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Lecture_11 - Lecture Note 11 Dr Jeff Chak-Fu WONG...

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Lecture Note 11 Dr. Jeff Chak-Fu WONG Department of Mathematics Chinese University of Hong Kong [email protected] MAT 2310 Linear Algebra and Its Applications Fall, 2007 Produced by Jeff Chak-Fu WONG 1
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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 R EAL V ECTOR S PACES 2
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H OMOGENEOUS S YSTEMS H OMOGENEOUS S YSTEMS 3
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Our aim is to solve several problems involving homogeneous systems that will be fundamental in “Eigenvalues, Eigenvectors, and Diagonalization”. Here we able to focus our attention on these problems without distracted by the additional material in “Eigenvalues, Eigenvectors, and Diagonalization” H OMOGENEOUS S YSTEMS 4
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Consider the homogeneous system A x = 0 , where A is an m × n matrix. As we have already observed in Example 9 of Lecture note 9, 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 Gauss-Jordan reduction presented in Lecture note 3. Thus we transform the augmented matrix [ A | 0 ] of the system to a matrix [ B | 0 ] 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. H OMOGENEOUS S YSTEMS 5
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If r = n , then [ B | 0 ] = 1 0 · · · 0 0 0 0 1 · · · 0 0 0 . . . . . . . . . . . . . . . . . . 0 0 · · · 0 1 mn = rr 0 r × 1= n × 1( r = n ) 0 0 · · · 0 0 . . . . . . . . . . . . 0 0 · · · 0 0 m × 1 and the only solution to A x = 0 is the trivial one. The solution space has no basis (sometimes referred to as an empty basis ) and its dimension is zero . H OMOGENEOUS S YSTEMS 6
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If r < n , then [ B | 0 ] = 1 0 0 · · · 0 b 1 r +1 · · · b 1 n 0 0 1 0 · · · 0 b 2 r +1 · · · b 2 n 0 0 0 1 · · · 0 . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 · · · 1 b rr +1 · · · b rn 0 r × 1 0 0 0 · · · 0 0 · · · 0 0 . . . . . . . . . . . . . . . . . . . . . 0 0 0 · · · 0 · · · 0 mn 0 m × 1 H OMOGENEOUS S YSTEMS 7
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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 . . . x r = - b rr +1 x r +1 - b rr +2 x r +2 - · · · - b rn x n , where x r +1 , x r +2 , . . . , x n can be assigned arbitrary real values s j , j = 1 , 2 , . . . , p and p = n - r ( dimension of solution space ). Thus x = x 1 x 2 .
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