§ Four fundamental subspaces
- Column space / Image: C(A), since it corresponds to C(A)≡{y:∃x,y=Ax}
- Null space N(A)≡{k:Ak=0}.
- Row space: row spans the row space, so it's all linear combinations of the rows of A. This is the same as all combinations of the columns of AT. Row space is denoted by C(AT).
- Null space of AT: N(AT), also called as the "left null-space of A".
Let A be m×n. The Null space of A is in Rn. The column
space is in Rm. The rows of A are in Rn. The nullspace
of AT is in Rm.
We want a basis for each of those spaces, and what are their dimensions?
- The dimension of the column space is the rank r.
- The dimension of the row space is also the rank r.
- The dimension of the nullspace is n−r.
- Similarly, the left nullspace must be m−r.
§ Basis for the column space
The basis is the pivot columns, and the rank is r.
§ Basis for the row space
C(R)=C(A). Row operations do not preserve the column space, though
they have the same row space. The basis for the row space of A and R
since they both have the space row space, we just read off the first r
rows of R.
§ Basis for null space
The basis will be the special solutions. Lives in Rn
§ Basis for left null space
It has vectors y such that ATy=0. We can equally write this as
yTA=0. Can we infer what the basis for the left null space is
from the process that took us from A to R? If we perform gauss-jordan,
so we compute the reduced row echelon form of [Am×nIm×m],
we're going to get [RE] where E is whatever the identity matrix became.
Since the row reduction steps is equivalent to multiplying by some matrix M,
we must have that:
M[AI]=[RE]MA=R;MI=E⟹M=E
So the matrix that takes A to R is E! We can find the basis for the left
nullspace by lookinag at E, because E gives us EA=R.
§ Reference