\(T : C^3 \rightarrow C^4 , T\)
\[\begin{bmatrix} x\\ y\\ z \end{bmatrix}\] = \[\begin{bmatrix} 3x + 2y +z\\ x+y+z\\ x-3y\\ 2x+3y+z \end{bmatrix}\])
Using Theorom MLTCV - Matrix Linear Transformation, Column Vectors we can find the matrix representation
\(C_1 = T(e_1) =\)
[ \[\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}\]=
\[\begin{bmatrix} 3*1 + 2*0 + 0\\ 1 + 0 + 0\\ 1 - (3 * 0)\\ (2*1) + (3 *0) + 0 \end{bmatrix}\] = \[\begin{bmatrix} 3\\ 1\\ 1\\ 2 \end{bmatrix}\]]
\(C_2 = T(e_2) =\)
\[\begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}\]=
\[\begin{bmatrix} (3*0) + (2*1) + 0\\ 0 + 1 + 0\\ 0 - (3 * 1)\\ (2*0) + (3*1) + 0 \end{bmatrix}\] = \[\begin{bmatrix} 2\\ 1\\ -3\\ 3 \end{bmatrix}\]\(C_3 = T(e_3) =\)
\[\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}\]=
\[\begin{bmatrix} 3*0 + 2*0 + 1\\ 0 + 0 + 1\\ 0 - (3 * 0)\\ (2*0) + (3 *0) + 1 \end{bmatrix}\] = \[\begin{bmatrix} 1\\ 1\\ 0\\ 1 \end{bmatrix}\]\(C_4 =\)
\[\begin{bmatrix} 3 & 2 & 1\\ 1 & 1 & 1\\ 1 & -3 & 0\\ 2 & 3 & 1 \end{bmatrix}\]Check that \(T(x) = Cx\)
C <- matrix(c(3,2,1,1,1,1,1,-3,0,2,3,1), nrow = 4, byrow = T)
v <- matrix(c(-2, 1, 4), nrow = 3, byrow = T)
C %*% v
## [,1]
## [1,] 0
## [2,] 3
## [3,] -5
## [4,] 3
T_function <- function(x){
m <- matrix(c(
(3 * x[1,]) + (2 * x[2,]) +x[3,],
x[1,] + x[2,] + x[3,],
x[1,] - (3 * x[2,]),
(2 * x[1,]) + (3 * x[2,]) + x[3,]
))
}
print(T_function(v))
## [,1]
## [1,] 0
## [2,] 3
## [3,] -5
## [4,] 3
T_function(v) == C %*% v
## [,1]
## [1,] TRUE
## [2,] TRUE
## [3,] TRUE
## [4,] TRUE