Question

Help with Linear transformations, Pleasee?

Answer 1

Here's the question.

Answer 2

For the first part, do you know how to show a transformation is linear? Given some transformation \(T(x)\), you have to show that \(T(a+b)=T(a)+T(b)\) and \(T(cx)=cT(x)\) for some vectors \(a,b\) in the domain set and scalar \(c\).

Answer 3

so pick some numbers for x, y, z?

Answer 4

Just picking some numbers is not enough. You have to show that the transformation is linear for any vector you plug in. Basically, you have to be very general.

\[T:\mathbb{R}^3\to\mathbb{R}^2,\\
T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2x+y+z\\y-3z\end{pmatrix}\]
Let \(\begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix},\begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}\in\mathbb{R}^3\). Then,
\[\begin{align*}T\begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix}+T\begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}&=\begin{pmatrix}2x_1+y_1+z_1\\y_1-3z_1\end{pmatrix}+\begin{pmatrix}2x_2+y_2+z_2\\y_2-3z_2\end{pmatrix}\\
&=\begin{pmatrix}2(x_1+x_2)+(y_1+y_2)+(z_1+z_2)\\(y_1+y_2)-3(z_1+z_2)\end{pmatrix}\\
&=T\left(\begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix}+\begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}\right)
\end{align*}\]
Thus vector addition is preserved. You would do something similar to show that scalar multiplication is preserved.

Answer 5

Unfortunately, it's been a while since I've taken linear algebra and not much else has stuck. So I'm afraid I won't be able to help you with the rest. sorry

Answer 6

Thank you... Is this the answer for part a?

Answer 7

It's not so much an answer as it is a proof. Once you show that addition and scalar multiplication is preserved, you're done with (a).

While I'm still here, I'll take a look at my lin alg textbook for some review.

Answer 8

Thank you so much!

Answer 9

You're welcome.

Answer 10

So according to an example in my book, part (b) involves examining the effects \(T\) has on the standard basis of the domain set:
\[\left\{\begin{pmatrix}1\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\end{pmatrix},\begin{pmatrix}0\\0\\1\end{pmatrix}\right\}\]
\[T\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}2\\0\end{pmatrix},~T\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}1\\1\end{pmatrix},~T\begin{pmatrix}0\\0\\1\end{pmatrix}=\begin{pmatrix}1\\-3\end{pmatrix}\]
These vectors will be the columns of the matrix that describe the transformation:
\[C=\begin{pmatrix}2&1&1\\0&1&-3\end{pmatrix}\\
T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2&1&1\\0&1&-3\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}\]

Answer 11

Part (c) looks like it uses the same steps as in (b), but this time you swap the standard \(\mathbb{R}^3\) basis with \(B_1\). I'm not sure what \(B_2\) is used for, thought, so I may be wrong.