Question

Test Test 3 (Chapters 5-6, and Cumulative) 3 of 30 (0 complete) Time Remaining : 01 25:53 S Matrix A is factored in the form PDP. Use the Diagonalization Theorem to find the eigenvalues of A and a basis for each eigenspace. 1 1 1 4. 2 2 1 2 1 2 500 A= 1 3 1 = 2 0 1 1 1 3 0 1 0 Select the correct choice below and fill in the answer boxes to complete your choice. (Use a comma to separate vectors as needed.) O A. There is one distinct eigenvalue, 2 = A basis for the corresponding eigenspace is O B. In ascending order, the two distinct eigenvalues are 1and ,= Bases for the corresponding eigenspaces are and respective Click to select and enter your answer(s). (hp)

Answer 1

Option B is correct. The detailed solution of the subparts is given in the image below.

2 А 1 5 o 3 2 4 2 2 2 2 2 I 2 2 -1 -2 -2 24 6 2 1 0 놀 3 4 -8 -2 4 pe [༠༧༧ 2 -1 o p! 호 3 4 4 A = P 00 pal AP = 2 2 A له به ام 2 2 = 5 NNN Alo 1 A 2 1 1-1 0 2 0 ( .. A has 2 eigen values ( Just do the matrix multiplication). 5 d 1. A₂ = 5. (Ang) * = 1 eigenspace of Bases of t, =B 2 2 2 Bez (Bases of *2) [!], []} (Ana). -> This since 2. Vecitors P is are linearly independen invertible.) B is the correct option There does not and other as P is Note :- exist eigen values Spon (B.,) + span (B) = TR3 invertible TH there distinct exist any its other, eigen value then eigen vector linearly independent with Bx, UBA which contradicts that the maximal Linearly independent set 3 is 3. in R