17.1 Determine if T is a linear transformation

[karush]

nmh{2000}
17.1 Let $T: \Bbb{R}^2 \to \Bbb{R}^2$ be defined by
$$T \begin{bmatrix}
x\\y
\end{bmatrix}
=
\begin{bmatrix}
2x+y\\x-4y
\end{bmatrix}$$
Determine if $T$ is a linear transformation. So if
$$T(\vec{x}+\vec{y})=T(\vec{x})+T(\vec{y})$$
Let $\vec{x}$ and $\vec{y}$ be vectors in $\Bbb{R}^2$ then we can write them as
$$\vec{x}
=\begin{bmatrix}
x_1\\x_2
\end{bmatrix}
, \vec{y}
=\begin{bmatrix}
y_1\\y_2
\end{bmatrix}$$
By definition, we have that
$$T(\vec{x}+\vec{y})
=\begin{bmatrix}
x_1+y_1 \\
x_2+y_2
\end{bmatrix}
=\begin{bmatrix}
2(x_1+y_1)+x_2+y_2\\
x_1+y_1-4(x_2+y_2)
\end{bmatrix}$$

OK just seeing if this is developing as it should
hopefully the next few steps will be an addition property
and this is a linear transformation

 

[topsquark]

17.1 Let $T: \Bbb{R}^2 \to \Bbb{R}^2$ be defined by
$$T \begin{bmatrix}
x\\y
\end{bmatrix}
=
\begin{bmatrix}
2x+y\\x-4y
\end{bmatrix}$$
Determine if $T$ is a linear transformation. So if
$$T(\vec{x}+\vec{y})=T(\vec{x})+T(\vec{y})$$
Let $\vec{x}$ and $\vec{y}$ be vectors in $\Bbb{R}^2$ then we can write them as
$$\vec{x}
=\begin{bmatrix}
x_1\\x_2
\end{bmatrix}
, \vec{y}
=\begin{bmatrix}
y_1\\y_2
\end{bmatrix}$$
By definition, we have that
$$T(\vec{x}+\vec{y})
= \begin{bmatrix}
x_1+y_1 \\
x_2+y_2
\end{bmatrix}
=\begin{bmatrix}
2(x_1+y_1)+x_2+y_2\\
x_1+y_1-4(x_2+y_2)
\end{bmatrix}$$

OK just seeing if this is developing as it should
hopefully the next few steps will be an addition property
and this is a linear transformation

Typo alert!
\(\displaystyle T(\vec{x}+\vec{y})
= T \begin{bmatrix}
x_1+y_1 \\
x_2+y_2
\end{bmatrix}
=\begin{bmatrix}
2(x_1+y_1)+x_2+y_2\\
x_1+y_1-4(x_2+y_2)
\end{bmatrix}\)

Good so far. Now what is T(x) + T(y)? And how do you verify T(cx) = cT(x)?

-Dan

 

[karush]

ok sorry I got to finish this later
but where is the typo...


actually I would say that could be determined just by observation... or not?

 

[topsquark]

ok sorry I got to finish this later
but where is the typo...


actually I would say that could be determined just by observation... or not?

It's a little bit of nothing. You left the T off in front of the \(\displaystyle \left [ \begin{matrix} x_1 + y_1 \\ x_2 + y_2 \end{matrix} \right ]\) in the second step of the last equation. (I know there's a way to change the color in LaTeX to highlite it but I'm too lazy to look it up right now.)

-Dan

 

[karush]

$\displaystyle T(\vec{x}+\vec{y})
= T \begin{bmatrix} x_1+y_1 \\ x_2+y_2 \end{bmatrix}
=T\begin{bmatrix} 2(x_1+y_1)+x_2+y_2\\ x_1+y_1-4(x_2+y_2) \end{bmatrix}$
So then
$T(\vec{x})+T(\vec{y})
= T\begin{bmatrix} x_1\\x_2 \end{bmatrix}
+T\begin{bmatrix} y_1\\y_2 \end{bmatrix}
=\begin{bmatrix} 2(x_1+x_2)\\x_1+x_2\end{bmatrix}+\begin{bmatrix} y_1+y_2\\-4(y_1+y_2 )\end{bmatrix}
=\begin{bmatrix} 2(x_1+y_1)+x_2+y_2\\ x_1+y_1-4(x_2+y_2) \end{bmatrix}$
and
$T(c\vec{x})=\begin{bmatrix}c x_1\\cx_2 \end{bmatrix}
= \begin{bmatrix} cx_1+cy_1 \\ cx_2+cy_2 \end{bmatrix}$
and
$cT(\vec{x})=c\begin{bmatrix} x_1\\x_2 \end{bmatrix}
= c\begin{bmatrix} x_1+y_1 \\ x_2+y_2 \end{bmatrix}
= \begin{bmatrix} cx_1+cy_1 \\ cx_2+cy_2 \end{bmatrix}$


ok typos for sure

 

[topsquark]

$\displaystyle T(\vec{x}+\vec{y})
= T \begin{bmatrix} x_1+y_1 \\ x_2+y_2 \end{bmatrix}
=T\begin{bmatrix} 2(x_1+y_1)+x_2+y_2\\ x_1+y_1-4(x_2+y_2) \end{bmatrix}$

Only one typo: Top line of equations, third term doesn't need the T. You've already taken the transform.

The second line of equations needs some work. You have the correct answer, but the intermediate step is wrong. Is this also a typo?

It should be
\(\displaystyle T \left [ \begin{matrix} x_1 \\ x_2 \end{matrix} \right ] + T \left [ \begin{matrix} y_1 \\ y_2 \end{matrix} \right ] = \left [ \begin{matrix} 2x_1 + x_2 \\ x_1 - 4x_2 \end{matrix} \right ] + \left [ \begin{matrix} 2y_1 + y_2 \\ y_1 - 4y_2 \end{matrix} \right ] = \left [ \begin{matrix} 2(x_1 + y_1) + (x_2 + y_2) \\ (x_1 + y_1) - 4(x_2 + y_2) \end{matrix} \right ] \)

-Dan

 

[karush]

ok really appreciate the help.

I still need more practice on these...

sorry for the wait on reply but I am limited to when the computer labs are open.

Mahalo...