Math 307 Spring, 2005 Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141 E-mail: hentzel@iastate.edu http://www.math.iastate.edu/hentzel/class.307.05 ============================================================ April 6 5.5 Complex Eigen Values. -1 Assignment. Find a matrix P such that P A P is in Jordan Canonical Form. | 3 3 2 2 | A = | 0 2 1 0 | | 1 1 3 2 | |-1 -2 -2 0 | ------------------------------------------------------ Previous Assignment Find a matrix P such that -1 | 1 1 0 | P A P = | 0 1 0 | | 0 0 1 | | 5 2 -1 | Where A = | -12 -5 3 | | -8 -4 3 | | 5-x 2 -1 | 3 Det[A-x I] = | -12 -5-x 3 | = -(x-1) | -8 -4 3-x | The only eigen value is x=1 x y=a z=b | 4 2 -1 | | 1 1/2 -1/4 | A-I = |-12 -6 3 | == | 0 0 0 | | -8 -4 2 | | 0 0 0 | | x | | -1/2 | | 1/4 | | y | = a| 1 | + b | 0 | | z | | 0 | | 1 | | -1 | | 1 | Use | 2 |, | 0 | | 0 | | 4 | | 5 2 -1 | | -1 1 | | -1 1 | Check: A[V1V2] = | -12 -5 3 | | 2 0 | = | 2 0 | | -8 -4 3 | | 0 4 | | 0 4 | | 4 2 -1 -x+y | | 1 1/2 -1/4 -x/4+y/4 | [A-I | xV1+yV2] = |-12 -6 3 2x | === | 0 0 0 -x+3y | | -8 -4 2 4y | | 0 0 0 -2x+6y | The only thing which extends is when x = 3y Let x=3,y=1 and the vector which extends is | -2 | V1 = | 6 | | 4 | The solution |-1/2| V2 = | 0 | | 0 | | 1 | V3 = | 0 | | 4 | -1 [V1 V2 V3] A [V1 V2 V3] = -1 | -2 -1/2 1 | | 5 2 -1 | | -2 -1/2 1 | | 1 1 0 | | 6 0 0 | |-12 -5 3 | | 6 0 0 | = | 0 1 0 | | 4 0 4 | | -8 -4 3 | | 4 0 4 | | 0 0 1 | ====================================================================== New Material: A Jordan Block looks like this. | x 1 0 0 0 0 0 | | 0 x 1 0 0 0 0 | | 0 0 x 1 0 0 0 | | 0 0 0 x 1 0 0 | | 0 0 0 0 x 1 0 | | 0 0 0 0 0 x 1 | | 0 0 0 0 0 0 x | The diagonal is all the same value, and the super diagonal is all 1's. Given a matrix A, we can always find a matrix P such that -1 P A P is a matrix with Jordan Blocks down its diagonal. Indicate the Jordan blocks in the following matrix. | 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0| | 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0| | 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0| | 0 0 0 3 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0| | 0 0 0 0 3 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0| | 0 0 0 0 0 3 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0| | 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0| | 0 0 0 0 0 0 0 3 1 0 0 0 0 0 0 0 0 0 0 0 0 0| | 0 0 0 0 0 0 0 0 3 1 0 0 0 0 0 0 0 0 0 0 0 0| | 0 0 0 0 0 0 0 0 0 3 1 0 0 0 0 0 0 0 0 0 0 0| | 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0| | 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0| | 0 0 0 0 0 0 0 0 0 0 0 0 4 1 0 0 0 0 0 0 0 0| | 0 0 0 0 0 0 0 0 0 0 0 0 0 4 1 0 0 0 0 0 0 0| | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 1 0 0 0 0 0 0| | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 1 0 0 0 0 0| | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 0 0 0 0 0| | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 1 0 0 0| | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 1 0 0| | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 1 0| | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 1| | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4| ------------------------------------------------------ | 2 1 | | 0 2 | | 2 | | 3 1 0 0 | | 0 3 1 0 | | 0 0 3 1 | | 0 0 0 3 | | 3 1 0 0 | | 0 3 1 0 | | 0 0 3 1 | | 0 0 0 3 | | 3 | | 4 1 0 0 0 | | 0 4 1 0 0 | | 0 0 4 1 0 | | 0 0 0 4 1 | | 0 0 0 0 4 | | 4 1 0 0 0| | 0 4 1 0 0| | 0 0 4 1 0| | 0 0 0 4 1| | 0 0 0 0 4| The order of the blocks is arbitrary. Traditionally, they are arranged by increasing diagonal element and among the blocks with the same diagonal element, by decreasing size of the block. Write all of the Possible Jordan Canonical forms for the 5x5 matrix A with characteristic polynomial (x-2)^5. J (2) J (2)(+)J (2) J (2)(+)J (2) J (2)(+)J (2)(+)J (2) 5 4 1 3 2 3 1 1 J (2)(+)J (2)(+)J (2) J (2)(+)J (2)(+)J (2)(+)J (2) 2 2 1 2 1 1 1 J (2)(+)J (2)(+)J (2)(+)J (2)(+)J (2) 1 1 1 1 1 Where J (x) is the nxn Jordan block with diagonal x. n ---------------------------------------------------- The individual Jordan blocks. If we have a nxn matrix A whose characteristic polynomial is (x-c)^n, then there exists a basis V1, V2, V3, ..., Vn such that under the action of A-cI 0<--- V1 <--- V2 <--- V3 <--- V4 <--- ... Vn-1 <---Vn If we write the matrix for A-cI we have V1 V2 V3 V4 ... Vn-1 Vn V1 0 0 0 0 0 0 V2 1 0 0 0 0 0 V3 0 1 0 0 0 0 . . . . . . . . . . . . . . . . . . . . . Vn 0 0 0 0 1 0 The matrix for A - cI is the transpose of this or 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 Matrix of A-cI is . . . . . . . . . . . . . . . 0 0 0 0 1 0 0 0 0 0 c 1 0 0 0 0 c 1 0 0 0 0 c 1 0 Matrix of A is . . . . . . . . . . . . . . . 0 0 0 . 1 0 0 0 0 c ----------------------------------------------------- Notice that if we write the basis elements backwards, we end up with a Jordan Canonical form with the ones below the diagonal. Thus you can see Jordan Canonical Form defined with ones on the bottom of the diagonal as well. ------------------------------------------------------- Write the solution for the differential equation | 2 1 | / | 2 1 | X = | 2 1 | X | 2 1 | | 2 | Look at the Taylor polynomials V0 = 1 V1 = 1 + t V2 = 1 + t + t^2/2! V3 = 1 + t + t^2/2! + t^3/3! V4 = 1 + t + t^2/2! + t^3/3! + t^4/4! Notice that under differentiation 0 <---- V0 <-----V1 <----- V2 <----- V3 <------ V4 The solutions are | V0 | |V0| |V0| |V0| |V0| 2t | V1 | 2t |V1| 2t |V1| 2t |V1| 2t | 0| e | V2 |, e |V2|, e |V2|, e | 0|, e | 0| | V3 | |V3| | 0| | 0| | 0| | V4 | | 0| | 0| | 0| | 0| This is a basis of all the solutions. ----------------------------------------------------------- Find the Jordan Canonical form for the matrix | 4 4 4 4 | A = | -1 1 -1 -2 | | 1 1 3 2 | | -1 -2 -2 0 | Hint: the characteristic polynomial is (x-2)^4. [ A-2I | RHS] | 2 4 4 4 x | | -1 -1 -1 -2 y | | 1 1 1 2 z | | -1 -2 -2 -2 w | | -x - 4 y | | -------- | | 1 0 0 2 2 | | | | x | | - + y | | 0 1 1 0 2 | | | | | | | | 0 0 0 0 y + z | | | | x | | w + - | | 0 0 0 0 2 | Thus y = -z and x = -2w x1 x2 x3 = a x4 = b RHS | 1 0 0 2 w + 2z | | 0 1 1 0 -w - z | | 0 0 0 0 0 | | 0 0 0 0 0 | This is the solution to | x1 | |-2w| A | x2 | = |-z | | x3 | | z | | x4 | | w | | x1 | | w+2z | | 0 | | -2 | | x2 | = | -w-z | + a |-1 | + b | 0 | | x3 | | 0 | | 1 | | 0 | | x4 | | 0 | | 0 | | 1 | w=0 z=1 w=1 z=0 | 0 | | 2 | | -2 | | 1 | 0 <--- |-1 | <---- |-1 | 0 <----- | 0 | <----- | -1 | | 1 | | 0 | | 0 | | 0 | | 0 | | 0 | | 1 | | 0 | | 0 2 -2 1 | P = |-1 -1 0 -1 | | 1 0 0 0 | | 0 0 1 0 | | 4 4 4 4 | A = | -1 1 -1 -2 | | 1 1 3 2 | | -1 -2 -2 0 | -1 | 2 1 0 0 | P A P = | 0 2 0 0 |. | 0 0 2 1 | | 0 0 0 2 | -1 Assignment. Find a matrix P such that P A P is in Jordan Canonical Form. | 3 3 2 2 | A = | 0 2 1 0 | | 1 1 3 2 | |-1 -2 -2 0 |