Question

Show that any two eigenvectors of the symmetric matrix corresponding to distinct eigenvalues are orthogonal. -1 0 -1 0-1 0 - 1 0 5 Find the characteristic polynomial of A. - A - Find the eigenvalues of A. (Enter your answers from smallest to largest.) (1, 12, 13) = ]) Find the general form for every eigenvector corresponding to N. (Uses as your parameter.) X1 = Find the general form for every eigenvector corresponding to 12. (Use t as your parameter.) X2 = Find the general form for every eigenvector corresponding to 13. (Use u as your parameter.) X3 = Find X1 X2 X1 X2 = Find X1 X3 X1 X3 = Find X2 X2 X2 X3 Determine whether the eigenvectors corresponding to distinct eigenvalues are orthogonal. (Select all that apply.) X1 and x2 are orthogonal. X1 and x3 are orthogonal. X2 and X3 are orthogonal. O

Answer 1

Ans: Given A = 0 o -1 o o 5 . This is a symmetric matrix Since A=AT. Any two eigen vectors of the this matrix Ĉi e of the symmetric matrix conrespone onding to distinct eigenvalues are orthogonal.on not, we can see this by the example given. For this, we need to find chanacteristic polynomial of A. 121 - A) eo => 1a 1699)- 0 -1 -1 O 5 ao -1 sy gti O o atl O 2-5 o co Chanacteristic polynomial of A 2 o - 7 (2+1){(2+i)(2-5)-0}+d {0+0}+1 {o. 6a+)3= 77 (+1) { 22-49-5}-(+1) -> (211) { 42-6? -> (2+1) or 72-49-6-0 + 4 +16 +24 2 => 2 = 27vo 2 o or . x 2-1

A are ,2 +000) =(2, 22, 23) eigen vectors will be esponding Eigenvalues of (-1,2-10 Now connes FOM 2,5-1, (A-2,5). 717- ni o . O di O 0 -1 - 0 - 5 O © -) mi o nu - = O 6 My » Có -n, +673 Solving these two, nz - O 스 ni Taking dz CI, X, :) 0

Similarly, for for 72 2 2-rio eigenvectore x2 will be (A-21)8 20 > ñ -1-2+ rio al 0 -1-2+vio ng m3 co 5-2 trio => 1-3 trio o - 1 mi 0 - 3+rio o \น nz 0 3 trio => [-stro)x, -23=0 (-3+010)M2 +3+010) 92²0 solving these nel these ,

- O син ni =(3 tro) na t Hence, 3 trio X2 al O Again for ng=atrio Eigenvector xz will bes (A-21) a=0 ay -3-vio 0 o -3-ro o m2 3-rio mz - =7 (-3-810) n, - My (-3-N10) R2 t(3-1)^2 = 0 Solving those ; al n n2 M (3-810) M3 Hences. X3 = 3-rio 0

XI. X2 3 trio (8 169 © 013+ r10)+1.0 + 0.1 =0 X, 4 X2 ane om rithogonal to each other. Hence xi. X3 3- rio - a o 6.(3-r10 ) +1.0 tool Hence; & Xz care otthogonal to each other X₂ X3 =/3+rom 3-rio XI o 《 O (3 tro)(3-710) +0.0 +1.1 9-10+! Hences X2& Xz are arthogonal to each other. So eigenvectors for any symmetric matrix Correspond to different eigenvalues must be onthogonal. Hence, all the cheekbox will be fick manked. All the eigenvalues are on thogonal to each others Ans.