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Unformatted text preview: Lecture Note 6: Oct 10  Oct 13, 2006 Dr. Jeff ChakFu WONG Department of Mathematics Chinese University of Hong Kong jwong@math.cuhk.edu.hk MAT 2310 Linear Algebra and Its Applications Fall, 2006 Produced by Jeff ChakFu WONG 1 R EAL V ECTOR S PACES 1. Vector Spaces 2. Subspaces 3. Linear Independence 4. Basis and Dimension 5. Homogeneous Systems 6. The Rank of a Matrix and Applications 7. Coordinates and Change of Basis 8. Orthonormal Bases in R n 9. Orthogonal Complements REAL VECTOR SPACES 2 H OMOGENEOUS S YSTEMS HOMOGENEOUS SYSTEMS 3 Our aim is to solve several problems involving homogeneous systems that will be fundamental in “Eigenvalues, Eigenvectors, and Diagonalization”. Here we abe able to focus our attention on these problems without distracted by the additional material in “Eigenvalues, Eigenvectors, and Diagonalization” HOMOGENEOUS SYSTEMS 4 Consider the homogeneous system A x = , where A is an m × n matrix. As we have already observed in Example 22 of Lecture note 41, the set of all solutions to this homogeneous system is a subspace of R n . An extremely important problem, which will occur repeatedly in “Eigenvalues, Eigenvectors, and Diagonalization”, is that of finding a basis for this solution space. To find such a basis , we use the method of GaussJordan reduction presented in Lecture note 11. Thus we transform the augmented matrix [ A  ] of the system to a matrix [ B  ] in reduced row echelon form, where B has r nonzero rows, 1 ≤ r ≤ m . Without loss of generality we may assume that the leading 1 s in the r nonzero rows of B occur in the first r columns. HOMOGENEOUS SYSTEMS 5 If r = n , then [ B  ] = 1 ··· 1 ··· . . . . . . . . . . . . . . . . . . ··· 1 mn = rr r × 1= n × 1 ( r = n ) ··· . . . . . . . . . . . . ··· m × 1 and the only solution to A x = is the trivial one. The solution space has no basis (sometimes referred to as an empty basis ) and its dimension is zero . HOMOGENEOUS SYSTEMS 6 If r < n , then [ B  ] = 1 ··· b 1 r +1 ··· b 1 n 1 ··· b 2 r +1 ··· b 2 n 1 ··· . . . . . . . . . . . . . . . . . . . . . . . . ··· 1 b rr +1 ··· b rn ··· ··· . . . . . . . . . . . . . . . . . . . . . ··· ··· mn HOMOGENEOUS SYSTEMS 7 Solving for the unknown corresponding to the leading 1 s, we have x 1 = b 1 r +1 x r +1 b 1 r +2 x r +2··· b 1 n x n x 2 = b 2 r +1 x r +1 b 2 r +2 x r +2··· b 2 n x n ....
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 Spring '09
 JeffWong
 Linear Algebra, Algebra

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