1) = (-3, 14, 7)
  • Create three equations:
    • -s + 0t = -3
    • 3s - t = 14
    • 4s + t = 7
  • From the first equation, s = 3.
  • Substitute s into the second equation:
    3(3) - t = 14
    9 - t = 14
    t = -5
  • Check if s and t satisfy the third equation:
    4(3) + (-5) = 7
    12 - 5 = 7
    7 = 7
  • Since all three equations are satisfied, the vectors are coplanar.
  • Concluding statement:
    3(-1, 3, 4) - 5(0, -1, 1) = (-3, 14, 7)

Example: Show Vectors Do Not Lie on the Same Plane

  • Prove that the vectors cannot be written as a linear combination.

  • Given vectors u, v, and w, try to express w as a linear combination of u and v.

  • Set up equations and use elimination.

  • Solve the resulting equations for the coefficients.

  • If the resulting coefficients do not satisfy all three equations, the vectors are not coplanar.

  • It will be shown through the derivation (seen in the video) that \frac{88}{9} \neq 16

  • Therefore the vectors are not coplanar.