Question

Could you please just solve Question (i) A: Thanks

3. For each of the following matrices, a. Determine the characteristic polynomial corresponding to the matrix. b. Find the eigenvalues of the matrix. c. For each eigenvalue, determine the corresponding eigenspace as a span of vectors. d. Determine an eigenvector corresponding to each eigenvalue. e. Pick one eigenvalue of each matrix and the corresponding eigenvector chosen in part (d) and verify that they are indeed an eigenvalue and eigenvector of the matrix using the definition of what it means for 2 to be an eigenvalue of a matrix A, with corresponding eigenvector x. (2 -3 1) (i) A=1 -2 1 11 -3 2) (-2 0 1 ) (ii) B= 1 0 -1 10 1 -1) (120) (iii) C = 0 1 0 (001

Answer 1

/ 2 -3 : A= 11 - 2! - 3 2 1 @ the charac tenistic polynomial is - JA-113!=0 | A- #131-0 og Dahi ie (2-2 -3 na -0 1 11 -2-3 -3 1 2-| ie (2-1)(-4+22-24+2+3) +3 {2-1-1] + '{-3+2+2) =0 ie -x3 +22²_2=0 ie x3_22²70=0 - 0 0 Now, 23, 2x²+2=0 . DAN let À be eigen value of A 13 - 212+1=0 ie (4² - 2x+1)=0 the ied (4-1)²=0 wire ne ie d = Oo lol. - Hence the eigen values are 0,791

© Let Eo be the eigen space Corzesponding to the e. v o ie null space of the matrix Eo = K (A-0.1) =K-A we reduce the matrix A by elementanu now openation as follows - w [ 2 -3 17 VRI [ -3/427 A = -2 1 -2 1 Li -32] Li -3 2 1 1. J R-Ra 1-3/227(-2)22/-30 1727 R-R3 ri -372 127 To 1 - Ao -1/2 12 / F lo -12 Y LÖ -3/2 39 Lo -3/2 3/2 L1 -3 21 1 % R2 123 ri -312 72 Il 0 -17 34R2+ Rilo 1 Lo o o Looo Thus, the solutions a of Ax=0 satisfy a x1=43 and 22=23 The nue space k(A-0.1) consist of vetons -- [13] =[:] 3 5o = K (9–6.2) –spon [11] Cok S L 233

Similarly o for Ei = K (A-1.I) =k ( A-I) wall -3 17 - A-1 = -3.|| Clearly, the now neduced form is L 0 0 0 Thus, the solution x of (A-I)x=0 satisfy ni=342-23 aw The Nuie space consists of vectors 5 x= 322 - 3137 1 3 212 o L = 22 . U2 23 . Fi = k (-) = som { [2],[73] @ Let dal Then Xi be the eigen veeton Cormesponding to dal ie AX,= 1X, [2 -3 17 7:17. 1 - T 712 21 [23 rai 1.212 &m a To E baie 2- 372+23 7. | 21-322+ 213 I no-372 +23] 200-342+ 213=0

Let 20= C 22= d. 23 = -2+3d. men x. = 1 -- []*[i] whene, cand d ame Constant @ we choose, del on The condition for eigen value and eigen veeloos is AX =AX 2 -3 IT -Oī 1 -3 2 Hese, xize []+[:] 13:18] - [] asso (9] = [] (verify). !][%]-[:] Again, wrw Hence, AX=dX verified. (no