§ Four fundamental subspaces

Let AA be m×nm \times n. The Null space of AA is in Rn\mathbb R^n. The column space is in Rm\mathbb R^m. The rows of AA are in Rn\mathbb R^n. The nullspace of ATA^T is in Rm\mathbb R^m. We want a basis for each of those spaces, and what are their dimensions?
  • The dimension of the column space is the rank rr.
  • The dimension of the row space is also the rank rr.
  • The dimension of the nullspace is nrn - r.
  • Similarly, the left nullspace must be mrm - r.

§ Basis for the column space

The basis is the pivot columns, and the rank is rr.

§ Basis for the row space

C(R)C(A)C(R) \neq C(A). Row operations do not preserve the column space, though they have the same row space. The basis for the row space of AA and RR since they both have the space row space, we just read off the first rr rows of RR.

§ Basis for null space

The basis will be the special solutions. Lives in Rn\mathbb R^n

§ Basis for left null space

It has vectors yy such that ATy=0A^T y = 0. We can equally write this as yTA=0y^T A = 0. Can we infer what the basis for the left null space is from the process that took us from AA to RR? If we perform gauss-jordan, so we compute the reduced row echelon form of [Am×nIm×m][A_{m\times n} I_{m \times m}], we're going to get [RE][R E] where EE is whatever the identity matrix became. Since the row reduction steps is equivalent to multiplying by some matrix MM, we must have that:
M[AI]=[RE]MA=R;MI=E    M=E \begin{aligned} &M [AI] = [RE] \\ &MA = R; MI = E \implies M = E \end{aligned}
So the matrix that takes AA to RR is EE! We can find the basis for the left nullspace by lookinag at EE, because EE gives us EA=REA = R.

§ Reference