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17.1 Determine if T is a linear transformation

karush

Well-known member
Jan 31, 2012
2,657
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

Well-known member
MHB Math Helper
Aug 30, 2012
1,120
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

Well-known member
Jan 31, 2012
2,657
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

Well-known member
MHB Math Helper
Aug 30, 2012
1,120
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

Well-known member
Jan 31, 2012
2,657
$\displaystyle T(\vec{x}+\vec{y})
= T \begin{bmatrix} x_1+y_1 \\ x_2+y_2 \end{bmatrix}
=T\begin{bmatrix} 2(x_1+y_1)+x_2+y_2\\ x_1+y_1-4(x_2+y_2) \end{bmatrix}$
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

Well-known member
MHB Math Helper
Aug 30, 2012
1,120
$\displaystyle T(\vec{x}+\vec{y})
= T \begin{bmatrix} x_1+y_1 \\ x_2+y_2 \end{bmatrix}
=T\begin{bmatrix} 2(x_1+y_1)+x_2+y_2\\ x_1+y_1-4(x_2+y_2) \end{bmatrix}$
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

Well-known member
Jan 31, 2012
2,657
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....