Results 1 to 3 of 3

Thread: A mathematical problem related to standard basis vector and standard matrix

  1. #1
    Senior Member x3bnm's Avatar
    Joined
    Nov 2009
    Posts
    308
    Thanks
    16

    A mathematical problem related to standard basis vector and standard matrix

    I am having difficulty in understanding the standard basis and standard matrix.

    I'll explain my problem through a math problem. (My question and all the geometric diagrams at the end of the post)


    The problem:
    Suppose let l be the line in the xy-plane that passes through the origin and makes an angle $\displaystyle \theta$ with
    the positive x-axis, where $\displaystyle 0 \leq \theta < \pi$. As illustrated in Figure a. , let $\displaystyle T:R^2 \rightarrow R^2$
    be a linear operator that maps each vector into its orthogonal projection on l.

    Find standard matrix for T.
    -------------------------------------------------------------------------------------

    The book i am reading did it like this:

    We know that:

    $\displaystyle [T] = [T(e_1) | T(e_2)] $

    Where $\displaystyle e_1$ and $\displaystyle e_2$ are the standard basis vectors for $\displaystyle R^2$. We consider the case where
    $\displaystyle 0 \leq \theta \leq \frac{\pi}{2}$; the case where $\displaystyle \frac{\pi}{2} < \theta < \pi$ is similar.
    Referring to Figure b. we have $\displaystyle \left | \left | T(e_1) \right| \right | = \cos \theta$, so

    $\displaystyle T(e_1) = \left[ {\begin{array}{c}
    \left | \left | T(e_1) \right| \right| \cos(\theta) \\
    \left | \left | T(e_1) \right| \right| \sin(\theta) \\
    \end{array} } \right]

    = \left[ {\begin{array}{c}
    \cos^{2}(\theta) \\
    \sin(\theta) \cos(\theta)\\
    \end{array} } \right]
    $

    And referring to Figure c. we have $\displaystyle \left | \left | T(e_2) \right| \right | = \sin(\theta)$, so:

    $\displaystyle T(e_2) = \left[ {\begin{array}{c}
    \left | \left | T(e_2) \right| \right| \cos(\theta) \\
    \left | \left | T(e_2) \right| \right| \sin(\theta) \\
    \end{array} } \right]

    = \left[ {\begin{array}{c}
    \sin(\theta) \cos(\theta)\\
    \sin^{2}(\theta) \\
    \end{array} } \right]
    $


    Thus the standard matrix for T is:

    $\displaystyle
    \left[T\right] =
    \left[ {\begin{array}{cc}
    \cos^{2}(\theta) & \sin(\theta) \cos(\theta)\\
    \sin(\theta) \cos(\theta) & \sin^{2}(\theta) \\
    \end{array} } \right]
    $





    Now my question: According to my text,
    $\displaystyle [T] = [T(e_1) | T(e_2)] $

    I understand this. But then how did the author come up with the following lines?
    (The author said nothing about how he got it)

    $\displaystyle T(e_1) = \left[ {\begin{array}{c}
    \left | \left | T(e_1) \right| \right| \cos(\theta) \\
    \left | \left | T(e_1) \right| \right| \sin(\theta) \\
    \end{array} } \right]

    = \left[ {\begin{array}{c}
    \cos^{2}(\theta) \\
    \sin(\theta) \cos(\theta)\\
    \end{array} } \right]
    $


    And also why???


    $\displaystyle T(e_2) = \left[ {\begin{array}{c}
    \left | \left | T(e_2) \right| \right| \cos(\theta) \\
    \left | \left | T(e_2) \right| \right| \sin(\theta) \\
    \end{array} } \right]

    = \left[ {\begin{array}{c}
    \sin(\theta) \cos(\theta)\\
    \cos^{2}(\theta) \\
    \end{array} } \right]
    $

    If you look at the top the author says that "We consider the case where$\displaystyle 0 \leq \theta \leq \frac{\pi}{2}$;
    the case where $\displaystyle \frac{\pi}{2} < \theta < \pi$ is similar." why is that?


    Can anyone kindly explain these questions? Thanks.


    A mathematical problem related to standard basis vector and standard matrix-a_svg.jpg
    Figure a.

    A mathematical problem related to standard basis vector and standard matrix-b_svg.jpg
    Figure b.

    A mathematical problem related to standard basis vector and standard matrix-c_svg.jpg
    Figure c.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,798
    Thanks
    3035
    T maps $\displaystyle e_1$, the vector from (0,0) to (1,0) into its projection on the line through the origin with angle $\displaystyle \theta$. drawing a perpendicular from (1, 0) to that line gives a right triangle with hypotenuse 1 and angle $\displaystyle \theta$. That is, the distance from (0, 0) to point of intersection is $\displaystyle cos(\theta)$. If you now drop a perpendicular to the x-axis, you have a right triangle with angle $\displaystyle \theta$ and hypotenuse $\displaystyle cos(\theta)$. The x and y coordinates of that point are the lengths of the "opposite" and "near" sides of that right triangle. That is, we have $\displaystyle cos(\theta)= \frac{x}{cos(\theta)}$ and $\displaystyle sin(\theta)= \frac{y}{cos(\theta)}$. From that the (x,y) coordinates of the point are $\displaystyle \left(cos^2(\theta), cos(\theta)sin(\theta)\right)$.

    Now, do the same with the projection of (0, 1) onto that line.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member x3bnm's Avatar
    Joined
    Nov 2009
    Posts
    308
    Thanks
    16
    Thank you sir for your kind help. That's the answer I was looking for. My problem was I was thinking this problem in a erroneous way. Again thanks for clearing that up.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] M 2x2 matrices of C with standard basis.
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Nov 30th 2011, 12:25 AM
  2. standard basis
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Oct 19th 2010, 07:12 AM
  3. standard matrix problem
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Sep 27th 2009, 07:22 PM
  4. Standard matrix problem, please help
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Sep 8th 2009, 04:59 AM
  5. Matrix with respect to the standard basis
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Apr 26th 2009, 01:31 PM

/mathhelpforum @mathhelpforum