e2 Show that S and T are both linear transformations

Nov 2009
717
133
Wahiawa, Hawaii
$S:\mathbb{R}^2\to \mathbb{R}^2$ 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 $\mathbb{R}^2$.
 

Walagaster

MHF Helper
Apr 2018
223
144
Tempe, AZ
It's OK as far as it goes. Don't forget for linearity you also need $T(ax) = aT(x)$.
 
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Nov 2009
717
133
Wahiawa, Hawaii
ok thanks I thot addition was enough ....
got a take a break then back in the saddle
 

Plato

MHF Helper
Aug 2006
22,474
8,643
$S:\mathbb{R}^2\to \mathbb{R}^2$ and $T:\mathbb{R}^2 \to \mathbb{R}^2$ be transformations defined by
To show that a mapping $S$ is a linear transformation does $S(X+\alpha\cdot Y)=S(X)+\alpha\cdot S(Y)~?$
 
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Nov 2009
717
133
Wahiawa, Hawaii
I would have to follow an example to see how to do that
 

Plato

MHF Helper
Aug 2006
22,474
8,643
To show that a mapping $S$ is a linear transformation does $S(X+\alpha\cdot Y)=S(X)+\alpha\cdot S(Y)~?$
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 .
 
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Nov 2009
717
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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 .
$T\left( {\alpha \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right]} \right)
= T\left( {\left[ {\begin{array}{*{20}{c}}{\alpha x}\\{\alpha y}\end{array}} \right]} \right)
= \left[ {\begin{array}{*{20}{c}}{\alpha x - 4\alpha y}\\{3\alpha x}\end{array}} \right]$\\
$\text{so then}\\$
$S\left( {\alpha \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right]} \right)
= S\left( {\left[ {\begin{array}{*{20}{c}}
{\alpha x}\\{\alpha y}\end{array}} \right]} \right)
= \left[ {\begin{array}{*{20}{c}}
{2\alpha x+\alpha y}\\{\alpha x-\alpha y}\end{array}} \right]$
 
Last edited:
Nov 2009
717
133
Wahiawa, Hawaii
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
 

Plato

MHF Helper
Aug 2006
22,474
8,643
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.
 
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Nov 2009
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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?
 
Last edited: