# A mathematical problem related to standard basis vector and standard matrix

• Mar 10th 2011, 10:56 AM
x3bnm
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.
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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.

Attachment 21105
Figure a.

Attachment 21104
Figure b.

Attachment 21106
Figure c.
• Mar 10th 2011, 12:11 PM
HallsofIvy
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.
• Mar 10th 2011, 01:39 PM
x3bnm
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.