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In English please?

  1. Apr 21, 2004 #1
  2. jcsd
  3. Apr 22, 2004 #2
    I don't think you really want to know.

    cookiemonster
     
  4. Apr 22, 2004 #3

    ahrkron

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    A bilinear form is basically an object that "eats" two vectors and spits out a number. Imagine that the number produced is a "grade" that Mr. Bilinear Form gives to the taste of the vector combination.

    The kernel is formed by all those vectors that totally kill the taste of the combination: when Mr. Form eats one of those, the combination gets a zero, no matter what the second vector is.
     
  5. Apr 22, 2004 #4
    I don't get what u just said ahrkron, but thanks anyway.
     
  6. Apr 22, 2004 #5
    Well, I'll try to simplify Ahrkron's "food" analogy:

    Suppose I'm called a "billinear form". What I do is eat two different bits of food (these are the vectors) at the same time and then decide how much I like that particular combination and give it a rating (these are the real numbers). So, suppose I eat sushi and schnitzel at the same time. I like it a lot, so I give it a 106. Another time I eat pretzels and haggis at the same, but it's disgusting, so I give it a -34560. So those combinations I like better get a rating closer to positive infinity, and those combinations I like worse get a rating closer to negative infinity. The combinations that are right in the middle, the ones that I don't like or dislike, I give a rating of zero.

    Suppose what I want to do now is find all those food combinations which I don't really like, but I don't really dislike either. I look through the ratings I gave the combinations and pick out the ones which have got a rating of 0. This is what the "kernel" is.
     
  7. Apr 22, 2004 #6

    chroot

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    And no, it's definitely not Algebra 2.

    - Warren
     
  8. Apr 22, 2004 #7

    honestrosewater

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    Just because I like making trouble for myself, what makes the combination=0? From your analogy, the output of the bilinear form is a scalar- not a vector, so the kernel is not just vectors that cancelled from addition.?
    Guess I should look this up, but I wonder if my reasoning is right...
    Rachel
     
  9. Apr 22, 2004 #8

    matt grime

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    you aren't adding together the vectors in the algebraic sense of addition of vectors

    A bilinear form on R^n, say, is given by an nxn matrix A, and the operation is b(x,y) = x^tAy for x,y column vectors.

    stevo's interpretation of the kernel using the food analogy (which I dont' think I like, as it happens) is not accurate. If we must, then sometihng is in the kernel iff when we combine it with any other food substance, the answer is zero.

    something, v say, is in the kernel of b(?,?) iff b(v,?) is zero for all possible choices of ? (note we are assuming symmetric so that this definition yields the same set as b(?,v) would do.

    by fixing v we turn b into a linear map. the kernel is then the v that make this linear map the zero map.
     
    Last edited: Apr 22, 2004
  10. Apr 22, 2004 #9

    honestrosewater

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    Okay, I've jumped too far ahead. I understand part of what you're saying. I think I get the kernel part, but not the bilinear form part ;) I haven't read enough about vector spaces, matrices, or rings (for shame) to even think of an example of a bilinear form kernel- I can't see how the kernel can contain more than one v! Oh, so many questions, but, whatever, I'll work on it.
    Thanks for trying
    Rachel
     
  11. Apr 22, 2004 #10

    matt grime

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    Let b be the form on R^n given by B(v,w) = 0 for all W. The kernel is then all of R^n. Referred to the standard basis the form given by A = diag{1,1,1,1...1,0,0..,0} with p ones and q zeros has q-dimensional kernel. (you don't need to learn anything about rings or modules, it's simple (bi)linear algebra).
     
  12. Apr 22, 2004 #11

    honestrosewater

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    Okay, if you're willing, I'm able. Of the above, "= 0" was about all I understood. I just "learned" today what a vector space is (for shame;), so I will have to look up all of this and get past the language barrier.

    Matrix A has only p 1s, and all the rest of the elements, (p*q)-p, are 0s.
    The "n" in R^n and "q" in q-dimensional refer to the number of rows in the matrix.
    If both of those statements are correct, I kind of get what you're saying. Even if "n" and "q" refer to "n*n" and "q*q" matrices I still kind of get what you're saying. (Which shows how much I get it.)

    The biggest problem I have now is that I can't figure out how to transform a ?linear map? into a matrix- or whatever gets transformed into a matrix.
    I see the relationship between
    1)the ordered pair (x, y) from the cartesian product of the two domain vector spaces, say X and Y, AND the (row, column) coordinates of the matrix.
    2)the value of the function at (x, y) AND the element at (row, column).
    But my grasp of everything else is so tenuous, I can't get much of anywhere. For instance, I read that the range must be the base field of the two vector spaces, and I can recall the definitions, but the statement has no significance yet- it's of no consequence. Perhaps that's my biggest problem.
    I'm still working
    Rachel
     
  13. Apr 22, 2004 #12

    matt grime

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    diag{ } is a standard way of saying diagonal matrix with these diagonal entries and zero elsewhere.

    the n means the dimension of the vector space, and the linear maps are then nxn matrices.

    the kernerl of a bilinear map is not a subset of nxn matrices, it is a subset of the vector space the bilinear form acts on.
     
  14. Apr 22, 2004 #13

    honestrosewater

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    Okay, I need something more meaningful.
    So, vector space R^3 is the set of all 3-tuples (x, y, z) where x, y, z, are in R?
    Do I need to say R^3 over R?
    The linear maps are then
    x, 0, 0,
    0, y, 0,
    0, 0, z,
    or is this the basis? I have no idea.
    Rachel
     
  15. Apr 22, 2004 #14

    honestrosewater

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    Okay, must call it a day. But I'm not giving up.
    Rachel
     
  16. Apr 22, 2004 #15

    matt grime

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    those are the diagonal linear maps wrt that basis. the linear maps are all 3x3 matrices.
     
  17. Apr 22, 2004 #16

    honestrosewater

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    As in
    x_1, x_2, x_3,
    y_1, y_2, y_3,
    z_1, z_2, z_3,
    or
    x, y, z,
    z, x, y,
    y, z, x,
    or am I thinking of this the wrong way? I really should just sit down with a book and start from vector spaces ;)
    Rachel

    EDIT- What I mean is, the n-tuples are ordered, how does that order transfer to the matricies?

    EDITEDIT- oh, I think I see, x is a column, y is a column, z is a column?
     
    Last edited: Apr 22, 2004
  18. Apr 22, 2004 #17

    matt grime

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    it doesn't appear to be correct. you really ought to stop writing x,y,z for everything in sight as that is suggestive in the wrong direction.
     
  19. Apr 23, 2004 #18

    honestrosewater

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    If X and Y are vector spaces over R, b(x,y)=x*y, then the kernel of B:X*Y->R is x=y=0, b(0,y) and b(x,0), i.e., the x and y axes of the cartesian plane?
     
  20. Apr 24, 2004 #19

    matt grime

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    That is only true if X and Y are 1-dimensional, where did they come from, as you don't use them in the second part of the post? How do you mulitiply two vectors x and y using a * operation?

    a bilinear form on V (one vector space, say) is a pairing from VxV to R (if they are real vector spaces, let's not get too complicated).


    so it takes two vectors and in a bilinear way yields a real number.

    bilinear means b(x+y,z) = b(x,z)+b(y,z) and b(kx,y)=kb(x,y) for all vectors x,y,z in V and all real numbers k, and this is true in second component too.

    Examples: the trivial bilinear form b(x,y) = (x^t)y (think of them as column vectors). The kernel of this is 0.

    this is just the usual inner product on V. inner products are the most obvious kind of bilinear form.

    Here's another one.

    b(x,y) = x_1y_1, that is just the product of the first components of the vectors. the kernel of this is then the span of the other basis vectors (all but the first one).

    Any bilinear form can be represented as a matrix, A with b(x,y) = (x^t)Ay

    write that out and do a few examples by picking V as R^2 and just choosing A at random.
     
  21. Apr 24, 2004 #20

    honestrosewater

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    I am making a mistake somewhere- perhaps this will help show where.

    I meant them to be one-dimensional, I thought that was understood. ?b(x,y) x is in X and y is in Y?

    I meant XxY, the same thing you mean by "VxV"- I wasn't sure what notation to use.
    Oh, you say "vectors" so you mean how do I do b(x,y)=x*y? I don't know why that's wrong.

    Here's what I don't get. I have the rules for vector spaces, and *think* I understand them- vector addition and scalar multiplication.
    Now, (s is scalar, v is vector)
    s_1*v_1 + s_2*v_2 + ... + s_n*v_n
    is a linear combination, right?
    But I don't see how or why you combine them in this way. Why do you add all the scalar products together? :confused: :mad: :confused:
    I know I need to read more, and I'm getting a book today.
    Rachel
     
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