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Homework Help: Prove ||B(x,y)|| = ||(x,y)|| for all x,y in R^2 (Rotations in R^2)

  1. Feb 24, 2008 #1
    B = [cos[tex]\theta[/tex] -sin[tex]\theta[/tex]]
    ......[sin[tex]\theta[/tex] cos[tex]\theta[/tex]]

    for some [tex]\theta[/tex] in R[tex]^{2}[/tex].

    (a) Prove that || B(x,y) || = || (x,y) || for all (x,y)[tex]\in[/tex]R[tex]^{2}[/tex]

    Question: What does B(x,y) and (x,y) notation mean?
    I have a result that says

    Let B=[b[tex]_{ij}[/tex]] be an mxn matrix whose entries
    are real numbers and let e[tex]_{1}[/tex],...,e[tex]_{n}[/tex] represent the usual basis of R^n. If T(x) = Bx, x[tex]\in[/tex]R^n , then T is a linear function from R^n to R^m and T(e[tex]_{j}[/tex])=(b[tex]_{1j}[/tex],b[tex]_{2j}[/tex],...,b[tex]_{mj}[/tex], j = 1,2,...n

    Warning: Superscripts are not superscipts. They are supposed to be SUBSCRIPTS. Sigh.

    Can I use this?

    1. I am very new to this material
    2. I am stuck with the notation.
    3. Please answer my first question carefully. I can't answer the question unless I know what they are asking. :)

    Please help me. Thank You,
  2. jcsd
  3. Feb 25, 2008 #2
    Not to be a jerk, but I'm not sure you should be doing problems like this if you don't even know what a vector or at least an ordered pair is. Regardless, here's a short introduction to vectors via Wikipedia.

    http://en.wikipedia.org/wiki/Vector_(spatial [Broken])
    Last edited by a moderator: May 3, 2017
  4. Feb 25, 2008 #3


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    Better yet, read your text book. As a "last resort"(!) ask your teacher what those things mean. Surely whoever gave you that problem was assuming you already knew that B(x,y) means to multiply the matrix B by the (column) vector (x, y).
    Last edited by a moderator: Feb 25, 2008
  5. Feb 25, 2008 #4

    I know what a vector and ordered pair is. So what they are really saying is this:
    Bx=x, right? If so, that makes alot more since then the (x,y) notation to me.
  6. Feb 26, 2008 #5

    This was easy! Thanks for your kind replies. My problem was the notation, after that, it just follows very quickly.

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