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Math Help - Scalars

  1. #1
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    Scalars

    Having some LaTeX issues so this might look a bit off

    let e_1= (1,0) and e_2 = (0,1) Show that for every X  \in R^2, there exist unique scalars

    \alpha and \beta such that X = \alpha e_1 + \beta e_2

    would this be \alpha(1,0) + \beta(0,1) = ( \alpha, 0) + (0, \beta) = ( \alpha, \beta) = X

    I'll start there, there are two more parts to the question but I will see if I have this part first.
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  2. #2
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    Re: Scalars

    Quote Originally Posted by Jonroberts74 View Post
    Having some LaTeX issues so this might look a bit off

    let e_1= (1,0) and e_2 = (0,1) Show that for every X  \in R^2, there exist unique scalars

    \alpha and \beta such that X = \alpha e_1 + \beta e_2

    would this be \alpha(1,0) + \beta(0,1) = ( \alpha, 0) + (0, \beta) = ( \alpha, \beta) = X

    I'll start there, there are two more parts to the question but I will see if I have this part first.
    ok there's a quick answer but this is important enough for a detailed answer

    Your e_1 and e_2 form a basis for R^2

    given any point in R^2 how do you compute what it's coordinates are with respect to this basis. You take the inner product of it and each of the basis vectors.

    in this case <X, e_1> and <X, e_2> = \alpha and \beta

    then

    X = \alpha e_1+ \beta e_2
    Thanks from Jonroberts74
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  3. #3
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    Re: Scalars

    okay the second part is

    Let Y = (3, 0) and Z = (2, 0). Is it possible , for each vector X \in R^2, to find \alpha and \beta such that X equals \alphaY + \betaZ
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    Re: Scalars

    Quote Originally Posted by Jonroberts74 View Post
    okay the second part is

    Let Y = (3, 0) and Z = (2, 0). Is it possible , for each vector X \in R^2, to find \alpha and \beta such that X equals \alphaY + \betaZ
    ok, you don't just get an answer for this.

    why or why not would those two vectors serve as a basis for R2 ?
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  5. #5
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    Re: Scalars

    would it be No, because they are not linearly independent? they both run directly along the horizontal axis. Y = 2/3(z) They'd be in R^1
    Last edited by Jonroberts74; December 15th 2013 at 07:00 PM.
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  6. #6
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    Re: Scalars

    Yes. They aren't linearly independent so they cannot span R2 so the answer is no, you can't find alpha and beta for each vector in R2 such that X = alpha Y + beta Z

    err no about the Y = 2/3 Z axis.

    Both Y and Z are spanned by {1,0}. So any linear combination of Y and Z will still be spanned by {1,0}. These are the set of vertical vectors.
    Last edited by romsek; December 15th 2013 at 07:18 PM.
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  7. #7
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    Re: Scalars

    wait, I am confused now. My book is showing a cartesian plane but showing the first coordinate is the horizontal axis point and the second coordinate is the point of the vertical axis. In an illustration it is show x_1 and y_1 on the horizontal and x_2 and y_2 on the vertical. so by that, wouldn't Y then be equal to 2/3Z?

    and the last part is

    For Y = (1,2) and Z = (-4, -4) Find scalars alpha and beta such that X = alphaY +betaZ. Wouldn't they again just be scalar multiples of each other? so there would be none. Also does the fact that the second coordinate is merely the first doubled play any part into that?
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  8. #8
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    Re: Scalars

    the set of horizontal vectors I mean.
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  9. #9
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    Re: Scalars

    Quote Originally Posted by Jonroberts74 View Post
    wait, I am confused now. My book is showing a cartesian plane but showing the first coordinate is the horizontal axis point and the second coordinate is the point of the vertical axis. In an illustration it is show x_1 and y_1 on the horizontal and x_2 and y_2 on the vertical. so by that, wouldn't Y then be equal to 2/3Z?
    Yes ok, Y = 3/2 Z but that only further shows you that the two vectors don't span R2. I guess I'm not sure what's confusing you.
    Last edited by romsek; December 15th 2013 at 10:10 PM.
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  10. #10
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    Re: Scalars

    sorry

    meant Y=\frac{3}{2}Z \Rightarrow (3,0) = \frac{3}{2}(2,0) was getting myself all mixed up
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  11. #11
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    Re: Scalars

    Quote Originally Posted by Jonroberts74 View Post
    and the last part is

    For Y = (1,2) and Z = (-4, -4) Find scalars alpha and beta such that X = alphaY +betaZ. Wouldn't they again just be scalar multiples of each other? so there would be none. Also does the fact that the second coordinate is merely the first doubled play any part into that?
    I'm a bit pressed for time at the moment so I'm going to refer you to a page that is on what this problem is about.

    Change of Basis - HMC Calculus Tutorial
    Thanks from Jonroberts74
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  12. #12
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    Re: Scalars

    Quote Originally Posted by romsek View Post
    I'm a bit pressed for time at the moment so I'm going to refer you to a page that is on what this problem is about.

    Change of Basis - HMC Calculus Tutorial
    what your problem is about is changing the basis from the column vectors e1={1,0} and e2={0, 1} to the column vectors v1={1, 2} and v2={-4, -4}.

    I guess the first thing to do is ensure that we can use v1 and v2 as a basis. These two vectors must span R2 and thus be linearly independent.

    The easiest way to verify this is to create a matrix of the column vectors v1 and v2 and find it's determinant. If it is non-zero then the two column vectors are linearly independent.

    Here we get V = {{1,-4},{2,-4}} and det(V) = 4. Thus v1 and v2 are linearly independent and can be used as a basis for R2

    Your original X was given in the first basis as X=a e1 + b e2

    for the last part of your problem you want to express X in the new basis {1,2} {-4,-4} as some c v1 + d v2

    You can write this all up using matrices of column vectors.

    let E = {{1,0},{0,1}} i.e. made up of the column vectors e1 and e2. And let u = {a, b}

    Then X = a e1 + b e2 = Eu

    Similarly we can express the new basis as

    V={{1,-4},{2,-4}}

    and we'd like to find the components of w = {c, d} such that X = Vw, i.e. we want to express X in this new basis.

    X = Vw = Eu

    w = V^{-1}Eu=V^{-1}u

    we can find the inverse of V easily enough

    for a 2x2 matrix M={{m11, m12}, {m21, m22}} the inverse is given by 1/det(M){m22, -m12},{-m21,m11}}

    so the inverse of V is 1/4{{-1,1},{-1/2,1/4}}

    and our w = V^{-1}u is given by

    1/4{{-1,1},{-1/2,1/4}}{a,b} = {-a+b, -a/2+b/4} and so c = -a+b, d = -a/2 + b/4

    In other words X =c v1 + d v2 as you were tasked to find.

    You can verify this

    (-a+b) v1 = {(-a+b),2(-a+b)} = {-a+b, -2a+2b }

    (-a/2 + b/4) v2 = {-4(-a/2+b/4),-4(-a/2+b/4)} = {2a-b, 2a-b}

    and summing we get

    {-a+b, -2a+2b } + {2a-b, 2a-b} = {a, b} as expected (or hoped for anyway )

    If your original basis wasn't E, and thus couldn't be just discarded in that matrix multiply to find w, then you'd just include it in that multiplication. You'd have it, it's just the column vectors of whatever basis you are going to originally describe X in.

    See if any of this makes sense.
    Last edited by romsek; December 15th 2013 at 11:26 PM.
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  13. #13
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    Re: Scalars

    sorry meant (1,2) and (-4,-8). I was clearly not at my best last night.

    also e_1 and e_2 are not related to (1,2) and (4,-8) in terms of the question. it was really more like three separate questions in one
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  14. #14
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    Re: Scalars

    well it's a quick write up on change of basis anyway. You can just apply everything in here to (1,2) (-4,8)
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