8 - Complete solution of Ax = b, rank of matrix
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add image of Ax = b
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For xparticular => set all free variables to
0 - Rank =>
- Number of pivot
- Number of independent columns/rows
- Dimension of column space
- r <=
max(m, n)- m = rows, n = columns
| Condition | Solution | Comment |
|---|---|---|
| r = n < m | 0 or 1 | No free variables. Hence, null space is empty |
| r = m < n | Infinite | Every row has pivot, n-r free variables |
| r = m = n | Unique | Invertible matrix |
| r < m && r < n | 0 or Infinite | depends on b |
9 - Linear Independence, Span, Basis
- Independent vectors => Linear combination is 0 only if all scalars are 0
- Span => vectors v1, v2, v3, v4, v5…., vn span a space if whole space can be generated using linear combinations of these vectors
- Basis => Set of vectors that span a space and are linearly independent
- right enough vector to span a space
- for a given space there exists infinite number of basis, but all basis consists of same number of vectors (equal to dimension of space)
10 - Four fundamental subspaces
For a matrix A
| Column Space C(A) | combination of columns of A | Rr in Rm |
| Null Space N(A) | all solution of Ax = 0 |
Rn-r in Rn |
| Row Space C(AT) | combination of rows of A | Rr in Rn |
| Left Null Space N(AT) | all solution of ATy = 0 | Rm-r in Rm |
add image showing othogonality between spaces
add image of matrix R with elimination steps
- Basis of
rowspace of A ==> firstrrows of R or A - Basis of
columnspace of A ==> pivot columns ofA - Basis of
nullspace of A ==> special solution of A - Basis of
left nullspace of A ==> transforming[A | I] ---> [R | E], look for combination of rows which give zero row