Linear transformation

**jackie** · February 8th 2009, 03:35 PM

Will someone help me with this problem about linear transformation. Let $T:R^2 \rightarrow R^2$ defined to be the transformation of rotation of the plane by angle $\theta$ around 0.
1) Show that T is linear.
2) Compute $[T]_{\beta}$ given $\beta=\{(1,0),(0,1)\}$
3) Consider the standard inner product <u,v> on $R^2$ . Show that it is invariant under the transformation T.

For part 1) I tried to formulate T after drawing the picture I have something like $T(a_1,a_2)=(rcos(\alpha+\theta),rsin(\alpha+\theta ))$ , here $r=\sqrt{{a_1}^2+{a_2}^2}$ then I rewrite this in term of $\theta, a_1,a_2$ , but my professor told me this is not how he wants it. He wants me to draw some paralellogram and do it like a geometric proof, which I have no idea. I tried to formulate this because I want to show T(cx+y)=cT(x)+T(y).
For part 2) he also wants me to do it geometrically. I really don't know how to di it. Can someone help please?

**ThePerfectHacker** · February 8th 2009, 06:08 PM

Originally Posted by jackie

1) Show that T is linear.

Draw vector $u,v$ in the plane.
Form a parallelogram to draw $u+v$ .
Now, let $\hat u$ be $u$ rotated by $\theta$ - draw it.
Now, let $\hat v$ be $v$ rotated by $\theta$ - draw it.
And let $\widehat{u+v}$ be the vector $u+v$ rotated by $\theta$ - draw it.
Argue that $\hat u + \hat v = \widehat{u+v}$ by looking at it geometrically.

2) Compute $[T]_{\beta}$ given $\beta=\{(1,0),(0,1)\}$

$T(1,0)$ is the point $(1,0)$ rotated by $\theta$ . Therefore, $T(1,0) = (\cos \theta,\sin \theta)$ .
$T(0,1)$ is the point $(0,1)$ rotated by $\theta$ . Therefore, $T(0,1) = (\cos ( \theta + \tfrac{\pi}{2}), \sin ( \theta + \tfrac{\pi}{2})) = ( - \sin \theta, \cos \theta)$ .
Now you should be able to form the matrix.

3) Consider the standard inner product <u,v> on $R^2$ . Show that it is invariant under the transformation T.

Remember that $\left<u,v\right> = |u||v|\cos \phi$ where $\phi$ is angle between them. But $Tu,Tv$ have the same length and the same angle between themselves because $T$ is rotating both of them by the same number of degrees. Thus, $\left<Tu,Tv\right> = \left<u,v\right>$ .

**jackie** · February 8th 2009, 09:12 PM

Thank you very much for your help, ThePerfectHacker. You helped me many times. Do you think if there is a reason that my professor wants us to do this problem geometrically?

**ThePerfectHacker** · February 9th 2009, 05:57 PM

Originally Posted by jackie

Do you think if there is a reason that my professor wants us to do this problem geometrically?

There is a nice connection between math (like linear algebra and complex analysis) and geometry. Sometimes problems in geometry can nicely be expressed in math. And so many geometry problems can be solved with non-geometric methods and vice-versa. That is what this problem was about.

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