Consider the subspace \(W = \langle \{ \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix} , \begin{bmatrix} 4 & 0 \\ 2 & 3 \end{bmatrix} , \begin{bmatrix} -3 & 1 \\ 2 & 1 \end{bmatrix} \} \rangle\)
of the vector space of \(2x2\) matrices, \(M_{22}\). Is \(C = \begin{bmatrix} -3 & 3 \\ 6 & -4 \end{bmatrix}\) an element of W?
To answer this, we need to determine if C can be written as a linear combination of the 3 matrices in W. So we are looking for 3 scalars, \(\alpha_1, \alpha_2, \alpha_3\) that make the following equation true.
\[\begin{bmatrix} -3 & 3 \\ 6 & -4 \end{bmatrix} = \alpha_1 \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix} + \alpha_2 \begin{bmatrix} 4 & 0 \\ 2 & 3 \end{bmatrix} + \alpha_3 \begin{bmatrix} -3 & 1 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 2\alpha_1+4\alpha_2-3\alpha_3 & 2\alpha_1+\alpha_3 \\ 3\alpha_1+2\alpha_2+2\alpha_3 & -\alpha_1+3\alpha_2+\alpha_3 \end{bmatrix} \] We equate the leftmost and rightmost matrices of the above to create a system of equations:
\[\begin{array} \ 2\alpha_1 + 4\alpha_2 - 3 \alpha_3 = -3 \\ \alpha_1 + 0 \alpha_2 + \alpha_3 = 3 \\ 3\alpha_1 + 2\alpha_2 + 2 \alpha_3 = 6 \\ -\alpha_1 + 3\alpha_2 + \alpha_3 = -4 \end{array} \]
We represent this system as the augmented matrix on the left which row-reduces to the matrix on the right:
\[\begin{bmatrix} 2 & 4 & -3 & -3 \\ 1 & 0 & 1 & 3 \\ 3 & 2 & 2 & 6 \\ -1 & 3 & 1 & -4 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]
See the Row Reduction section for the steps of the computation of the above using the solve function I wrote for the first assignment.
We find the solution \(\begin{bmatrix} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}\) and then conclude that C is an element of W since it can be written as the following linear combination:
\[ 2 \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix} + -1 \begin{bmatrix} 4 & 0 \\ 2 & 3 \end{bmatrix} + 1 \begin{bmatrix} -3 & 1 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} -3 & 3 \\ 6 & -4 \end{bmatrix} \] It can also be determined from the row reduced matrix that the system is consistent. Therefore a solution exists, making C an element of W.
A <- matrix(c(2, 4, -3, 1, 0, 1, 3, 2, 2, -1, 3, 1), nrow=4, ncol=3, byrow = TRUE)
b <- matrix(c(-3, 3, 6, -4), nrow=4, ncol=1, byrow = TRUE)## [,1] [,2] [,3] [,4]
## [1,] 2 4 -3 -3
## [2,] 1 0 1 3
## [3,] 3 2 2 6
## [4,] -1 3 1 -4
## Scale A[ 1 , 1 ] to leading one
## [,1] [,2] [,3] [,4]
## [1,] 1 2 -1.5 -1.5
## [2,] 1 0 1.0 3.0
## [3,] 3 2 2.0 6.0
## [4,] -1 3 1.0 -4.0
## Make A[ 2 , 1 ] zero
## [,1] [,2] [,3] [,4]
## [1,] 1 2 -1.5 -1.5
## [2,] 0 -2 2.5 4.5
## [3,] 3 2 2.0 6.0
## [4,] -1 3 1.0 -4.0
## Make A[ 3 , 1 ] zero
## [,1] [,2] [,3] [,4]
## [1,] 1 2 -1.5 -1.5
## [2,] 0 -2 2.5 4.5
## [3,] 0 -4 6.5 10.5
## [4,] -1 3 1.0 -4.0
## Make A[ 4 , 1 ] zero
## [,1] [,2] [,3] [,4]
## [1,] 1 2 -1.5 -1.5
## [2,] 0 -2 2.5 4.5
## [3,] 0 -4 6.5 10.5
## [4,] 0 5 -0.5 -5.5
## [,1] [,2] [,3] [,4]
## [1,] 1 2 -1.5 -1.5
## [2,] 0 -2 2.5 4.5
## [3,] 0 -4 6.5 10.5
## [4,] 0 5 -0.5 -5.5
## Scale A[ 2 , 2 ] to leading one
## [,1] [,2] [,3] [,4]
## [1,] 1 2 -1.50 -1.50
## [2,] 0 1 -1.25 -2.25
## [3,] 0 -4 6.50 10.50
## [4,] 0 5 -0.50 -5.50
## Make A[ 1 , 2 ] zero
## [,1] [,2] [,3] [,4]
## [1,] 1 0 1.00 3.00
## [2,] 0 1 -1.25 -2.25
## [3,] 0 -4 6.50 10.50
## [4,] 0 5 -0.50 -5.50
## Make A[ 3 , 2 ] zero
## [,1] [,2] [,3] [,4]
## [1,] 1 0 1.00 3.00
## [2,] 0 1 -1.25 -2.25
## [3,] 0 0 1.50 1.50
## [4,] 0 5 -0.50 -5.50
## Make A[ 4 , 2 ] zero
## [,1] [,2] [,3] [,4]
## [1,] 1 0 1.00 3.00
## [2,] 0 1 -1.25 -2.25
## [3,] 0 0 1.50 1.50
## [4,] 0 0 5.75 5.75
## [,1] [,2] [,3] [,4]
## [1,] 1 0 1.00 3.00
## [2,] 0 1 -1.25 -2.25
## [3,] 0 0 1.50 1.50
## [4,] 0 0 5.75 5.75
## Scale A[ 3 , 3 ] to leading one
## [,1] [,2] [,3] [,4]
## [1,] 1 0 1.00 3.00
## [2,] 0 1 -1.25 -2.25
## [3,] 0 0 1.00 1.00
## [4,] 0 0 5.75 5.75
## Make A[ 1 , 3 ] zero
## [,1] [,2] [,3] [,4]
## [1,] 1 0 0.00 2.00
## [2,] 0 1 -1.25 -2.25
## [3,] 0 0 1.00 1.00
## [4,] 0 0 5.75 5.75
## Make A[ 2 , 3 ] zero
## [,1] [,2] [,3] [,4]
## [1,] 1 0 0.00 2.00
## [2,] 0 1 0.00 -1.00
## [3,] 0 0 1.00 1.00
## [4,] 0 0 5.75 5.75
## Make A[ 4 , 3 ] zero
## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 2
## [2,] 0 1 0 -1
## [3,] 0 0 1 1
## [4,] 0 0 0 0## $matrix
## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 2
## [2,] 0 1 0 -1
## [3,] 0 0 1 1
## [4,] 0 0 0 0
##
## $solution
## [1] 2 -1 1 0