$Post:=If[MatrixQ[#],TableForm[#],#]&

Exercise 2.5.1

A=Array[Random[Integer,{-100,100}]&,{5,7}]

RowReduce[A]//TableForm

RowReduce[Transpose[A]]

By inspection, one sees that the row rank above is
equal to the column rank (and therefore the dimensions
of row space and column space are the same).

Exercise 2.5.2

f[i_,j_]:=i+j

a)

A=Array[f,{5,13}]

NullSpace[A]

RowReduce[A]

RowSpaceBasis={%[[1]],%[[2]]}

RowReduce[Transpose[A]]

ColumnSpaceBasis={%[[1]],%[[2]]}

b)

A=Array[f,{11,7}]

NullSpace[A]

RowReduce[A]

RowSpaceBasis={%[[1]],%[[2]]}

RowReduce[Transpose[A]]

ColumnSpaceBasis={%[[1]],%[[2]]}

c)

A=Array[f,{16,8}]

NullSpace[A]

RowReduce[A]

RowSpaceBasis={%[[1]],%[[2]]}

RowReduce[Transpose[A]]

ColumnSpaceBasis={%[[1]],%[[2]]}

Exercise 2.5.3

Yes, it generalizes to the fact that the rank of the
matrix ( = dimension of the column space = dimension
of the row space) and the dimension of the null space
sum up to the number of columns. Reason being that each
column will either
            - be linearly independent of the preceding,
              and therefore able to be part of a basis
              for the column space,
            or
            - be a linear combination of the preceding,
              in which case the corresponding x[i] it
              multiplies in Ax=0 is arbitrary, therefore
              it generates one more dimension for the
              null space.

Exercise 2.5.4

h[i_,j_]:=1/(i+j-1)

a)

A=Array[h,{13,15}]

NullSpace[A]

RowSpaceBasis=RowReduce[A]

RowReduce[Transpose[A]]

ColumnSpaceBasis=Array[%[[#]]&,13]

b)

A=Array[h,{14,3}]

NullSpace[A]

RowSpaceBasis=Array[RowReduce[A][[#]]&,3]

ColumnSpaceBasis=RowReduce[Transpose[A]]

c)

A=Array[h,{27,5}]

NullSpace[A]

RowSpaceBasis=Array[RowReduce[A][[#]]&,5]

ColumnSpaceBasis=RowReduce[Transpose[A]]