Problem Set 1:

  1. What is the rank of the matrix A?

To find the rank of matrix A, we need to reduce the matrix into reduced row-echelon form and see how many pivot columns there are. We can get the rank from the pivot variables because every pivot variable is associated with a linearly independent vector in the column space.The number of basis vectors required to span the column space is equal to the number of pivot variables in a matrix. I’ll now reduce matrix A using the echelon function from the matlib library.

library(matlib)
A <- matrix(c(1,-1,0,5,2,0,1,4,3,1,-2,-2,4,3,1,-3), nrow = 4)
matlib::echelon(A)
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0    0
## [2,]    0    1    0    0
## [3,]    0    0    1    0
## [4,]    0    0    0    1

Now that matrix A has been reduced to row-echelon form, we can count the pivot variables, which will be the counting the 1’s along the diagonal. In our case, there are 4 pivot columns, meaning the rank of this matrix is 4.

  1. Given an mxn matrix where m > n, what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero?

Given the information that the rows are greater than the columns, our maximum rank can only be less than or equal to n - the number of columns (rank <= n). This calculation is telling us the number of linearly independent column (or row) vectors in the matrix. Had the question not said that m > n, than rank would have to be less than or equal the min(m,n). This is because you can’t have more than min(m,n) row/column vectors to span the space - for example if you have a 3 x 5 matrix, your max rank would be 3, meaning that two of the column vectors would not be linearly independent. Assuming that a matrix is non-zero, the minimum rank of a matrix would be 1. This is because if you even have one non-zero value, you are creating single vector that would be linearly independent within the matrix.

  1. What is the rank of matrix B?

As we did in question 1, we’ll reduce the following matrix to reduced row-echelon form. Once this is complete, we will look at the pivot entries or non-zero rows to determine the rank. This will tell us how many linearly independent rows are in the matrix.

B <- matrix(c(1,3,2,2,6,4,1,3,2), nrow = 3)
matlib::echelon(B)
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    0    0    0
## [3,]    0    0    0

Based on the reduced matrix above, the rank is 1 for matrix B.

Problem Set 2: