Question LISS.C20

In the vector space of \(2 \times 2\) matrices, \(M_{22}\), determine if the set \(S\) below is linearly independent.

\[\begin{align} \ S= \begin{bmatrix} 2 & -1\\ 1 & 3 \end{bmatrix}, \begin{bmatrix} 0 & 4\\ -1 & 2 \end{bmatrix}, \begin{bmatrix} 4 & 2\\ 1 & 3 \end{bmatrix} \end{align}\]



First, I created the three matrices and one solution matrix with zeros. Then I augmented them together.

v1 <- matrix(c(2,-1,1,3), ncol=2, nrow=2)
v2 <- matrix(c(0,4,-1,2), ncol=2, nrow=2)
v3 <- matrix(c(4,2,1,3), ncol=2, nrow=2)

sol <- matrix(c(0,0,0,0), ncol=2, nrow=2)

v <- matrix(c(v1, v2, v3, sol), ncol=4, nrow=4)

v
##      [,1] [,2] [,3] [,4]
## [1,]    2    0    4    0
## [2,]   -1    4    2    0
## [3,]    1   -1    1    0
## [4,]    3    2    3    0



I applied the Gaussian elimination method (by hand) to reduce the matrix to row-echelon form.


First round:

v[1,] <- v[1,]/v[1,1]

v[2,1] <- v[2,1] - (v[2,1] * v[1,1])
v[2,2] <- v[2,2] - (v[2,1] * v[1,2])
v[2,3] <- v[2,3] - (v[2,1] * v[1,3])
v[2,4] <- v[2,4] - (v[2,1] * v[1,4])

v[3,1] <- v[3,1] - (v[3,1] * v[1,1])
v[3,2] <- v[3,2] - (v[3,1] * v[1,2])
v[3,3] <- v[3,3] - (v[3,1] * v[1,3])
v[3,4] <- v[3,4] - (v[3,1] * v[1,4])

v[4,1] <- v[4,1] - (v[4,1] * v[1,1])
v[4,2] <- v[4,2] - (v[4,1] * v[1,2])
v[4,3] <- v[4,3] - (v[4,1] * v[1,3])
v[4,4] <- v[4,4] - (v[4,1] * v[1,4])

v
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    2    0
## [2,]    0    4    2    0
## [3,]    0   -1    1    0
## [4,]    0    2    3    0


Second round:

v[2,] <- v[2,]/v[2,2]

v[3,2] <- v[3,2] - (v[3,2] * v[2,2])
v[3,3] <- v[3,3] - (v[3,2] * v[2,3])
v[3,4] <- v[3,4] - (v[3,2] * v[2,4])

v[4,2] <- v[4,2] - (v[4,2] * v[2,2])
v[4,3] <- v[4,3] - (v[4,2] * v[2,3])
v[4,4] <- v[4,4] - (v[4,2] * v[2,4])

v
##      [,1] [,2] [,3] [,4]
## [1,]    1    0  2.0    0
## [2,]    0    1  0.5    0
## [3,]    0    0  1.0    0
## [4,]    0    0  3.0    0


Third round:

v[3,] <- v[3,]/v[3,3]

v[4,3] <- v[4,3] - (v[4,3] * v[3,3])
v[4,4] <- v[4,4] - (v[4,3] * v[3,4])

v
##      [,1] [,2] [,3] [,4]
## [1,]    1    0  2.0    0
## [2,]    0    1  0.5    0
## [3,]    0    0  1.0    0
## [4,]    0    0  0.0    0



This results in the following system of equations:

\[\begin{align} a + b &= 0\\ b + 0.5c &= 0\\ c &= 0 \end{align}\]

.

When we solve, we get:

\[\begin{align} a &= 0\\ b &= 0\\ c &= 0 \end{align}\]

.

When we set the system equal to zero, the solution was all zeros. This does not meet the requirements for linear dependence, so the system is linearly independent.