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Unformatted text preview: P r e l i m i n a r y d r a f t o n l y : p l e a s e c h e c k f o r fi n a l v e r s i o n ARE211, Fall 2007 LECTURE #14: TUE, OCT 16, 2007 PRINT DATE: AUGUST 21, 2007 (LINALGEBRA4) Contents 3. Linear Algebra (cont) 1 3.11. The graph of a linear function from R 2 to R 2 (cont) 1 3.12. Computational details 3 3.13. Determinants, Rank and volume 4 3.14. Solving linear equation systems and Cramers Rule 6 3. Linear Algebra (cont) 3.11. The graph of a linear function from R 2 to R 2 (cont) So far, we have been considering only symmetric matrices. Reason is that we were talking about eigenvalues, eigenvectors and definiteness. For symmetric matrices, these relationships are very clearcut. The waters are substantially muddied with nonsymmetric matrices. For our purposes, we are only interested definiteness of matrices when the matrices are Hessians, and these matrices are always symmetric, i.e., the crosspartials are identical. So the restriction is harmless. Specifically, the following facts are true: For any symmetric n n matrix, there exists a set of eigenvectors which each have unit length and are pairwise orthogonal, i.e., for any two elements in the set, v 1 and v 2 , the inner product v 1 v 2 is zero. 1 2 LECTURE #14: TUE, OCT 16, 2007 PRINT DATE: AUGUST 21, 2007 (LINALGEBRA4) For any symmetric n n matrix, the matrix is positive ( negative ) definite if and only if all of its eigenvalues are positive (negative). For any symmetric n n matrix, the rank of the matrix is equal to the number of nonzero eigenvalues. More precisely, let A be a symmetric n n matrix, let V = braceleftbig v 1 ,...,v n bracerightbig be a set of pairwise orthogonal eigenvectors for A and let = { 1 ,..., n } be a set of corresponding eigenvalues, i.e., for each i , Av i = i v i . Then the rank of A is equal to the number of nonzero elements of . Semidefinite matrices Take the following symmetric matrix: D = 1 2 2 4 . Note that both vectors are pointing in the same direction, so that we know its singular, and that we are going to get something degenerate Indeed, look at what happens to the unit circle: any vector ( x 1 ,x 2 ) gets mapped into ( x 1 + 2 x 2 ) 1 2 (because the second vector in the matrix is in fact 2 1 2 ), i.e., the unit circle collapse to a line which is of course the subspace of R 2 that is spanned by the columns of D Note that it does indeed have two orthogonal eigenvectors: v 1 = parenleftBig 1 5 , 2 5 parenrightBig and v 2 = parenleftBig 2 5 , 1 5 parenrightBig . v 2 gets mapped to (0 , 0), i.e., ( v 2 1 ,v 2 2 ) gets mapped into ( v 2 1 + 2 v 2 2 ) 1 2 = ( 2 + 2 * 1) 1 2 = 0 Corresponding eigenvalues are: (5,0) The first eigenvector gets pulled out along the line The second goes nowhere Similarly, a 3 3 symmetric matrix which is semidefinite will take the unit sphere either to an...
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This note was uploaded on 08/01/2008 for the course ARE 211 taught by Professor Simon during the Fall '07 term at University of California, Berkeley.
 Fall '07
 Simon

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