LTE alll 77%. 5:47 PM 4 (i) The characteristic polynomial of A is Press the \"Su
ID: 3169809 • Letter: L
Question
LTE alll 77%. 5:47 PM 4 (i) The characteristic polynomial of A is Press the "Submit Answers" button before attempting part (ii) The blanks for entering answers to (ii) and (ii) will appear only if your answers to previous parts were correct. In that case, please ignore the message stating that some of your answers were incorrect! (ii) The distinct eigenvalues of A are with an algebraic multiplicty of and a geometric multiplicty of with an algebraic multiplicty of and a geometric multiplicty of Press the "Submit Answers" button before attempting part (ii) (iii) The eigenspaces of A are (You'll have to solve parts (i) and (ii) correctly before you can attempt part (ii)Explanation / Answer
(i). The characteristic polynomial of A is p(x) = det(A-xI3)= 0 or, x3+(-5)x2+7x +(-3 )=0 or, (x-1)(x-1)(x-3) = 0.
(ii). The distinct eigenvalues of A are 1 =1 with an algebraic multiplicity of 2 ( and geometric multiplicity 1, as only one distinct eigenvector is associated with it; Please see part(iii) below) and 2 =3 with an algebraic multiplicity of 1 (and geometric multiplicity 1, as only one distinct eigenvector is associated with it; Please see part(iii) below).
(iii). The eigenvectors of A associated with its eigenvalue 1 =1 are the solutions to the equation (A-I3)X = 0. To solve this equation, we will reduce the matrix A-I3 to its RREF, which is
1
0
1
0
1
2/3
0
0
0
Now, if X = (x,y,z)T, then the equation (A-I3)X = 0 is equivalent to x+z = 0 or, x = -z and y+2z/3 = 0 or, y = -2z/3. Then X = (-z,-2z/3,z)T. Now, let z = 3t. Then X = (-3t,-2t,3t)T = t (-3,-2,3)T. Thus, the only eigenector of A associated with its eigenvalue 1 =1 is (-3,-2,3)T.
Similarly, the eigenvectors of A associated with its eigenvalue 2 =3 are the solutions to the equation (A-3I3)X = 0. To solve this equation, we will reduce the matrix A-3I3 to its RREF, which is
1
0
1
0
1
0
0
0
0
Now, if X = (x,y,z)T, then the equation (A-3I3)X = 0 is equivalent to x+z = 0 or, x = -z and y = 0 so that X = (-z,-0,z)T = z(-1,0,1)T. Thus, the only eigenector of A associated with its eigenvalue 2 =3 is (-1,0,1)T.
The eigenspaces of A are E1 = span{(-3,-2,3)T} and E2 = span{(-1,0,1)T}.
1
0
1
0
1
2/3
0
0
0
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