# Thread: e2 Show that S and T are both linear transformations

1. ## e2 Show that S and T are both linear transformations

$S:\mathbb{R}^2\to \mathbb{R}^2$ and $T:\mathbb{R}^2 \to \mathbb{R}^2$ be transformations defined by
$S\begin{bmatrix}x\\y \end{bmatrix}= \begin{bmatrix} 2x+y \\x-y \end{bmatrix}, \quad T \begin{bmatrix}x\\y \end{bmatrix}= \begin{bmatrix}x-4y\\3x\end{bmatrix}$
(a) Show that S and T are both linear transformations

ok my first attempt would be to see if S preserves addition so

$S \left(\begin{bmatrix}x_1\\y_1 \end{bmatrix} +\begin{bmatrix}x_2\\y_2 \end{bmatrix}\right)= \begin{bmatrix}x_1+x_2 \\y_1+y_2 \end{bmatrix} =\begin{bmatrix} 2(x_1+x_2)+(y_1+y_2) \\(x_1+y_1)-(y_1+y_2) \end{bmatrix}$
and
$S\begin{bmatrix}x_1\\y_1 \end{bmatrix}+S\begin{bmatrix}x_2\\y_2 \end{bmatrix} =\begin{bmatrix} 2x_1+y_1 \\x_1-y_1 \end{bmatrix}+\begin{bmatrix} 2x_2+y_2 \\x_2-y_2 \end{bmatrix} =​\begin{bmatrix} 2(x_1+x_2)+(y_1+y_2) \\(x_1+y_1)-(y_1+y_2) \end{bmatrix}$

maybe typos but think steps were ok for S still have T and there is
(b) Find $ST \begin{bmatrix} x\\y \end{bmatrix} \textit{ and } T^2 \begin{bmatrix} x\\y \end{bmatrix}$
(c) Find the matrices of S and T with respect to the standard basis for $\mathbb{R}^2$.

2. ## Re: e2 Show that S and T are both linear transformations

It's OK as far as it goes. Don't forget for linearity you also need $T(ax) = aT(x)$.

3. ## Re: e2 Show that S and T are both linear transformations

ok thanks I thot addition was enough ....
got a take a break then back in the saddle

4. ## Re: e2 Show that S and T are both linear transformations

Originally Posted by bigwave
$S:\mathbb{R}^2\to \mathbb{R}^2$ and $T:\mathbb{R}^2 \to \mathbb{R}^2$ be transformations defined by
To show that a mapping $S$ is a linear transformation does $S(X+\alpha\cdot Y)=S(X)+\alpha\cdot S(Y)~?$

5. ## Re: e2 Show that S and T are both linear transformations

I would have to follow an example to see how to do that

6. ## Re: e2 Show that S and T are both linear transformations

Originally Posted by Plato
To show that a mapping $S$ is a linear transformation does $S(X+\alpha\cdot Y)=S(X)+\alpha\cdot S(Y)~?$
Originally Posted by bigwave
I would have to follow an example to see how to do that
Why not at least try it?
$T\left( {\alpha \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right]} \right) = T\left( {\left[ {\begin{array}{*{20}{c}}{\alpha x}\\{\alpha y}\end{array}} \right]} \right) = \left[ {\begin{array}{*{20}{c}}{\alpha x - 4\alpha y}\\{3\alpha x}\end{array}} \right]$

ALSO
${T^2}\left( {\left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right]} \right) = T \circ T\left( {\left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right]} \right) = \left[ {\begin{array}{*{20}{c}} {(x - 4y) - 12x}\\ {3x - 12y} \end{array}} \right]$ WHY & HOW?
Post your work so others may see .

7. ## Re: e2 Show that S and T are both linear transformations

Originally Posted by Plato
Why not at least try it?
$T\left( {\alpha \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right]} \right) = T\left( {\left[ {\begin{array}{*{20}{c}}{\alpha x}\\{\alpha y}\end{array}} \right]} \right) = \left[ {\begin{array}{*{20}{c}}{\alpha x - 4\alpha y}\\{3\alpha x}\end{array}} \right]$

ALSO
${T^2}\left( {\left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right]} \right) = T \circ T\left( {\left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right]} \right) = \left[ {\begin{array}{*{20}{c}} {(x - 4y) - 12x}\\ {3x - 12y} \end{array}} \right]$ WHY & HOW?
Post your work so others may see .
$T\left( {\alpha \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right]} \right) = T\left( {\left[ {\begin{array}{*{20}{c}}{\alpha x}\\{\alpha y}\end{array}} \right]} \right) = \left[ {\begin{array}{*{20}{c}}{\alpha x - 4\alpha y}\\{3\alpha x}\end{array}} \right]$\\
$\text{so then}\\$
$S\left( {\alpha \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right]} \right) = S\left( {\left[ {\begin{array}{*{20}{c}} {\alpha x}\\{\alpha y}\end{array}} \right]} \right) = \left[ {\begin{array}{*{20}{c}} {2\alpha x+\alpha y}\\{\alpha x-\alpha y}\end{array}} \right]$

8. ## Re: e2 Show that S and T are both linear transformations

ok finally Find $ST\begin{bmatrix}x\\y\end{bmatrix} =S\left(\begin{bmatrix}x-4y\\3x\end{bmatrix}\right) =\left[\begin{array}{c}2(x-4y)+3x \\ x-4y-3x \end{array}\right]$
and
$T^2\left(\left[\begin{array}{c}x \\ y \end{array}\right]\right) =T\left(T\left(\left[\begin{array}{c}x \\ y \end{array}\right]\right)\right) =T\left(\left[\begin{array}{c}x-4y\\3x \end{array}\right]\right) =\left[\begin{array}{c}x-4y-4(3x) \\ 3(x-4y) \end{array}\right]$
hopefully

9. ## Re: e2 Show that S and T are both linear transformations

Originally Posted by bigwave
ok finally Find $ST\begin{bmatrix}x\\y\end{bmatrix} =S\left(\begin{bmatrix}x-4y\\3x\end{bmatrix}\right) =\left[\begin{array}{c}2(x-4y)+3x \\ x-4y-3x \end{array}\right]$
and
$T^2\left(\left[\begin{array}{c}x \\ y \end{array}\right]\right) =T\left(T\left(\left[\begin{array}{c}x \\ y \end{array}\right]\right)\right) =T\left(\left[\begin{array}{c}x-4y\\3x \end{array}\right]\right) =\left[\begin{array}{c}x-4y-4(3x) \\ 3(x-4y) \end{array}\right]$
hopefully
Good Job.
However, it should be $S{\circ}T\begin{bmatrix}x\\y\end{bmatrix}$ because $ST$ indicates multiplication and not function composition.

10. ## Re: e2 Show that S and T are both linear transformations

Originally Posted by Plato
Good Job.
However, it should be $S{\circ}T\begin{bmatrix}x\\y\end{bmatrix}$ because $ST$ indicates multiplication and not function composition.
$$ST\begin{bmatrix}x\\y\end{bmatrix} =S\left(T\begin{bmatrix}x-4y\\3x\end{bmatrix}\right) =\left[\begin{array}{c}2(x-4y)+3x \\ x-4y-3x \end{array}\right]$$

do you mean this?

11. ## Re: e2 Show that S and T are both linear transformations

Originally Posted by bigwave
$$ST\begin{bmatrix}x\\y\end{bmatrix} =S\left(T\begin{bmatrix}x-4y\\3x\end{bmatrix}\right) =\left[\begin{array}{c}2(x-4y)+3x \\ x-4y-3x \end{array}\right]$$

do you mean this?
Yes. The term ST is, in fact, a multiplication as Plato pointed out. You really do need to write it as $\displaystyle S \circ T$.

On the other hand I've never seen the composition symbol used in any of the Physics classes I've taken. So if you are doing Math do it as Plato (or your instructor) says. Otherwise ask your instructor.

-Dan