How do you calculate the dimensions of the null space and column space of the following matrix? 
I understand you are supposed to get the reduced row echelon form, which I did, and this is what I came up with:

1   -2  0   19  -6  0   -37
0   0   1   -6  2   0   6
0   0   0   0   0   1   3
0   0   0   0   0   0   0

From here, I know you're supposed to put it in equations, which I also did, and this is what I got:
x1 – 2x2 + 19x4 – 6x5 – 37x7 = 0
x3 – 6x4 + 2x5 + 6x7 = 0
x6 + 3x7 = 0
x1 = 2x2 – 19x4 + 6x5 + 37x7
x3 = 6x4 – 2x5 – 6x7
x6 = -3x7
From here I know you make the columns, but what I don't know is if I'm supposed to also solve the equations for x2, x4, x5, and x7, and make columns for those as well, which would give me a different dimension for the column space.  Do I do that or do I stick with the current equations only and end up with a column space of 4?
Thank you for your help.

Because the matrix is already in row-echelon form:


*

*The number of leading $1$'s (three) is the rank; in fact, the columns containing leading $1$'s (i.e., the first, third, and sixth columns) form a basis of the column space.

*The number of columns not containing leading $1$'s (four) is the dimension of the null space (a.k.a. the nullity).
