In the vector space C^3, compute the vector representation ρB (v) for the basis B and vector v below.

\[ B = \left(\begin{array}{cc} 2 & 1 & 3\\ -2 & 3 & 5\\ 2 & 1 & 2 \end{array}\right) \] \[ v = \left(\begin{array}{cc} 11 \\ 5 \\ 8 \end{array}\right) \]

Suppose that V is a vector space with a basis B = {v1, v2, v3, …, vn}. Define a function ρB : V → Cn as follows. For w ∈ V define the column vector ρB (w) ∈ Cn by w = [ρB (w)]1 v1 + [ρB (w)]2 v2 + [ρB (w)]3 v3 + · · · + [ρB (w)]n vn

Given w ∈ V , we write w as a linear combination of the basis elements of B.

Let us express v as a linear combination of the vectors in B.

\[ a1 \left(\begin{array}{cc} 2 \\ -2 \\ 2 \end{array}\right) + a2 \left(\begin{array}{cc} 1 \\ 3 \\ 1 \end{array}\right) + a3 \left(\begin{array}{cc} 3 \\ 5 \\ 2 \end{array}\right) = \left(\begin{array}{cc} 11 \\ 5 \\ 8 \end{array}\right) \]

Which is equal to

\[ B = \left(\begin{array}{cc} 2 & 1 & 3 & 11\\ -2 & 3 & 5 & 5\\ 2 & 1 & 2 & 8 \end{array}\right) \]

To find ρB (y), we need to find scalars, a1, a2, a3, a4 such that v=a1u1 +a2u2 +a3u3 +a4u4

A <- matrix(c(2, -2, 2, 1, 3, 1, 3, 5, 2, 11, 5, 8), ncol=4, nrow=3)
A
##      [,1] [,2] [,3] [,4]
## [1,]    2    1    3   11
## [2,]   -2    3    5    5
## [3,]    2    1    2    8
value <- (A[,1]/(-1*A[1,1]))
multiply <- matrix(c(1, value[2], value[3], 0, 1, 0, 0, 0, 1), nrow=3, ncol=3) 
transformation_1 <- multiply %*% A

value_2 <- -1*(transformation_1[3,2]/transformation_1[2,2])
multiply_2 <- matrix(c(1, 0, 0, 0, 1, value_2, 0, 0, 1), nrow=3, ncol=3) 
lower_triangular_matrix <- multiply_2 %*% transformation_1
lower_triangular_matrix
##      [,1] [,2] [,3] [,4]
## [1,]    2    1    3   11
## [2,]    0    4    8   16
## [3,]    0    0   -1   -3

When we solve the system

\[ \left(\begin{array}{cc} 2 & 1 & 3\\ 0 & 4 & 8\\ 0 & 0 & -1 \end{array}\right) \left(\begin{array}{cc} a1\\ a2\\ a3 \end{array}\right) = \left(\begin{array}{cc} 11\\ 16\\ -3 \end{array}\right) \]

\[ -a3 = -3 \\ a3 = 3 \\ \\ 4a2 + 8a3 = 16 \\ a2 = -2 \\ 2a1 + a2 + 3a3 = 11 \\ a1 = 2 \]

Therefore, a1 = 2, a2 = -2, a3 = 3 and

\[ ρB (v) = ρB \left(\begin{array}{cc} 11 \\ 5 \\ 8 \end{array}\right) = ρB( 2 \left(\begin{array}{cc} 2 \\ -2 \\ 2 \end{array}\right) + (-2) \left(\begin{array}{cc} 1 \\ 3 \\ 1 \end{array}\right) + 3 \left(\begin{array}{cc} 3 \\ 5 \\ 2 \end{array}\right)) \]