Chapter 4 General Vector Spaces Face Recognition ( á¹¢å ± 彐嬀)

Rank• Theorem 4.13: The row space and the column space of amatrix A have the same dimension. See the book for a proof The dimension is called the rank of A, denoted rank(A) rank(A) min (m, n) for A R mnLA 04_37• Example:— Row space 1 2 0 00 0 1 01 412 0 0 , 0 0 1 0 R — Column space: 1 2 0 0 1 02 0 , 0 , 1 , 0 0 , 1 R1Theorem 4.15Let A and B be row equivalent matrices. Then A and Bhave the same row space and rank(A) = rank(B).• Idea of Proof:Theorem 4.16BB 110010010100001001AAAAAA123123AAAA1213AA, 13A2BLA 04_39 100010004AAA1234AA12A3Let E be the reduced echelon form of a matrix A. Thenonzero row vectors of E form a basis for the row space ofA. The rank of A is the number of nonzero row vectors in E.Proof: A and E and row equivalent and Theorem 4.15.Properties of the Rank 1 2 3 Example 2 A 0 1 2 2 5 8 — (2, 5, 8) = 2(1, 2, 3) + (0, 1, 2) LA 04_38— (1, 2, 3) and (0, 1, 2) linearly independent rank(A) = 2. Theorem 4.14: The nonzero row vectors of a matrix A thatis in reduced echelon form are a basis for the row space ofA.— Example 3: Three nonzero row vectors Rank(A) = 3.A 1000200001000010A SolutionExample 4Use elementary row operations to find a reduced echelon form121251345LA 04_40121251345 10021 13 22 1000107 20• The two vectors (1, 0, 7) and (0, 1, 2) form a basis for the rowspace of A Rank(A) = 2.• (1, 2, 3) = (1, 0, 7) + 2 (0, 1, 2).• (2, 5, 4) = 2 (1, 0, 7) + 5 (0, 1, 2).• (1, 1, 5) = (1, 0, 7) + (0, 1, 2).