L95 - Hand Examples Fall 2003 Math 308\/501502 9 Matrix Methods for Linear Systems 541\/2 9.5 Homogeneous Linear Systems with Constant Coefficients Find

L95 - Hand Examples Fall 2003 Math 308/501502 9 Matrix...

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Fall 2003 Math 308/501–502 9 Matrix Methods for Linear Systems 9.5 Homogeneous Linear Systems with Constant Coefficients Fri, 14/Nov c 2003, Art Belmonte Summary Today we’ll take a broad view of how to solve a linear system x 0 = Ax with constant coefficients. That is, the n × n coefficient matrix A has constant elements. Definitions An eigenvalue of A is a number r such that Av = r v for some nonzero vector v , which is called an eigenvector associated with the eigenvalue r . Collectively, r and v are called an eigenpair . Equivalently, we have ( A - r I ) v = 0 , where I is the n × n identity matrix and 0 the n × 1 column vector of zeros. Thus v is a nontrivial vector in nullspace of A - r I , a.k.a, the eigenspace of r , a subspace of R n or C n . Observe that any nonzero multiple of an eigenvector v is also an eigenvector associated with eigenvalue r . The characteristic polynomial of A is p ( r ) = det ( A - r I ) . The characteristic equation of A is p ( r ) = 0. Its roots (i.e., the values for which the polynomial is zero) are precisely the eigenvalues of A . These are also known as characteristic roots . Theorem If r is an eigenvalue of a matrix A with assoicated eigenvector v , then the function x ( t ) = e rt v is a solution of the system x 0 = Ax satisfying the vector initial condition x ( 0 ) = v . Roots various 1. If the eigenvalues r k , k = 1 , . . . , n , of a matrix A are distinct real roots , then a fundamental solution set for x 0 = Ax is given by x k ( t ) = e r k t v k , k = 1 , . . . , n , where v k is an eigenvector associated with r k . 2. If some of the eigenvalues are complex numbers, exponential solutions given by the preceding theorem are complex- valued. We’ll learn how to extract real-valued solutions from these in Section 9.6. 3. If the eigenvalues (be they real or complex) are not distinct (i.e., there are repeated roots), then this may present complications. The ultimate resolution to this difficulty is found in the notion of generalized eigenvectors . But that will come later in the chapter . . . Hand Examples 541/2 Find the eigenvalues and eigenvectors of A = 6 - 3 2 1 . Solution To find the characteristic polynomial p ( r ) , we compute the determinant of the matrix A - r I , where I is the identity matrix having the same size as A . p ( r ) = det 6 - r - 3 2 1 - r = 6 - 7 r + r 2 + 6 = r 2 - 7 r + 12 The eigenvalues of A are roots of the characteristic equation, 0 = p ( r ) = r 2 - 7 r + 12 = ( r - 3 )( r - 4 ) , whence r = 3 , 4. For r = 3 , compute the reduced row echelon form of A - r I . 3 - 3 2 - 2 rref 1 - 1 0 0 In order for ( A - r I ) v = 0 , we must have v 1 - v 2 = 0 or v 1 = v 2 . Hence 3 1 1 is an eigenpair. Similarly, for r = 4, we have A - r I = 2 - 3 2 - 3 rref 1 - 3 2 0 0 , whence 4 3 2 is an eigenpair. We thus have a full set of linearly independent eigenvectors. 541/6 Find the eigenvalues and eigenvectors of A = 0 1 1 1 0 1 1 1 0 . Solution Solve 0 = p ( r ) = det ( A - r I ) = det - r 1 1 1 - r 1 1 1 - r = - r ( r 2 - 1 ) - ( - r - 1 ) + ( 1 + r ) = ( r + 1 )( 2 + r - r 2 ) = ( 1 + r ) 2 ( 2 - r ) to obtain eigenvalues r = - 1 , - 1 , 2.

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