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 = ± 62 0 - 2 ² As we know from linear algebra, the eigenvalues λ of A are the roots of the characteristic polynomial de±ned by ± A - λ · I ± = ± λ · I - A ± = 0. In this problem, that polynomial is: ± 6 - λ 2 0 - 2 - λ ² =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 de±ned 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 - 0 - 2 - 6 ² · ± v x v y ² = ± 0 0 ² which de±nes 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 - 8 v y 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 = ³ α 0 ´ T for any nonzero number α . The eigenvectors for λ = - 2 can be obtained by creating a similar pair of equations: ± 6 -- 22 0 - 2 2 ² · ± v x v y ² = ± 0 0 ² or: 8 v x +2 v y 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 · ³ 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 = 12 3 02 - 1 00 1 First, we notice that the matrix equation B x 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 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 · ± 210 ² T for any constant α . Note that technically, α =0 does satisfy the equation. .. it’s just not all that interesting. 11-2
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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 = ± 62 0 - 2 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
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This note was uploaded on 05/08/2008 for the course M&AE 326 taught by Professor Psiaki during the Spring '08 term at Cornell University (Engineering School).

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

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