{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

sol2.09 - ECE521 Linear Systems Fall 2007 Homework 2...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
ECE521 Linear Systems Fall 2007 Homework 2 Solutions Problem 1 : Consider a 2 matrix A ( t ) which is continuous for all real t but is not invertible for any t . 1. As an example of such A ( t ) , show that A ( t ) = a ( t ) b ( t ) T where a ( t ) , b ( t ) R 2 are continu- ous for all real t is not invertible for any t . The matrix A ( t ) has rank one so it cannot be invertible for any t . The other way of establishing this fact is by calculationg the determinant of A ( t ) which clearly is equal to zero. 2. Is it necessary the case that integraltext T 0 A ( τ ) is not invertible for any t ? Explain. The answer is no. As a counterexample consider a ( t ) = b ( t ) = [1 t ] T and calculat integraltext T 0 A ( τ ) . We have integraldisplay T 0 A ( τ ) = integraldisplay T 0 parenleftbigg 1 τ τ τ 2 parenrightbigg = parenleftbigg T 1 2 T 2 T 1 3 T 3 parenrightbigg which is nonsingular for all T negationslash = 0. Problem 2: 1. Let columns of X i , i = 1 , 2 , span invariant subspaces of A . Do the columns of X = [ X 1 , X 2 ] also span an invariant subspace of A ? Say that X 1 R n × k 1 and X 2 R n × k 2 . Hence X R n × k 1 + k 2 . Let x span( X ). Then there exists α R n × k 1 + k 2 s.t. x = . Let α = [ α T 1 α T 2 ] T where α i R n × k i . Then x = [ X 1 X 2 ][ α T 1 α T 2 ] T = X 1 α 1 + X 2 α 2 = x 1 + x 2 where x i span( X i ), i = 1 , 2. Thus Ax i span( X i ). Consequently Ax = Ax 1 + Ax 2 span( X 1 ) span( X 2 ) = span( X ). 2. Let S = 1 2 9 10 0 1 2 3 0 0 1 2 0 0 0 1 Find a generalized eigenvector of S of maximal order. Show your work.
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
September 18, 2009 2 There are at least two ways of approching this problem. One is by selecting a vector that we think might be a generalized eigenvector of maximal order and then proving that it indeed is a generalized eigenvector. The other is by solving a sequence of linear systems ( S λI 4 ) x i = x i - 1 , i = 1 , ..., 4 , where λ = 1 is the eigenvalue of S with the multiplicity 4 and x 0 = [1 0 0 0] T being the only eigenvector of S . The latter method will always give a generalized eigenvector of maximal order. However, as the matrix in question is upper triangular then the (educated) guess of x = [0 0 0 1] T will do. It is easy to check that S 3 x negationslash = 0 = S 4 x which shows that x is a generalized
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern