Question

Q4. Let 1.01 0.99 0.99 0.98 (a) Find the eigenvalue decomposition of A. Recall that λ is an eigenvalue of A if for some u1],u2 (entries of the corresponding eigenvector) we have (1.01 u0.99u20 99u [1] + (0.98-A)u[2] = 0. Another way of saying this is that we want the values of λ such that A-λ| (where I is the 2 x 2 identity matrix) has a non-trivial null space there is a nonzero vector u such that (A-AI)u = 0. Yet another way of saying this is that we want the values of λ such that det(A-λ/) = 0, Once you have found the two eigenvalues, you can solve the 2 x 2 systems of equations Aul-λυί and At42 λ2u2 for ul and u2. (b) I determine the solution to Az -y. (e) Now let y-.1 1] and solve Ax y. Comment on how the solution changed. (d) Suppose we observe with 1. We form an estimate A1y. Which vector e yields the maximum error lli-II (e) Which vector e yields the minimum error?

Answer 1

Q 4 o.9909 1.99ャV ζ.9601-0.0S 88 1-99さ 1. 990又 98S) and Aa 0.0049 0.791'ス -0.975) Taki -) we get t,1000 o 9 849

1.0051 0.99-1 (-ズ7 0.99 .97513 Takı n -loisas Ficgnvaliu clecomposit ion o A 1000 10000G985 349IoisasL004 Oo0o 100000 849-OUSaS To Hird o soluton to Ax y. 1.01 0.99- C

าเ2-3 .06186 7(1 =-1.0309 χ=(%)-(-ane 186 ,'. Solution is to tivd Hu solution to A ey 0.001772= -6.079

0.79기 = g.q8l4 = 9.072) /1,072) isブr(,): solution "Xi つ( -8 1443 The sohtion has a major ditjevene (a haso dih-vevre of 10.30616 This ditfevenre is cbtained ky bu chamging he ist component of yhy