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**Unformatted text preview: **7.3: LINEAR INDEPENDENCE, EIGENVALUES, AND EIGENVECTORS KIAM HEONG KWA We will outline the following linear algebraic notions: The linear independence of a finite set of vectors; The determinant of a square matrix; The eigenvalues and eigenvectors of a square matrix. During lectures, only the ones needed for this class will be highlighted. 1. Systems of Linear Algebraic Equations A set of n simultaneous linear algebraic equations in n variables a 11 x 1 + a 12 x 2 + + a 1 n x n = b 1 , (1.1) a 21 x 1 + a 22 x 2 + + a 2 n x n = b 2 , . . . a n 1 x 1 + a n 2 x 2 + + a nn x n = b n can be expressed economically in matrix notation as follows. Set (1.2) A = ( a ij ) = a 11 a 12 a 1 n a 21 a 22 a 2 n . . . . . . . . . a n 1 a n 2 a nn , x = x 1 x 2 . . . x n , and b = b 1 b 2 . . . b n . Then (1.1) is equivalent to (1.3) Ax = b . This system is said to be homogeneous if b = ; otherwise, it is called non-homogeneous . If A is nonsingular 1 , then x = I n x = A- 1 Ax = A- 1 b . Date : February 27, 2011. 1 Recall that a matrix A is nonsingular or invertible if and only if it is a square matrix and there is a square matrix A- 1 such that AA- 1 = A- 1 A = I n , where I n is the identity matrix of order n and n is the order of A . Otherwise, A is said to be singular or non-invertible. 1 2 KIAM HEONG KWA Hence if A is nonsingular, then (1.3) has only the solution A- 1 b . How- ever, if A is singular, then (1.3) may either have no solution at all or have infinitely many solutions. It is usually more efficient to solve (1.3) by performing elementary row operations on the augmented matrix (1.4) A | b = a 11 a 12 a 1 n | b 1 a 21 a 22 a 2 n | b 2 . . . . . . . . . | . . . a n 1 a n 2 a nn | b n The goal is to transform the part in A | b which corresponds to A into a lower triangular matrix, i.e., a square matrix whose elements below the main diagonal are all zeros: c 11 c 12 c 13 c 1 n- 1 c 1 n c 22 c 23 c 2 n- 1 c 2 n c 33 c 3 n- 1 c 3 n . . . . . . . . . . . . . . . . . . c n- 1 n- 1 c n- 1 n c nn . If (1.4) is transformed into (1.5) c 11 c 12 c 13 c 1 n- 1 c 1 n | d 1 c 22 c 23 c 2 n- 1 c 2 n | d 2 c 33 c 3 n- 1 c 3 n | d 3 . . . . . . . . . . . . . . . . . . | . . . c n- 1 n- 1 c n- 1 n | d n- 1 c nn | d n through a sequence of elementary row operations, then c nn x n = d n , c n- 1 n- 1 x n- 1 + c n- 1 n x n = d n- 1 , etc. It is then easy to see whether the system has solutions, and to calculate them if they do exist....

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