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Baisc linear algebra

  1. Aug 13, 2014 #1
    Why do we need fields (Why do we define fields?)for linear algebra?
  2. jcsd
  3. Aug 13, 2014 #2


    Staff: Mentor

    Here's a discussion on fields:


    The short answer is that in mathematics, mathematical systems are classified as groups, rings and fields. Knowing what a mathematical system is helps us understand how to work with it. Is is distributive? Is it commutative? Is it closed?

    In mathematics it helps to understand how mathematical systems behave under addition and multiplication knowing that its a field means knowing the rules that apply like whether addition or multiplication are commutative or distributive or associative...

    In the case of linear algebra we are working with matrices of numbers adding and multiplying them in specific ways and so its important to know how they will work under commutativity for example.


    and matrix multiplication is anti-commutative vs real number multiplication which is commutative.
  4. Aug 13, 2014 #3
    We don't. You can formulate linear algebra perfectly without ever using the notion or even the concept of a field. Most introductory courses do things this way and assume that all scalars are in ##\mathbb{R}##.

    However, this becomes awkward in the long run. Indeed, we can also do linear algebra with scalars in ##\mathbb{C}## and this is incredibly useful. And almost all of the theorems that hold in the ##\mathbb{R}## case hold also in the ##\mathbb{C}## case (with some notable exceptions!).

    And then if you go further in math, you might want to restrict the scalars to ##\mathbb{Q}##, to ##\mathbb{Q}[\sqrt{2}]## or even to finite sets such as ##\{0,1\}## with some weird addition and multiplication. A lot of theorems in ##\mathbb{R}## still hold in these cases with virtually the same proofs!

    So instead of giving the same results and proofs over and over again, it is easier to just abstract things and work in a general field instead. In practical applications however, we will always make the field concrete and choose to work in ##\mathbb{R}##, ##\mathbb{C}## or something else.
  5. Aug 13, 2014 #4
  6. Aug 13, 2014 #5


    Staff: Mentor

    Yes, my mistake its just not commutative ie if A and B are two matrices then AB =/= BA unless one is the identity matrix.

    I guess I'm getting too smart for my own good.
  7. Aug 13, 2014 #6


    User Avatar
    Science Advisor

    That happened to me long ago. Although, for some reason, my friends don't see it that way!
  8. Aug 13, 2014 #7


    Staff: Mentor

    Must be an age thing. I retired once maybe I need to retire again. :-)
  9. Aug 13, 2014 #8
    Sorry to bother you again, but I think this statement is also not accurate. Surely it is possible for two matrices to commute even if both are not the identity?
  10. Aug 13, 2014 #9


    Staff: Mentor

    Thanks again...

    Now I know why I liked Physics and computers... in physics you can be fuzzy with your math and in computers any mistakes can be blamed on them.
  11. Aug 13, 2014 #10


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    Science Advisor
    Gold Member

    Sorry, but diagonal matrices actually commute with all matrices.
  12. Aug 14, 2014 #11
    I was just reading some coding theory (has to do with some very tenuous job lead that I'm pursuing), and apparently, it's useful to make your codes into a vector space over finite fields. So, there are actually practical applications for other fields.

    Here's the reference, if you're interested.


    Also, there's probably some use in cryptography, I would guess. Various different subfields of the complex numbers were historically the reason why they were introduced, in Galois theory, which answers questions about the solubility of polynomial equations. In that context, it's a little more natural because you just want to consider sets that are closed under multiplication, addition, and inverses.
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