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Math Help - A mathematical problem related to standard basis vector and standard matrix

  1. #1
    Senior Member x3bnm's Avatar
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    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 \theta with
    the positive x-axis, where 0 \leq \theta < \pi. As illustrated in Figure a. , let 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:

    [T] = [T(e_1) | T(e_2)]

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

     T(e_1) = \left[ {\begin{array}{c}<br />
 \left | \left | T(e_1) \right| \right| \cos(\theta)  \\<br />
 \left | \left | T(e_1) \right| \right| \sin(\theta)  \\<br />
 \end{array} } \right]<br /> <br />
= \left[ {\begin{array}{c}<br />
\cos^{2}(\theta) \\<br />
 \sin(\theta) \cos(\theta)\\<br />
 \end{array} } \right]<br />

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

     T(e_2) = \left[ {\begin{array}{c}<br />
 \left | \left | T(e_2) \right| \right| \cos(\theta)  \\<br />
 \left | \left | T(e_2) \right| \right| \sin(\theta)  \\<br />
 \end{array} } \right]<br /> <br />
= \left[ {\begin{array}{c}<br />
\sin(\theta) \cos(\theta)\\<br />
\sin^{2}(\theta) \\ <br />
 \end{array} } \right]<br />


    Thus the standard matrix for T is:

    <br />
\left[T\right] = <br />
\left[ {\begin{array}{cc}<br />
\cos^{2}(\theta)  & \sin(\theta) \cos(\theta)\\<br />
\sin(\theta) \cos(\theta) & \sin^{2}(\theta) \\<br />
 \end{array} } \right]<br />





    Now my question: According to my text,
    [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)

     T(e_1) = \left[ {\begin{array}{c}<br />
 \left | \left | T(e_1) \right| \right| \cos(\theta)  \\<br />
 \left | \left | T(e_1) \right| \right| \sin(\theta)  \\<br />
 \end{array} } \right]<br /> <br />
= \left[ {\begin{array}{c}<br />
\cos^{2}(\theta) \\<br />
 \sin(\theta) \cos(\theta)\\<br />
 \end{array} } \right]<br />


    And also why???


     T(e_2) = \left[ {\begin{array}{c}<br />
 \left | \left | T(e_2) \right| \right| \cos(\theta)  \\<br />
 \left | \left | T(e_2) \right| \right| \sin(\theta)  \\<br />
 \end{array} } \right]<br /> <br />
= \left[ {\begin{array}{c}<br />
\sin(\theta) \cos(\theta)\\<br />
\cos^{2}(\theta) \\ <br />
 \end{array} } \right]<br />

    If you look at the top the author says that "We consider the case where 0 \leq \theta \leq \frac{\pi}{2};
    the case where \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.
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  2. #2
    MHF Contributor

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    T maps e_1, the vector from (0,0) to (1,0) into its projection on the line through the origin with angle \theta. drawing a perpendicular from (1, 0) to that line gives a right triangle with hypotenuse 1 and angle \theta. That is, the distance from (0, 0) to point of intersection is cos(\theta). If you now drop a perpendicular to the x-axis, you have a right triangle with angle \theta and hypotenuse 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 cos(\theta)= \frac{x}{cos(\theta)} and sin(\theta)= \frac{y}{cos(\theta)}. From that the (x,y) coordinates of the point are \left(cos^2(\theta), cos(\theta)sin(\theta)\right).

    Now, do the same with the projection of (0, 1) onto that line.
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  3. #3
    Senior Member x3bnm's Avatar
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    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.
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