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Solve this Maths question, or else, you aren't good enough for Oxbridge Watch

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    (Original post by james22)
    I've just finished my second year in my maths degree, and I've done loads of linear algebra. I have never heard of what you are talking about.


    You've never heard the term 'matrix rank'????


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    (Original post by LightBlueSoldier)
    You've never heard the term 'matrix rank'????


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    Yes. I know that the rank of the matrix gives what you are saying, but I've always associate the number of equations with the dimension of the matrix.
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    (Original post by james22)
    Yes. I know that the rank of the matrix gives what you are saying, but I've always associate the number of equations with the dimension of the matrix.
    Who taught you that?


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    (Original post by LightBlueSoldier)
    Who taught you that?


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    LOL!!!!
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    (Original post by LightBlueSoldier)
    Who taught you that?


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    Don't know, it's just how we do it. If you have n equations with a parameter involved where for some values of the parameter the matrix has rant<n, you still model it with an nxn matrix.
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    (Original post by james22)
    Yes. I know that the rank of the matrix gives what you are saying, but I've always associate the number of equations with the dimension of the matrix.
    (Original post by LightBlueSoldier)
    Who taught you that?


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    "The rank of a matrix is defined as (a) the maximum number of linearly independent column vectors in the matrix or (b) the maximum number of linearly independent row vectors in the matrix. Both definitions are equivalent."

    I think james22 is correct, unless he wasn't specific enough for you.
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    (Original post by james22)
    Don't know, it's just how we do it. If you have n equations with a parameter involved where for some values of the parameter the matrix has rant<n, you still model it with an nxn matrix.
    Well yes but you also have to deal with the cases where the matrix is not full rank...viewing a matrix of rank m<n as a matrix of rank n is a bizarre and wrong way of thinking.


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    (Original post by rayquaza17)
    "The rank of a matrix is defined as (a) the maximum number of linearly independent column vectors in the matrix or (b) the maximum number of linearly independent row vectors in the matrix. Both definitions are equivalent."

    I think james22 is correct, unless he wasn't specific enough for you.
    I'm guessing you don't know what linearly independent means.

    In any case, the equations he gave were

    x+y = 1

    2(x+y)=2

    Which always has an underlying matrix of rank 1...


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    And the other example he gave was ax+y=1, x+y=1 which is either determined with (0,1) or represented by a matrix of rank 1 with a =1. Again not full rank in any case.


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    (Original post by LightBlueSoldier)
    Well yes but you also have to deal with the cases where the matrix is not full rank...viewing a matrix of rank m<n as a matrix of rank n is a bizarre and wrong way of thinking.


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    I'm not viewing it as a matrix of rank n, but a matrix of dimension n. You are the one who wants to view it as a matrix which is smaller than the number of equations.
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    (Original post by LightBlueSoldier)
    And the other example he gave was ax+y=1, x+y=1 which is either determined with (0,1) or represented by a matrix of rank 1 with a =1. Again not full rank in any case.


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    It is determined by a dimension 2 matrix, rank doesn't matter here.
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    (Original post by LightBlueSoldier)
    And the other example he gave was ax+y=1, x+y=1 which is either determined with (0,1) or represented by a matrix of rank 1 with a =1. Again not full rank in any case.


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    You're not even a mathematician anymore.

    You can't sit with us.

    PS: Actually I do know what it means!!
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    (Original post by james22)
    It is determined by a dimension 2 matrix, rank doesn't matter here.
    It is represented by a dimension 2 matrix, not determined.

    You can think of it however you want, as long as you get correct answers. However, if you ever want to do more advanced stuff you will run into serious problems with your way of thinking.


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    (Original post by rayquaza17)
    You're not even a mathematician anymore.

    !
    I actually do quite a lot of maths still. Mostly numerical solutions to PDEs and some stochastic stuff but it still counts!
    We also do stuff with matrices and vectors but mostly full rank stuff (or at least we assume things to be full rank and then abuse maths to make it work )

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    (Original post by LightBlueSoldier)
    It is represented by a dimension 2 matrix, not determined.

    You can think of it however you want, as long as you get correct answers. However, if you ever want to do more advanced stuff you will run into serious problems with your way of thinking.


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    I meant represented.

    How do you represent a problem that may have redundant equations? Do you consider various cases and have a different matrix for each one, or do you have 1 nxn matrix (which is what I do)?
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    (Original post by LightBlueSoldier)
    I actually do quite a lot of maths still. Mostly numerical solutions to PDEs and some stochastic stuff but it still counts!
    We also do stuff with matrices and vectors but mostly full rank stuff (or at least we assume things to be full rank and then abuse maths to make it work )

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    Who is we?

    TBH I haven't studied that much linear algebra and I haven't done any work on matrix rank at uni; my knowledge is entirely based on two webpages I visited today!
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    Silence speaks louder than words.
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    (Original post by james22)
    I meant represented.

    How do you represent a problem that may have redundant equations? Do you consider various cases and have a different matrix for each one, or do you have 1 nxn matrix (which is what I do)?
    Generally you start with a big matrix and then split it up case by case. I think it's useful to get into a mindset of rank=equations as when you get into some of the applications where it is unclear what the natural rank and dimension should be you can generate a rank in various ways (amongst other situations: this is the first that came to mind)


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    (Original post by rayquaza17)
    Who is we?

    TBH I haven't studied that much linear algebra and I haven't done any work on matrix rank at uni; my knowledge is entirely based on two webpages I visited today!
    My company (well my particular division).


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    (Original post by LightBlueSoldier)
    My company (well my particular division).


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    So you get to use maths in an actual job?? :zomg:
 
 
 
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