Let T : C3 C3 be the linear operator defined by Tx = Ax where 0 0 0.33 0 0.71 0.
ID: 3167456 • Letter: L
Question
Let T : C3 C3 be the linear operator defined by Tx = Ax where 0 0 0.33 0 0.71 0.94 A=10.18 0 0 a) Find the eigenvalues of A. b) Show T is a bounded operator. That is, show that there exists a constant C > 0 such that Ax3Cxll3 for all x E C3. where || Il3 is the norm in C3 defined as follows: for any x E C3, x (x1,x2, x3) where xk E C. Since xk (for k = 1, 2, 3) is com- plex, ak = ak + ide, and Ixkl /of + , Thus we define l 1x11,- c) Argue that the sequence given by x,,+,-Axn, where x0 E R3, is contained in R. Finally, what is linnx x,? Justify your answer.Explanation / Answer
(a) Find the eigenvalue of the given matrix
A = ( 0 0 0.33
0.18 0 0
0 0.71 0.94 )
Charecteristic equation is det(A-lamda I) =0
(-lamda 0 0.33
det 0.18 - lamda 0
0 0.71 0.94-lamda)
= - lamda (-0.94lamda +lamda^2)+0+0.33(0.18*0.71)
= - lamda^3+0.94*lamda^2+ 0.042174
Find the roots of the characteristic equation, those are the eigenalues.
lamda = 0.983593, -0.0217964 - 0.205918 i, -0.0217964 + 0.205918 i are the eigenvalus.
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