Question

| 0 14 147 Find the eigenvalues of the symmetric matrix 14 0 14 14 14 0 For each eigenvalue, find the dimension of the corresponding eigenspace. Selected Answer: 2.1 = 22; dimension of eigenspace = 1 2.2 = 14: dimension of eigenspace = 1 1x = -11; dimension of eigenspace=1 a.

Answer 1

Find the eigen values of the symmetric matria 0 14 14 14 O 14 ) for each eigen Xalue, find the dimension of the corresponding 14 14 0 eigen space. Sol:- Let A = O 14 14 0 14 14 14 Now for eigen kalues, we write characteristic equation | A-11=0 hare &ů eigen value and I ů 3x3 Identity matrix. -> 14 14 14. - 14 14 14 -4( 12196) -144-148-196) +14(196+144) = 0 -13 +1964 +1964 + 14x196 +14x196 +1960 = 0 -13+5880 + 5488=0 13-5884-5488=0 (1+1492 (1-28) = 6 = 28.-14, -14 so the eigenvalues of above matrix is A₂ = -14 A =-p4 d=28,

Now for eigen vector *-(3) AX=dx (A-11)X = 0 hese X, o eigen vector Corresponding eigen Kallied, Now for x=28 -28 (A-AI) - T-28 14 14 14 R₂ Ratry 14 -21 21 14 - 28 14 14 14 14 -28 Ryt htR/ 14 14 -28 1-28 14 14 R₂ Ryth -21 21 -21 21 21 21 0 0 R, - Fч - 2 RP2/21 0 0 D 0 Now EJ016 -H₂ taz -23, +72 +23 =0 * = * 3 x = ₂ of x,= d = &q=t

So X 1:( 23 ů So the eigen vector corresponding to eigen kalue 28 and eigen space {[:]} and dimension of eigen space is Now for x = -14 (A-AI) = Ausuar 14 14 14 BTR-fi 14 14 14 O 14 14 14 ry 14 14 Rz7R3-R, O 0 R, 0 0 O 0 Now [:::01:13 aita, +₂=0 let a = t x = 5 -t-5 SO -t-s X = t tt [1] S 0 So the eigen vector corresponding to eigen Xalue -14 is and [} and eigen space -{[:].0:1 and dimension of eigen space is Answer