Oxford Linear Algebra: Basis, Spanning and Linear Independence

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  • Опубликовано: 19 сен 2024
  • University of Oxford mathematician Dr Tom Crawford explains the terms basis, spanning and linear independence in the context of vectors and vector spaces. Check out ProPrep with a 30-day free trial to see how it can help you to improve your performance in STEM-based subjects: www.proprep.uk...
    Test your understanding of the content covered in the video with some practice exercises courtesy of ProPrep. You can download the workbooks and solutions for free here: api.proprep.co...
    And here: api.proprep.co...
    You can also find several video lectures from ProPrep explaining the content covered in the video at the links below.
    Basis for R^n: www.proprep.co...
    Basis: www.proprep.co...
    Spanning: www.proprep.co...
    Linear Independence: www.proprep.co...
    As with all modules on ProPrep, each set of videos contains lectures, worked examples and full solutions to all exercises.
    Watch the other videos from the Oxford Linear Algebra series at the links below.
    Solving Systems of Linear Equations using Elementary Row Operations (ERO’s): • Oxford Linear Algebra:...
    Calculating the inverse of 2x2, 3x3 and 4x4 matrices: • Oxford Linear Algebra:...
    What is the Determinant Function: • Oxford Linear Algebra:...
    The Easiest Method to Calculate Determinants: • Oxford Linear Algebra:...
    Eigenvalues and Eigenvectors Explained: • Oxford Linear Algebra:...
    Spectral Theorem Proof: • Oxford Linear Algebra:...
    Vector Space Axioms: • Oxford Linear Algebra:...
    Subspace Test: • Oxford Linear Algebra:...
    The video begins with an intuitive example of a basis via the vector space of polynomials up to degree n. We then give the formal definition of a basis as a spanning set of linearly independent vectors.
    The terms spanning and linear independence are then formally defined with examples given for each. We also show the definition of linear independence is equivalent to showing that the only solution to a linear combination of the vectors being equal to zero is for all of the coefficients to be zero. Linear dependence is defined as the lack of linear independence, or when a vector in a set can be written as a linear combination of the other vectors in the set.
    Finally, we move on to a series of worked examples, beginning with several possible bases for the Cartesian plane R^2. We then look at examples of linearly independent and linearly dependent sets of vectors, and how to show this is the case. Finally, we construct two possible bases for 3D space R^3.
    Produced by Dr Tom Crawford at the University of Oxford. Tom is an Early-Career Teaching and Outreach Fellow at St Edmund Hall: www.seh.ox.ac....
    For more maths content check out Tom's website tomrocksmaths....
    You can also follow Tom on Facebook, Twitter and Instagram @tomrocksmaths.
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    Get your Tom Rocks Maths merchandise here:
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Комментарии • 16

  • @TomRocksMaths
    @TomRocksMaths  Год назад +4

    Check out ProPrep with a 30-day free trial to see how it can help you to improve your performance in STEM-based subjects: www.proprep.uk/info/TOM-Crawford

  • @SerhiyS-j6v
    @SerhiyS-j6v 4 месяца назад

    Almost never comment on YT, but this is very helpful for anyone trying to not only understand the theory of Linear Algebra but also its applications to some basic problems. Keep up the good work, Dr. Crawford!

  • @chemiflask7692
    @chemiflask7692 Год назад +5

    I love the youtube play button in the back, nice!

  • @azhar4057
    @azhar4057 Год назад +5

    Sir, Your works are too great

  • @RC32Smiths01
    @RC32Smiths01 Год назад +4

    Appreciate the information as always. Learning something new and insightful everytime man

  • @Irfankhan-zl4tx
    @Irfankhan-zl4tx 6 месяцев назад

    Sir your deserve standing at this place. Appreciate

  • @willbutplural
    @willbutplural Год назад +2

    Beautiful explanation. 100x better than Axler's version thank you sir

  • @SuperKiko112
    @SuperKiko112 Год назад +5

    You are the living example that looks can be deceptive

  • @Thiago-qp9xs
    @Thiago-qp9xs Год назад +3

    you should do a ita or ime exam, 2 of the hardest exams to get into universities in brazil

  • @copiryte9535
    @copiryte9535 Год назад +1

    Hey at 14:31 when trying to suppose that αi is not equal to 0 going from αi*vi=-a1v1-...-αi-1*vi-1+αi+1*vi+1-...-αn*vn to vi
    shouldn't it be vi=-1/αi(a1v1+...+αi-1*vi-1+αi+1*vi+1+...+αn*vn) instead of vi=-1/αi(v1+...+vi-1+vi+1+...+vn)?

  • @giuseppedipellegrino9252
    @giuseppedipellegrino9252 2 месяца назад

    Great!

  • @wernerhartl2069
    @wernerhartl2069 Месяц назад

    It needs a definition of vector addition, and multiplication by a scalar at the very beginning. Otherwise very nice.

  • @maxmasterton3724
    @maxmasterton3724 Год назад

    Where did you get your blackboard from?

  • @alhafiz-hk2bl
    @alhafiz-hk2bl 3 месяца назад

    Greet

  • @BernhardRiemann966
    @BernhardRiemann966 Год назад +1

    I'm in love with the shape of you
    We push and pull like a magnet do
    Although my heart is falling too