Define the linear transformation
\[T:{ { \mathbb{C} }^{ 3 } }\xrightarrow [ ]{ } { \mathbb{C} }^{ 2 },\quad T\begin{pmatrix} \begin{bmatrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{bmatrix} \end{pmatrix}=\begin{bmatrix} 2{ x }_{ 1 }-{ x }_{ 2 }+5{ x }_{ 3 } \\ -4{ x }_{ 1 }+2{ x }_{ 2 }-10{ x }_{ 3 } \end{bmatrix}\]
Answer:
Knowing what we do from the reading (specifically the MLT Section) we can define the translation as a matrix such that:
\[A=Tx \\ \begin{bmatrix} 2{ x }_{ 1 }-{ x }_{ 2 }+5{ x }_{ 3 } \\ -4{ x }_{ 1 }+2{ x }_{ 2 }-10{ x }_{ 3 } \end{bmatrix}=T\begin{bmatrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{bmatrix}\]
Therefore, we know we can repesent &Tx& in matrix form as:
\[A=\begin{bmatrix} 2 & -1 & 5 \\ -4 & 2 & -10 \end{bmatrix}\begin{bmatrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{bmatrix}\]
Which can be simplified to:
\[A=\begin{bmatrix} 2 \\ -4 \end{bmatrix}{ x }_{ 1 } + \begin{bmatrix} -1 \\ 2 \end{bmatrix}{ x }_{ 2 } + \begin{bmatrix} 5 \\ -10 \end{bmatrix}{ x }_{ 3 }\]
Verify that T is a linear transformation.
Answer:
To verfiy, we must prove that adding two vectors then translating them is the same as adding two translated vectors:
\[T\begin{pmatrix} \begin{bmatrix} ({ x }_{ 1 }+{ y }_{ 1 }) \\ ({ x }_{ 2 }+{ y }_{ 2 }) \\ ({ x }_{ 3 }+{ y }_{ 3 }) \end{bmatrix} \end{pmatrix}=\begin{bmatrix} 2({ x }_{ 1 }+{ y }_{ 1 })-({ x }_{ 2 }+{ y }_{ 2 })+5({ x }_{ 3 }+{ y }_{ 3 }) \\ -4({ x }_{ 1 }+{ y }_{ 1 })+2({ x }_{ 2 }+{ y }_{ 2 })-10({ x }_{ 3 }+{ y }_{ 3 }) \end{bmatrix}\]
\[=\begin{bmatrix} 2{ x }_{ 1 }+2{ y }_{ 1 }-{ x }_{ 2 }-{ y }_{ 2 }+5{ x }_{ 3 }+5{ y }_{ 3 } \\ -4{ x }_{ 1 }-4{ y }_{ 1 })+2{ x }_{ 2 }+2{ y }_{ 2 }-10{ x }_{ 3 }-10{ y }_{ 3 }) \end{bmatrix}\]
\[=\begin{bmatrix} 2{ x }_{ 1 }-{ x }_{ 2 }+5{ x }_{ 3 }+2{ y }_{ 1 }-{ y }_{ 2 }+5{ y }_{ 3 } \\ -4{ x }_{ 1 }+2{ x }_{ 2 }-10{ x }_{ 3 }-4{ y }_{ 1 }+2{ y }_{ 2 }-10{ y }_{ 3 } \end{bmatrix}\]
\[=\begin{bmatrix} 2{ x }_{ 1 }-{ x }_{ 2 }+5{ x }_{ 3 } \\ -4{ x }_{ 1 }+2{ x }_{ 2 }-10{ x }_{ 3 }\end{bmatrix} + \begin{bmatrix} 2{ y }_{ 1 }-{ y }_{ 2 }+5{ y }_{ 3 } \\ -4{ y }_{ 1 }+2{ y }_{ 2 }-10{ y }_{ 3 }\end{bmatrix}\]
Now, we can check that scalar multiplcation holds true, which is pretty easy for matrices
\[T\begin{pmatrix} \begin{bmatrix} \alpha{ x }_{ 1 } \\ \alpha{ x }_{ 2 } \\ \alpha{ x }_{ 3 } \end{bmatrix} \end{pmatrix}=\begin{bmatrix} 2(\alpha{ x }_{ 1 })-(\alpha{ x }_{ 2 })+5(\alpha{ x }_{ 3 }) \\ -4(\alpha{ x }_{ 1 })+2(\alpha{ x }_{ 2 })-10(\alpha{ x }_{ 3 }) \end{bmatrix}\]
\[=\begin{bmatrix} \alpha(2{ x }_{ 1 }-{ x }_{ 2 }+5{ x }_{ 3 }) \\ \alpha(-4{ x }_{ 1 }+2{ x }_{ 2 }-10{ x }_{ 3 }) \end{bmatrix}\]
\[=\alpha \begin{bmatrix} 2{ x }_{ 1 }-{ x }_{ 2 }+5{ x }_{ 3 } \\ -4{ x }_{ 1 }+2{ x }_{ 2 }-10{ x }_{ 3 } \end{bmatrix}\]
This was further proven in the MOLT and ALT examples to be generally true for linear transformations to matrices.
Note, I initially had some trouble displaying the complex number symbol \(\mathbb{C}\), so I thought this link would help (http://milde.users.sourceforge.net/LUCR/Math/mathpackages/amsfonts-symbols.pdf). It looks like it’s just part of a non-standard font package, which is easy to fix. There are also many additional symbols which can now be shown in Rmarkdown files.