Chapter LT, Exercise LT.C20

Let \[ w = \begin{bmatrix} -3 \\ 1 \\ 4 \end{bmatrix} \]

compute
S(w) two different ways. First use the definition of S, then compute the matrix-vector product \[C_w\]

Suppose \[S:C_3→C_4\] is defined by

\[S(\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}) = \begin{bmatrix} 3x_1-2x_2+5x_3 \\ x_1 + x_2 + x_3 \\ 9x_1-2x_2+5x_3 \\ 0x_1 + 4x_2 + 0x_3\end{bmatrix}\]

Solution

Method1

W <- matrix(c(-3,1,4), nrow = 3)
W
##      [,1]
## [1,]   -3
## [2,]    1
## [3,]    4
S <- matrix(c(3,1,9,0,-2,1,-2,4,5,1,5,0), nrow = 4)
S
##      [,1] [,2] [,3]
## [1,]    3   -2    5
## [2,]    1    1    1
## [3,]    9   -2    5
## [4,]    0    4    0
S%*%W # Multiply both Matrices
##      [,1]
## [1,]    9
## [2,]    2
## [3,]   -9
## [4,]    4

Method2

\[ S = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -3 \\ 1 \\ 4 \end{bmatrix} \]

\[S(e_1)=3x_1−2x_2+5x_3\] \[= 3∗−3−2∗1+5∗4\]

\[= -9 -2 +20\] \[9\]

\[S(e_2)=x_1+x_2+x_3\]

\[=-3+1+4\]

\[=2\]

\[S(e_3)=9x_1−2x_2+5x_3\]

\[=9∗−3−2∗1+5∗4\]

\[=-27-2+20\]

\[=-9\]

\[S(e_4)=0x_1+4x_2+0x_3\] \[=4∗1\] \[=4\] Hence, With both Methods we get

\[S(w)= \begin{bmatrix} 9 \\ 2 \\ -9 \\4 \end{bmatrix}\]