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MAE326HW11Soln

# MAE326HW11Soln - MAE 326 Spring 2008 1 Review Problem 1...

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MAE 326 - Spring 2008 Homework 11 Solutions 1) Review Problem 1 For this problem we are to obtain the eigenvalues and eigenvectors of the matrix A : A = 6 2 0 - 2 As we know from linear algebra, the eigenvalues λ of A are the roots of the characteristic polynomial defined by A - λ · I = λ · I - A = 0. In this problem, that polynomial is: 6 - λ 2 0 - 2 - λ = 0 or ( λ - 6)( λ + 2) = 0. The roots of the polynomial are λ = { 6 , - 2 } , which are the eigenvalues of A . The eigenvectors v for a particular eigenvalue λ i are then defined as the basis for the null space of the matrix A - λ i · I . The eigenvectors may be found by solving the problem: ( A - λ · I ) v = 0 (1) for the eigenvector v . All solutions to equation 1 are eigenvectors associated with eigenvalue λ i . For λ = 6, we have: 6 - 6 2 0 - 2 - 6 · v x v y = 0 0 which defines a pair of algebraic equations for the components v x and v y of eigenvalue λ = 6 (one for each row of the matrix equation). These algebraic equations are: 2 v y = 0 - 8 v y = 0 The only useful information we get from these is the constraint v y = 0. The equations thus describe a family of eigenvectors v 6 for λ = 6: v 6 = α 0 T for any nonzero number α . The eigenvectors for λ = - 2 can be obtained by creating a similar pair of equations: 6 - - 2 2 0 - 2 - - 2 · v x v y = 0 0 or: 8 v x + 2 v y = 0 0 = 0 The only useful information we get from these equations is the constraint - 4 v x = v y . As before, this describes a family of eigenvectors v - 2 for eigenvalue λ = - 2: v - 2 = α · 1 - 4 T for any nonzero number α . 11 - 1

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MAE 326 - Spring 2008 Homework 11 Solutions 2) Review Problem 2 For this problem we are to solve the matrix equation B x = 2 x for the vector x , given the matrix B: B = 1 2 3 0 2 - 1 0 0 1 First, we notice that the matrix equation B x = 2 x is just an eigenvalue / eigenvector equation for the matrix B , provided that λ = 2 is an eigenvalue of B . Fortunately, we’re also told that λ = 2 is an eigenvalue of B , and we’re even given the associated eigenvector v 2 . We can therefore immediately write down the solution: x = α · 2 1 0 T for any constant α . Note that technically, α = 0 does satisfy the equation... it’s just not all that interesting. 11 - 2
MAE 326 - Spring 2008 Homework 11 Solutions 3) Review Problem 3 For this problem we are to sketch the solution trajectories of the linear matrix ordinary differential equation: ˙ x = A · x = 6 2 0 - 2 · x Since we already have the eigenvalues λ = { 6 , - 2 } and eigenvectors v 6 and v - 2 from Review Problem 1, we can just sketch the solution trajectories straight away. First, we know from linear algebra and ordinary differential equation techniques that the time solution x ( t ) will take the form: x ( t ) = x 0 · e A · t where x 0 is the initial condition x 0 = x ( t = 0). Using the eigenvalue / eigenvector decomposition to diagonalize the matrix A , the solution may also be written: x ( t ) = c 6 v 6 · e 6 t + c - 2 v - 2 · e - 2 t where c 6 and c - 2 are constant coefficients. From this solution form we know that as t → ∞ , all solutions for which c 6 = 0 will head toward a constant multiple of the vector v 6 . Conversely, as t → -∞ , all solutions for which c - 2

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