Question

Assume that the matrix A is row equivalent to B. Without calculations, list rank A and dim Nul A. Then find bases for Col A. Row A, and Nul A. 1 1 -2 0 -2 -2 1 1 0 1 -1 0 - 1 1 A= 1 - 1 1 -2 0 -2 -2 2-3 0-3 - 1 0 0 -3 -10 1 - 2 21 5 -2 1 - 1 00 B 1 1 4 3 0 0 00 0 0 1 2 2 0 0 0 0 0 1 rank A= dim Nul A= A basis for Col Ais (Use a comma to separate vectors as needed.) A basis for Row Ais ) (Use a comma to separate vectors as needed) A basis for Nul Ais (Use a comma to separate vectors as needed)

Answer 1

For the matrix B, observe that the submatrix obtained by considering the 1st, 2nd, 3rd , 5th and 6th columns is an upper traingular matrix. Since this submatrix only has 1s in its leading diagonal, hence its determinant is 1(product of the elements in the leading diagonal). Hence, the rank of B is at least 5. Further, since B has 5 rows and 6 columns, hence the rank of B $\leq$ 5 = min{5,6}. Thus, the rank of B is 5.
Now, since the rank of a matrix is preserved under elementary row transformations, hence the rank of A = 5


Observe that the matrix A can be considered as a linear transformation from $\mathbb{R}$ ⁶ to $\mathbb{R}$ ⁵ . Hence, by the rank nullity theorem, 6 = dim Nul A + rank(A) = dim Nul A + 5 which implies that, dim Nul A = 1



Observe that the column rank of B = rank of B = 5. Thus, some 5-element subset of the column vectors of B should be a basis for the column space of B. Observe that the 1st, 2nd and 4th columns are linearly independent. Also, the 1st, 2nd, 3rd and 4th columns are linearly dependent. Consequently, the 1st, 2nd, 4th, 5th and 6th columns should constitute a basis for the column space of B.
Since the column space of a matrix is invariant under elementary row transformations, hence a basis for Col A is
{ (1,1,1,1,1) , (1,2,-1,-2,-1) , (-2,-3,0,2,0) , (0,0,1,0,0) , (-2,-3,-3,5,2) , (-2,-1,-10,-2,3) }




Since the row rank of B = rank of B = 5, hence the set of all the 5 row vectors of B should be a basis for the row space of B.
Since the row space of a matrix is invariant under elementary row transformations, hence a basis for Row A is
{ (1,1,-2,0,-2,-2) , (1,2,-3,0,-3,-1) , (1,-1,0,0,-3,-10) , (1,-2,2,1,5,-2) , (1,-1,0,0,2,3) }




The null space Nul A has dimension 1. Hence, the null space Nul B has dimension 1. A basis for the null space of B is { (1,1,1,-1,0,0) }.
Since the Null space of a matrix is invariant under elementary row transformations, hence a basis for Nul A is
{ (1,1,1,-1,0,0) }