\[ B = \{ \begin{align} \begin{bmatrix} 2\\ -2\\ 3 \end{bmatrix}, \begin{bmatrix} 1\\ 3\\ 1 \end{bmatrix}, \begin{bmatrix} 3\\ 5\\ 2 \end{bmatrix}\} \end{align} \qquad \mathbf{v} = \begin{bmatrix} 11\\ 5\\ 8 \end{bmatrix} \]
The vector \(\mathbf{v}\) can be represented as a unique linear combination of the basis elements of \(B\). To do this, we will first create an augmented matrix with each basis element and the vector.
b1 <- matrix(c(2, -2, 2), nrow=3, ncol=1)
b2 <- matrix(c(1, 3, 1), nrow=3, ncol=1)
b3 <- matrix(c(3, 5, 2), nrow=3, ncol=1)
v <- matrix(c(11, 5, 8), nrow=3, ncol=1)
aug <- matrix(c(b1, b2, b3, v), nrow=3, ncol=4)
aug## [,1] [,2] [,3] [,4]
## [1,] 2 1 3 11
## [2,] -2 3 5 5
## [3,] 2 1 2 8Then, we need to find the unique solutions to the augmented matrix by changing the matrix into upper-triangular form.
a <- aug
# Replace row 2
a[2,] <- a[2,] + a[1,]
# Replace row 3
a[3,] <- a[3,] + (-1 * a[1,])
a## [,1] [,2] [,3] [,4]
## [1,] 2 1 3 11
## [2,] 0 4 8 16
## [3,] 0 0 -1 -3This results in the following system. Using backwards substitution, we find that \(x=2, \ y=-2,\) and \(z=3\)
\[ \begin{align} 2x + y + 3z &= 11 && \text{x = 2} \\ 4y + 8z &= 16 && \text{y = -2} \\ -z &= -3 && \text{z = 3} \end{align} \]
I double-checked my work using the
pracmapackage and found the same unique solutions when reduced to row-echelon form:
rref(aug)## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 2
## [2,] 0 1 0 -2
## [3,] 0 0 1 3Thus, the vector representation \(\rho _B\) of \(\mathbf{v}\) can be expressed as:
\[ \rho _B(\mathbf{v}) = \begin{align} 2 \begin{bmatrix} 2\\ -2\\ 2 \end{bmatrix}- 2 \begin{bmatrix} 1\\ 3\\ 1 \end{bmatrix}+ 3 \begin{bmatrix} 3\\ 5\\ 2 \end{bmatrix} = \begin{bmatrix} 2\\ -2\\ 3 \end{bmatrix} \end{align} \]