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Determinant as Dot Product in R^2 Question

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data

    In R^2, vectors x = (x1, x2) and a = (a1, a2). For fixed a, det(a, x) is a scalar-valued linear function of the vector x. Thus it can be written as the dot product of x with some fixed vector w. Explain why w is perpendicular to a. Do not use an expression of w in terms of the components of a.


    2. Relevant equations
    Anything involving the dot product, cross product, or determinants I suppose.


    3. The attempt at a solution

    If you simply calculate the determinant, it's clear to see that w = (-a2, a1), and that this is perpendicular to a.

    I know that in R^3 that det(a, b, x) = (a x b) o x, and (a x b) is perpendicular to a and b, so in the original problem, saying that w is perpendicular to a may be some analog for that. I tried to gain some insight about why (a x b) is perpendicular to a and b follows from the properties of the determinant, but couldn't find anything useful.

    I also noticed that if we assume that w is perpendicular to x then: w o x = det(a, b) = 0, so x must be parallel to a, which means that w is perpendicular to a. I tried to generalize this, but couldn't come up with anything.

    I'm stumped. Thanks for your help, and sorry I have no idea how to use LaTex.
     
    Last edited: Feb 4, 2012
  2. jcsd
  3. Feb 5, 2012 #2
    It's just a shot in the dark, but maybe Cramer's rule helps?
     
  4. Feb 5, 2012 #3

    Deveno

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    we are given that det(a,x) = La(x) for some linear function La.

    in turn, that implies La is a linear functional on R2 (since it has a scalar as output), which means that La = <_,w>, for some vector w in R2.

    that is: det(a,_) = La = <_,w>.

    now, calculate La(a), using the LHS and RHS of the equation above, and what do you get?
     
  5. Feb 5, 2012 #4

    tiny-tim

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    welcome to pf!

    hi sharpie! welcome to pf! :smile:
    hint: so what is the value of that function at a ? :wink:
     
  6. Feb 5, 2012 #5
    If I am understanding both of you correctly, you are saying that det(a,a) must equal 0, so a o w = 0 meaning a and w are perpendicular. Doesn't that only prove that a and w are perpendicular when x = a though? Thanks for your help.
     
  7. Feb 5, 2012 #6
    You have that L( x ) = det ( a, x ) = < x, z > for some fixed z and any x. What is the condition for which det( a , x ) = 0? It's not *only* when x = a
     
  8. Feb 5, 2012 #7
    I am pretty unfamiliar with determinants, so I must be missing something. det( a, x ) = 0 whenever x is a constant multiple of a (in other words parallel?), or either a or x is zero, right? So then it can be shown that whenever x is parallel to a, w is perpendicular to a. But what about when x is not a multiple of a? Sorry, I must be missing something. Thanks.

    EDIT: Oh wait. It's enough to show that whenever x is parallel with a, <x, w> = 0, because it shows that whenever w is dotted with something parallel to a you get zero, so w must be perpendicular to a. Is that correct?
     
    Last edited: Feb 5, 2012
  9. Feb 5, 2012 #8

    Deveno

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    one DEFINITION of perpendicularity between a and w is <a,w> = 0

    (here <x,y> means an inner product, which in your case you may take to be the dot product in R2).

    but what is <a,w> equal to?

    next question: why do we know that det(a,a) = ___?

    for a 2x2 matrix, this is very easy.

    there can be LOTS of vectors perpendicular to w. so what? we're not interested in ALL the x such that <x,w> = 0, we're just interested in whether or not <a,w> = 0.

    now, det(a,x) = 0 can be true for other values of x besides x = a. for example, x = 0 works, too. but that's besides the point. we're not trying to discover every vector x that makes det(a,x) = 0. we're interested in a particular vector and its relationship to w.
     
  10. Feb 5, 2012 #9

    tiny-tim

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    hi sharpie! :smile:
    yes … the question tells you …
    … ie there exists w such that det(a,x) = x.w for all x …

    you have to find w, and 0 = det(a,a) = a.w :wink:
     
  11. Feb 5, 2012 #10
    Thanks everyone. Grr, I even said in the original post:
    "I also noticed that if we assume that w is perpendicular to x then: w o x = det(a, b) = 0, so x must be parallel to a, which means that w is perpendicular to a. I tried to generalize this, but couldn't come up with anything. "
    But I didn't think this was enough for some reason. I'm dumb, haha. Thanks again.
     
  12. Feb 5, 2012 #11

    Deveno

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    i think this should be:

    <w,x> = det(a,x) (we don't need to introduce an extra letter).

    and you're almost correct. it's rather hard to think of the 0-vector as parallel to anything, but it IS perpendicular to everything (kind of by default).

    and sure, geometrically speaking, if you visualize det as giving you the area of the parallelogram spanned by a and x, this parallelogram will have area 0 only if 1 side has length 0, or both sides are "pointing in the same direction".

    the point of the problem is just to see what is happening conceptually, instead of relying on computation (because then you have to determine a coordinate system, but the geometric fact should be "coordinate-independent"), or as mathematicians like to say, without recourse to a basis.
     
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