If T : \(C^2\) → \(C^2\) satisfies
\[T(\left[ \begin{array}{cc} 2 \\ 1 \end{array} \right]) % =\left[ \begin{array}{cc} 3 \\ 4 \end{array} \right] \]
and
\[T(\left[ \begin{array}{cc} 1 \\ 1 \end{array} \right]) % =\left[ \begin{array}{cc} -1 \\ 2 \end{array} \right] \]
find \[T(\left[ \begin{array}{cc} 2 \\ 1 \end{array} \right])\]
\[We\ have\ to\ express \left[ \begin{array}{cc} 4 \\ 3 \end{array} \right] % as\ a\ sum\ of\ \left[ \begin{array}{cc} 2 \\ 1 \end{array} \right] % and \left[ \begin{array}{cc} 1 \\ 1 \end{array} \right] \]
\[\left[ \begin{array}{cc} 4 \\ 3 \end{array} \right] =% \left[ \begin{array}{cc} 2 \\ 1 \end{array} \right] % + 2\left[ \begin{array}{cc} 1 \\ 1 \end{array} \right] \]
\[T(\left[ \begin{array}{cc} 4 \\ 3 \end{array} \right]) =% T(\left[ \begin{array}{cc} 2 \\ 1 \end{array} \right] % + 2\left[ \begin{array}{cc} 1 \\ 1 \end{array} \right]) =% \left[ \begin{array}{cc} 3 \\ 4 \end{array} \right] % + 2\left[ \begin{array}{cc} -1 \\ 2 \end{array} \right] =% \left[ \begin{array}{cc} 3 \\ 4 \end{array} \right] % + \left[ \begin{array}{cc} -2 \\ 4 \end{array} \right] =% \left[ \begin{array}{cc} 1 \\ 8 \end{array} \right]\]
Comment to C25
Theorem MBLT(Matrices Build Linear Transformations) gives us a quicker way to construct linear transformations. We can grab an m×n matrix A and define T(x)=Ax
\[T(\left[ \begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array} \right]) =% T(\left[ \begin{array}{cc} 2*x_1-x_2+5*x_3 \\ -4*x_1+2*x_2-10*x_3 \end{array} \right]) =% x_1\left[ \begin{array}{cc} 2 \\ -4 \end{array} \right] % + x_2\left[ \begin{array}{cc} -1 \\ 2 \end{array} \right] % + x_3\left[ \begin{array}{cc} 5 \\ -10 \end{array} \right] =% \left[ \begin{array}{cc} 2 & -1 & 5 \\ -4 & 2 & -10 \end{array} \right] % * \left[ \begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array} \right]\]
\[\left[ \begin{array}{cc} 2 & -1 & 5 \\ -4 & 2 & -10 \end{array} \right]% is\ 3x2\matrix \ A \]
Theorem MBLT tells us that T is a linear transformation from \(C^n\) to \(C^m\) , without any further checking.