Exercise LT.C41



If T:C\(^2\) -> C\(^3\) satifies T(\(\left[ \begin{array}{ccc} 2\\ 3 \end{array} \right]\)) = \(\left[ \begin{array}{ccc} 2\\ 2\\ 1 \end{array} \right]\) and T(\(\left[ \begin{array}{ccc} 3\\ 4 \end{array} \right]\)) = \(\left[ \begin{array}{ccc} -1\\ 0\\ 2 \end{array} \right]\), find the matrix representation of T.

We can use -

Theorem MLTCV Matrix of a Linear Transformation, Column Vectors
Suppose that T:C\(^n\) -> C\(^m\) is a linear transformation. Then there is an
m × n matrix A such that T(x) = Ax.

Using Theorem MMIM for an idenity matirx, for I\(_n\) a linear transformation can be done such as

A\(_i\) = T(e\(_i\)), where e\(_i\) is a column of identity matrix I\(_n\).

Using the identity matrix I\(_n\), we can say that T(x) = AI\(_n\)x
= T(e\(_1\)|e\(_2\)|…|e\(_n\)|x)
= T(e\(_1\)x\(_1\)|e\(_2\)x\(_2\)|….e\(_n\)x\(_n\))
Since A\(_i\) = T(e\(_i\)), T(x) = T(A\(_1\)x\(_1\)|A\(_2\)x\(_2\)|….A\(_n\)x\(_n\))

For the exercise above,

T(e\(_1\)) = T (\(\left[ \begin{array}{ccc} 1\\ 0 \end{array} \right]\)) = T (x\(_1\) \(\left[ \begin{array}{ccc} 2\\ 3 \end{array} \right]\) + x\(_2\) \(\left[ \begin{array}{ccc} 3\\ 4 \end{array} \right]\))

A = matrix(c(2,3,3,4),nrow=2,byrow=TRUE)
B = matrix(c(1,0),nrow=2)
X = solve(A,B)
X
##      [,1]
## [1,]   -4
## [2,]    3

T(e\(_2\)) = T (\(\left[ \begin{array}{ccc} 0\\ 1 \end{array} \right]\)) = T (y\(_1\) \(\left[ \begin{array}{ccc} 2\\ 3 \end{array} \right]\) + y\(_2\) \(\left[ \begin{array}{ccc} 3\\ 4 \end{array} \right]\))

A = matrix(c(2,3,3,4),nrow=2,byrow=TRUE)
B = matrix(c(0,1),nrow=2)
Y = solve(A,B)
Y
##      [,1]
## [1,]    3
## [2,]   -2

Solving for matrix A -

A1 = matrix(c(2,2,1),nrow=3,byrow=TRUE)
A2 = matrix(c(-1,0,2),nrow=3,byrow=TRUE)
A = cbind((X[1] * A1 + X[2] * A2),(Y[1]* A1 + Y[2]*A2)) 
A
##      [,1] [,2]
## [1,]  -11    8
## [2,]   -8    6
## [3,]    2   -1