Notice that the matrix A is upper-triangle, so its determinant is a
product
of diagonal elements:
det(A) = 1*2*3 = 6.
Moreover, det(A) is
non-zero.
Therefore the columns of A are linearly independent and
therefore
they constitutive a basis for R^3, so R^3 = Col(A).
Hence
any base of R^3 fits to be a base for Col(A),
for instance the columns of
A:
(1,0,0), (2,2,0), (3,3,3)
or
(1,0,0), (0,1,0),
(0,0,1)
are the base.
The dimension of Col(A) is the dimension of
R^3 and therefore is equal
to 3.
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