In the vector space C3, compute the vector representation pb(v) for the basis B and vector v below. \[\begin{equation*}B = { \mathbf{} \left[ \begin{matrix} 2 \\ -2\\ 2\\ \end{matrix} \right] , \left[ \begin{matrix} 1\\ 3\\ 1\\ \end{matrix} \right] ,\left[ \begin{matrix} 3\\ 5\\ 2\\ \end{matrix} \right] } v = \left[ \begin{matrix} 11\\ 5\\ 8\\ \end{matrix} \right] \end{equation*}\]
A = matrix(c(2,1,3,11,-2,3,5,5,2,1,2,8),nrow=3,byrow=TRUE)
rref(A)## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 2
## [2,] 0 1 0 -2
## [3,] 0 0 1 3This means the following: \[\begin{equation*} \mathbf{} \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{matrix} \right] \left[ \begin{matrix} a1\\ a2\\ a3\\ \end{matrix} \right] = \left[ \begin{matrix} 2\\ -2\\ 3\\ \end{matrix} \right] \end{equation*}\]
From this we can see:
\[\begin{equation*} \mathbf{} \left[ \begin{matrix} a1\\ a2\\ a3\\ \end{matrix} \right] = \left[ \begin{matrix} 2\\ -2\\ 3\\ \end{matrix} \right] \end{equation*}\]
If we plug it back in we can verify they are equal
\[\begin{equation*} \mathbf{}2* \left[ \begin{matrix} 2\\ -2\\ 2\\ \end{matrix} \right] -2* \left[ \begin{matrix} 1\\ 3\\ 1\\ \end{matrix} \right] +3* \left[ \begin{matrix} 3\\ 5\\ 2\\ \end{matrix} \right] = \left[ \begin{matrix} 11\\ 5\\ 8\\ \end{matrix} \right] \end{equation*}\]