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

1. 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.
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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}
\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 $\left | \left | T(e_2) \right| \right | = \sin(\theta)$, so:

$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:

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

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}
\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???

$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 $0 \leq \theta \leq \frac{\pi}{2}$;
the case where $\frac{\pi}{2} < \theta < \pi$ is similar." why is that?

Can anyone kindly explain these questions? Thanks.

Figure a.

Figure b.

Figure c.

2. 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.

3. 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.