Overall Graph:
Lecture 9. Independence, Basis, and Dimension with Nullspace.
- A with m rows, n columns, m <<n. Means we have more unknown variables than data. There are non-zero solutions to Ax = 0. Because we have more free columns. And free variables.
- Independent vector: c1 * x1 + c2 * x3 + …
no c will give 0, all c will give non-zero c1 * x1 + c2 * x3 + … = 0 ,
except for all c=0.
In other words, to reach origin, all vectors need to contribute something, otherwise, the endpoint will not be the origin. That is the reason why NULLSPACE is used to study independence.
3. x1 and 0 are dependent, due to c1*0 + c2*x1 = 0
4. For A, m <<n, there is always a solution for nullspace. So always depend.
5. Assume for v1 … vn, are columns of A, they are independent if null space is zero vectors. Dependent otherwise Ac=0.
6. Span a space means the space consist of all combinations of those vectors.
7. A sequence of vectors span the space, and independent is called the basis.
8. Square matrix: invertible.
9. There will be no free column. All basis have the same number of vectors, the i.e dimension.
10. The rank of A is the number of pivot column, and is also the dimension of column space. Not dimension of A.
12. All in all
The dimension of column space = Number of columns — Number of Free variables.
Lecture 10. Four subspaces
- Column space C(A), Nullspace N(A), Row space R(A), Nullspace of Row space N(A^T).
- The nullspace of Row space N(A^T) it also can be called the left of the nullspace of A.
- Row space, all combination of rows of A, all combinations of columns of A^T — C(A^T).
- Column space C(A) in R^m, Nullspace N(A) is in R^n, Row space is in R^n.
Column Space C(A):
The number of pivots — Rank r decides how many columns are independent.
Dim(C(A)) = Rank = r.
Row Space C(A^T):
Dim(C(A^T)) = r.
Nullspace N(A): Solution space for Ax=0
The number of free variables = n — r = dimension fo N(A)
Nullspace N(A): Solution space for A^Tx=0
The number of free variables = m— r = dimension of N(A^T)
6. The relationship between Row and Column space.
When we do row elimination to get Echelon Matrix R, the column space changed. C(R)!=C(A). Row operations preserve the row space. They have the same row space.
7. Basis for row space is first r rows of R.
8. Why left nullspace N(A^T).
9. Review of Gaussian-Jorden:
Purpose, Record the steps (directly find a matrix from A to R).
Step 1: attach an Identity matrix.
Step 2: Do Row elimination to get Echelon R as usual.
Step 3: E recorded everything EA=R
Step 4: For invertible matrix A, R = I, EA=I, E = A^-1.
10. For A^Ty = 0, E*y = 0
Instead of looking for columns that give the zero. This time, looking for rows that gives zero 0 rows. Because if it gives 0, then that means they are dependent. The rest of the non-zero combinations will be the basis?
11. Assume a new vector space M, M is R^{3*3}. M is a collection of 3x3 matrices. Coz, do operations like adding, multiply by scalar will not change the property of it.
12. It is equivalent to say, instead of talking about the real numbers, now it is talking about the matrice. R ->R^{3*3}
These 3 matrices are independent, and they are the basis of the diagonal matrices.
Lecture 11. Push the boundary from R to High-dimension
- Instead of considering numbers, this lecture considers matrices M as entries of a big matrix.
- Basis of M: Consider the matrix as a ‘number’, how many numbers you need to decide an M? 3x3=9. (It is like a 9-dimensional space.)
3. The dimension of M is 9. The dimension of the symmetric matrix is 6 (6 entries at upper triangular or lower triangular). The dimensional of a diagonal matrix is 3( 6 entries on the diagonal)
4.
Dim(S) + Dim(U) = 12 = Dim(S and U) + Dim(S + U)
5. The idea of dimension, basis can be extended to wider applications(not just matrix). The only thing we need is that it is able to do addition and multiplication, i.e. like vector space.
6. Rank 1 matrices (Very important!)
Any rank 1 matrices has the form of some row multiplied by some column.
E.g. Any matrix (5x17) with rank=4, => 4 Rank 1 matrices.
All Rank 4 matrix M and subsets, form a subspace? i.e M + M’ =M? No, the rank will between 4–8. M * c= M? No.
7. vector v is in R⁴(space).
Subspace S(S * c = 0 * c = 0) of v that: Av=0. What the basis of S? The combination of the basis of S will make Av=0.
A: m = 1, n = 4, pivot = 1
Nullspace for A: Ax=0, No. Free column = N(A) = 3 = n-r.
The column space of A is Number of columns-Number of Free variables = 1.
The Nullspace of A^T is {0}, because only one column, and one pivot. zero free variables, only one particular solution, which is 0.
8. Graph
5 Nodes, 6 Edges. 5x6 matrix to describe it.
