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#### karush

##### Well-known member

- Jan 31, 2012

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- Thread starter karush
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- #1

- Jan 31, 2012

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- Jan 30, 2012

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The claim that $S$ is a linear transformation requires a proof, and a proof is not simply some collection of formulas. A proof is an argument that starts with assumptions and arrives and the desires conclusion. proofs are best expressed using text in a natural language (English) written in complete grammatical sentences.

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- #3

- Jan 31, 2012

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The claim that $S$ is a linear transformation requires a proof, and a proof is not simply some collection of formulas. A proof is an argument that starts with assumptions and arrives and the desires conclusion. proofs are best expressed using text in a natural language (English) written in complete grammatical sentences.

ok here is the example I am trying to follow

- Jan 30, 2012

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Yes, so far what you wrote is correct, and it follows the example.

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LetYes, so far what you wrote is correct, and it follows the example.

$S:\Bbb{R}^2\to \Bbb{R}^2$ and $T:\Bbb{R}^2 \to \Bbb{R}^2$ be transformations defined by

$\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}$

Show that S and T are both linear transformations

$\begin{align*}\displaystyle

S\left(\left[\begin{array}{} x_2 \\ y_2 \end{array}\right]

+\left[\begin{array}{} x_2\\y_2\end{array}\right]\right)

&=S\left[\begin{array}{}x_1+x_2\\y_1+y_2\end{array}\right]\\

&=\left[\begin{array}{c}2(x_1+x_2)+(y_1+y_2) \\ (x_1+x_2)-(y_1+y_2) \end{array}\right]\\

&=\left[\begin{array}{c} 2x_1+2x_2\\x_1+x_2 \end{array}\right]

+\left[\begin{array}{c}y_1+y_2\\-y_1-y_2) \end{array}\right]

\end{align*}$

ok for some reason I can't see how this is going to preserve addition

or is there another way to show transformaton?

- Jan 30, 2012

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\(\displaystyle \begin{bmatrix}2x_1+y_1\\x_1-y_1\end{bmatrix}+\begin{bmatrix}2x_2+y_2\\x_2-y_2\end{bmatrix}\).

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- #7

- Jan 31, 2012

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$S:\Bbb{R}^2\to \Bbb{R}^2$ and $T:\Bbb{R}^2 \to \Bbb{R}^2$ be transformations defined by

$\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}$

Show that S and T are both linear transformations

$\begin{align*}\displaystyle

S\left(\left[\begin{array}{} x_2 \\ y_2 \end{array}\right]

+\left[\begin{array}{} x_2\\y_2\end{array}\right]\right)

&=S\left[\begin{array}{}x_1+x_2\\y_1+y_2\end{array}\right]\\

&=\left[\begin{array}{c}2(x_1+x_2)+(y_1+y_2) \\ (x_1+x_2)-(y_1+y_2) \end{array}\right]\\

&=\begin{bmatrix}2x_1+y_1\\x_1-y_1\end{bmatrix}+\begin{bmatrix}2x_2+y_2\\x_2-y_2\end{bmatrix}\\

&=S \begin{bmatrix}x_1\\y_1\end{bmatrix} +S\begin{bmatrix} x_2\\y_2\end{bmatrix}\end{align*}$

S preserves addition, If c is any scalar.

$S\left(c\begin{bmatrix} x_1\\y_1\end{bmatrix}\right)

=S\begin{bmatrix} cx_2\\cy_2 \end{bmatrix}

=\begin{bmatrix} 2cx+cy \\ cx-cy \end{bmatrix}

=c\begin{bmatrix} 2x+y \\ x-y \end{bmatrix}

=cS\begin{bmatrix} x_1\\y_1\end{bmatrix}$

and consequently T preserves scalar multiplication.

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- #8

- Jan 31, 2012

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(b) Find $ST

\begin{bmatrix} x\\y

\end{bmatrix}$

$$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]$$

and $T^2

\begin{bmatrix} x\\y

\end{bmatrix}$

$$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]$$

(c) Find the matrices of S and T with respect to the standard basis for $\Bbb{R}^2$.

$$\displaystyle\left[S\right]_\infty^\infty

=\left[\begin{array}{cc}

2&1\\1&-1

\end{array}\right], \quad

\left[T\right]_\infty^\infty

=\left[\begin{array}{cc}

1&-4\\3&0

\end{array}\right]$$