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.
$\endgroup$ 12 Answers
$\begingroup$Dimension of the column space = number of linearly independent columns = column rank = row rank = number of linearly independent rows = rank.
By inspecting the original matrix, it should be apparent how many of the rows are linearly independent. Certainly the reduced row echelon form makes it clear that the rank is 3. Now apply the rank-nullity theorem to obtain the nullity (dimension of the null space):
$\text{dim}(\mathbb{R}^7) = \text{rank} + \text{nullity}$
So $7 = 3 + \text{nullity}$, whence $\text{nullity} = 4$.
$\endgroup$ $\begingroup$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).
