# Example 124 find the general solution of x 1 1 1 1 1

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Example 124. Find the general solution of x = 1 1 1 1 1 1 1 1 1 x . Solution. The characteristic polynomial is vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 λ 1 1 1 1 λ 1 1 1 1 λ vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = (1 λ ) vextendsingle vextendsingle vextendsingle vextendsingle 1 λ 1 1 1 λ vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 1 1 1 λ vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 1 λ 1 1 vextendsingle vextendsingle vextendsingle vextendsingle = (1 λ ) 3 3(1 λ ) 2 . Since x 3 3 x 2 = ( x + 1) 2 ( x 2) , the eigenvalues are λ = 1 x = 2 , 2 , 1 . Note that λ = 2 is repeated! We say that the eigenvalue λ = 2 has multiplicity 2 . λ = - 1 . 2 1 1 0 1 2 1 0 1 1 2 0 2 r 2 r 1 2 r 3 + r 1 2 1 1 0 0 3 3 0 0 3 3 0 2 1 1 0 0 1 1 0 0 0 0 0 Setting, v 3 = c , we find v 2 = c . Then, 2 v 1 + v 2 v 3 implies v 1 = c . Setting c = 1 , we find v = (1 , 1 , 1) T . λ =2 . 1 1 1 0 1 1 1 0 1 1 1 0 r 2 + r 1 r 3 r 1 1 1 1 0 0 0 0 0 0 0 0 0 The two zero rows are good news! It means that we will find two independent eigenvectors. Indeed, we are free to set v 3 = c and v 2 = d . Since v 1 + v 2 v 3 = 0 , it follows that v 1 = d c . Hence, the most general solution to the eigenvector equation is v = d c d c = c 1 0 1 + d 1 1 0 . Consequently, we have found solutions x 1 = 1 1 1 e t , x 2 = 1 0 1 e 2 t , x 3 = 1 1 0 e 2 t . The Wronskian at 0 is W (0) = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 1 1 1 0 1 1 1 0 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = 3 0 , which certifies that our three solutions are independent. Hence, the general solution is x = c 1 1 1 1 e t + c 2 1 0 1 e 2 t + c 3 1 1 0 e 2 t . Armin Straub [email protected] 30
Sketch of Lecture 31 Tue, 03/18/2014 Example 125. Consider x = parenleftbigg 1 3 3 7 parenrightbigg x . The characteristic polynomial (1 λ )(7 λ ) + 9 = λ 2 8 λ + 16 = ( λ 4) 2 has the double root λ = 4 . However, parenleftbigg 3 3 3 3 parenrightbigg v = 0 has solution only v = c parenleftbigg 1 1 parenrightbigg . We say that the eigenvalue 4 is defective with defect 1 (number of missing eigenvectors). So far, we have found the solution x 1 = parenleftbigg 1 1 parenrightbigg e 4 t but we are missing a second independent solution. We want to solve x = A x . Suppose that λ is a repeated and defective eigenvalue.