How to calculate the rank of a matrix?
rank: number of pivots in a matrix.
pivot position: the location that corresponds to a leading 1 in the rref of a matrix.
rref: reduced row echelon form.
Example:
A = 
continue the elimination, R3 = R3-1/2*R2, then the matrix is B=
We can easily find that there are 3 non-zero rows, so the rank of matrix A is 3. Pivots are 1,-2,1.
In matrix B, we have four columns. Column with pivot is called pivot column, and column without pivot value is a free column. In matrix B, there are 3 pivot variables and 1 free variable. That is to say, the column space is specified with only 3 pivot variables. We can use linear combinations of col1, col2, and col4 to get co3.
What is rref of matrix A? Now, we've got matrix B. rref is a matrix where the pivot is always one and other values in pivot column are 0.
Based on matrix B, R2 = R2/-2;R1 = R1-2*R2;R1 = R1+2*R3;R2 = R2-3*R3. rref(A) = 
Background knowledge:
1. Nullspace of A: all solutions of x in Ax = 0;
when doing elimination, nullspace doesn't change.
Ax=0 <=>rref(A) x = 0;
Besides, nullspace is a vector space.
// vector space: when v,w ∈ space n, cv+dw∈n (c,d ∈R).
Proof: v,w∈nullspace of A => Av=0, Aw = 0 => A(v+w) = 0, A(cv) = 0.
2. some definitions:
a: Vectors x1,x2,...,xn are independent iff no combinations give zero vector.
that is, c1*x1+c2*x2+...+cn*xn ≠ 0(except c1=c2=...=cn=0.)
b: Vectors v1,...vl span a space means: the space consists of all combinations of those vectors.
c: Basis for a space is a sequence of vectors, with the following two properties:
1.they are independent; 2. they span the space.
. Besides, given a space, every basis for the space has the same number of vectors, that is called the dimension of the space.
given a matrix A, rank(A) = # pivot columns = dimension(C(A)),
dimension(nullspace(A)) = # free variables(? if they are independent.)