Let \(w=\left[ \begin{matrix} -3 \\ 1 \\ 4 \end{matrix} \right]\).
Referring to Example MOLT, compute S(w) two different ways. First use the definition of S, then compute the matrix-vector product Cw (Definition MVP).
\[{ C }_{ 3 }\quad \rightarrow { C }_{ 4 }\quad is\quad defined\quad by\quad S\left( \left[ \begin{matrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{matrix} \right] \right) =\left[ \begin{matrix} 3{ x }_{ 1 }+{ 2 }_{ x2 }+{ x }_{ 3 } \\ { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } \\ { 2x }_{ 1 }+3{ x }_{ 2 }+{ x }_{ 3 } \\ 4{ x }_{ 2 } \end{matrix} \right] \]
\[{ C }_{ 1 }={ S }_{ ({ e }_{ 1 }) }=S\left( \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right] \right) =\left[ \begin{matrix} 3 \\ 1 \\ 2 \\ 0 \end{matrix} \right]\] \[{ C }_{ 2 }={ S }_{ ({ e }_{ 2 }) }=S\left( \left[ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right] \right) =\left[ \begin{matrix} 2 \\ 1 \\ 3 \\ 4 \end{matrix} \right] \] \[{ C }_{ 3 }={ S }_{ ({ e }_{ 3 }) }=S\left( \left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right] \right) =\left[ \begin{matrix} 1 \\ 1 \\ 1 \\ 0 \end{matrix} \right] \] So we can define it as: \[{ C }=\left[ { C }_{ 1 }|{ C }_{ 2 }|{ C }_{ 3 } \right] =\left[ \begin{matrix} 3 & 2 & 1 \\ 1 & 1 & 1 \\ 2 & 3 & 1 \\ 0 & 4 & 0 \end{matrix} \right] \]
Computing S(w)
\[S\left( \left[ \begin{matrix} { -3 } \\ { 1 } \\ 4 \end{matrix} \right] \right) =\left[ \begin{matrix} 3{ (-3 })+{ 2 }(1)+1(4) \\ -3+1+4 \\ 2(-3)+3(1)+1(4) \\ 4(1) \end{matrix} \right] =\left[ \begin{matrix} -3 \\ 2 \\ 1 \\ 4 \end{matrix} \right]\]
Computing Cw:
\[Cw=\left[ \begin{matrix} 3 & 2 & 1 \\ 1 & 1 & 1 \\ 2 & 3 & 1 \\ 0 & 4 & 0 \end{matrix} \right] \left[ \begin{matrix} -3 \\ 1 \\ 4 \end{matrix} \right] =\left[ \begin{matrix} 3{ (-3 })+{ 2 }(1)+1(4) \\ -3+1+4 \\ 2(-3)+3(1)+1(4) \\ 0(-3)+4(1)+0(4) \end{matrix} \right] =\left[ \begin{matrix} -3 \\ 2 \\ 1 \\ 4 \end{matrix} \right]\] Therefore \(S(w)=Cw\)
Additional Problem cause it was fun to do for an exerise:
C25 Define the linear transformation and Verify that T is a linear transformation.
\[T:\quad { C }^{ 3 }\rightarrow { C }^{ 2 },\quad T\left( \begin{matrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{matrix} \right) =\begin{bmatrix} 2{ x }_{ 1 } & { -x }_{ 2 } & +5{ x }_{ 3 } \\ -4{ x }_{ 1 } & { +2 }{ x }_{ 2 } & -10{ x }_{ 3 } \end{bmatrix}\]
This can be simplified down to:
\[{ x }_{ 1 }\begin{bmatrix} 2 \\ -4 \end{bmatrix}+{ x }_{ 2 }\begin{bmatrix} -1 \\ 2 \end{bmatrix}{ +x }_{ 3 }\begin{bmatrix} 5 \\ -10 \end{bmatrix}\]
Which is also equal to:
\[\begin{bmatrix} 2 & -1 & 5 \\ -4 & 2 & -10 \end{bmatrix}\begin{bmatrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{bmatrix}\]
Any function in this form is a linear transformation