Question

1. The matrix A is factored in the form PDP-1. USe the Diagonalization Theorem to find the eigenvalues of A and a basis for each eigenspace. 54 0 -2] -20 11 5 007 0 0 1 25 4 0 1 2 0 5 0 2 1 42 0 0 5 0 0 0 0 4 - 1 0 - 2 2. Diagonalize, if possible, the matrix A below, given that the eigenvalues are 1 = 2, 1. If not possible, explain why. 0 -4 A= 0 -3 2 5 1

Answer 1

Given, 0 A = popa where - 4 o - 2 2 O 2 5 4 - O 2 0 5 0 0 - 5 0 0 5 O O 4 theorem By Diagonalization Eigenvalues matnx of A are diagonal entries of are . Eigenvalues 554 Eigenvalues 54 bas multiplicity 2. are with eigenvalue 5 Now, Eigenspace Iis column in matrix p Basis for eigen space are - 2 EG) and o 2 o O

0 -4 0 2 5 To diagonalize need A علط find eigen vectors and eigenvalues A: consider, LA-d Il= 0 - 4 - 6 - 1 - 3 -O - 2 5-8 :. . . -> (12-51+6) + 4 (1-5 +3) – 6(-2+1) = 0 512 - 13 - 64 + 48-8 +12-60=0 - 13 + 512_88+ 4 = 0 13 – 5 12 + 81 -4 (1-1) (4- 2) x= 1,2,2 Eigen values .. . .. aze 12 A consider, (4-11)=0 .(A - I) = 0

-4 6 O - 3 34 H2 ak 0 1 2 4 0 Reduce the matrix we get 1 O 2 24 0 H2 0 o 0 313 0 .. 24+ 213 - 0 2 + 43 Take, 23 t 2 - - 2t 27 = -t 74 Ezt - 2 -t t 13 t - 2 For d=2 consider (A-dI) X=0 :: (A - 21) X=0 -4 -6 - 2 " -2 - 1 - 3 22 O 2 3 0

Reduce the matrix, we get - 2 3 O 0 O H - 0 O 0 .. 74 + 2*2 + 372 = 0 Take, 32 = t S 량 2 t - 35 - 2t-35 -2 - 2 24 ah t t + o 26 S 0 1 -2 eigen vectors are - 1 0 A matrix P is matrix eigenvectors, - 2 2 - 3 - - 1 0 1 D on the diagonal Matrix and is has diagonal eigenvalues matrix 0 - D 0 2 O 0 0 2

Now ps 3 - 2 - 1 -3 - 2 4 A= Popi is diagonalizable A O -6 - 2 0 - 2 0 1 - 2 -3 o 3 O 0 2. 0 - 3 01 2 0 - 0 4 0 2 O 2