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\textbf{Exercises on linear transformations and their matrices}
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\begin{problem}
Consider the transformation $T$ that doubles the distance between each point and the origin without changing the direction from the origin to the points.  In polar coordinates this is described by $$T(r, \theta) = (2r, \theta).$$
\begin{enumerate}[a)]
\item Yes or no: is $T$ a linear transformation?
\item Describe $T$ using Cartesian ($xy$) coordinates. Check your work by confirming that the transformation doubles the lengths of vectors.
\item If your answer to (a) was "yes", find the matrix of $T$. If your answer to (a) was "no", explain why the $T$ isn't linear. 
\end{enumerate}
\end{problem}

\solution{
\begin{enumerate}[a)]
\item Yes.  In terms of vectors, $T(\vb v) = 2\vb v$, so $T(\vb v_1 + \vb v_2) = 2\vb v_1 + 2\vb v_2 = T(\vb v_1) + T(\vb v_2)$ and $T(c \vb v) = 2 c \vb v = c T(\vb v)$.

\item $T(x,y) = (2x, 2y)$.  We know $\left| \small \twobyone xy
  \right| = \sqrt{x^2 + y^2}$ and can calculate $$\left| T\left(\small
  \twobyone xy\right) \right| = \left| \small \twobyone {2x}{2y}
  \right| = \sqrt{4(x^2 + y^2)} = 2\left| \small
  \twobyone xy \right|.$$  This confirms that $T$ doubles the lengths of vectors.

\item We can use our answer to (b) to find that the matrix of $T$ is $\small \twobytwo 2002$.
\end{enumerate}
}

\begin{problem}
Describe a transformation which leaves the zero vector fixed but which is not a linear transformation. 
\end{problem}

\solution{
If we limit ourselves to ``simple'' transformations, this is not an easy task! 

One way to solve this is to find a transformation that acts
differently on different parts of the plane. If $T\left( \small
\twobyone xy \right ) = \small \twobyonec x{|y|}$ then $$T\left( \twobyone 11  + \twobyone 1{-1} \right) = \twobyone 20$$
is not equal to 
$$T\left( \twobyone 11 \right) + T\left (\twobyone 1{-1}\right ) =
\twobyone 22.$$ 

Another approach is to use a nonlinear function to define the
transformation: if $T\left (\small \twobyone xy \right) = \small
\twobyonec x{y^2}$ then $T\left( c\small \twobyone xy \right) = \small
\twobyonec {cx}{c^2y^2}$ is not equal to $cT\left( \small \twobyone xy
\right) = \small \twobyonec{cx}{cy^2}$.
}

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