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

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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.

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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
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