Problem M11, page 270

Let \(V\) be the the set \(\mathbb{C}^2\) with the usual scalar multiplication, but with the vector addition defined by \[ \begin{bmatrix} x\\ y \end{bmatrix} + \begin{bmatrix} z\\ w \end{bmatrix} = \begin{bmatrix} y+w\\ x+z \end{bmatrix} \] Determine whether or not \(V\) is a vector space with these operations.

\(V\) is not a vector space because it does not have the property of addition being associative. Under this rule of addition: \[ \Bigg( \begin{bmatrix} x\\ y \end{bmatrix}+ \begin{bmatrix} z\\ w \end{bmatrix} \Bigg) + \begin{bmatrix} t\\u \end{bmatrix} = \begin{bmatrix} x+z+u\\ y+w+t \end{bmatrix} \] but \[ \begin{bmatrix} x\\ y \end{bmatrix}+ \Bigg( \begin{bmatrix} z\\ w \end{bmatrix} + \begin{bmatrix} t\\u \end{bmatrix} \Bigg) = \begin{bmatrix} y+z+t\\ x+w+u \end{bmatrix} \] In most cases \[ \begin{bmatrix} x+z+u\\ y+w+t \end{bmatrix} \neq \begin{bmatrix} y+z+t\\ x+w+u \end{bmatrix} \]

We can run an example of this in R to see how this works in practice:

m11add<- function(c1,c2){
  ans<-matrix(c(c1[2]+c2[2],c1[1]+c2[1]),nrow=2)
  return(ans)
}
v1<-c(1,2) 
v2<-c(3,4)
v3<-c(1,0)

m11add(m11add(v1,v2),v3)
##      [,1]
## [1,]    4
## [2,]    7
m11add(v1,m11add(v2,v3))
##      [,1]
## [1,]    6
## [2,]    5
m11add(m11add(v1,v2),v3) == m11add(v1,m11add(v2,v3))
##       [,1]
## [1,] FALSE
## [2,] FALSE