(a) Let be a real number. Compute A I. (b) Find the eigenvalues of A, that is, f
ID: 3169645 • Letter: #
Question
(a) Let be a real number. Compute A I.
(b) Find the eigenvalues of A, that is, find the values of for which the matrix A I is not invertible. (Hint: There should be exactly 2. Label the larger one 1 and the smaller 2.)
(c) Compute the matrices A 1I and A 2I.
(d) Find the eigenspace associated with 1, that is the set of all solutions v = v1 v2 to (A 1I)v = 0.
(e) Find the eigenspace associated with 2 similarly.
Repeat the process to find the eigenvalues and corresponding eigenspaces for
A = [ 0 2 0
2 0 0
1 1 4 ]
(Note that this matrix has three eigenvalues, not 2.)
Explanation / Answer
(a). We have A – I3 =
-
2
0
2
-
0
1
1
4-
(b). The characteristic equation of A is det(A – I3)= 0 or, 3-42-4+16 = 0 or, (-4)( +2)( -2)= 0. Thus, the eigenvalues of A are 1 =4, 2=-2 and 3 = 2.
(c). The matrix A – 1I3 is
-4
2
0
2
-4
0
1
1
0
The matrix A – 2I3 is
2
2
0
2
2
0
1
1
6
The matrix A – 3I3 is
-2
2
0
2
-2
0
1
1
2
(d). The RREF of the matrix A – 1I3 is
1
0
0
0
1
0
0
0
0
If v = (x,y,z)T, then the equation (A – 1I3)v = 0 is equivalent to x = 0 and y = 0 so that v = (0,0,z)T= z(0,0,1)T. Hence the eigenvector of A associated with the eigenvalue 1= 4 is v1 = (0,0,1)T.
The RREF of the matrix A – 2I3 is
1
1
0
0
0
1
0
0
0
If v = (x,y,z)T, then the equation (A – 2I3)v = 0 is equivalent to x+y = 0 or, x = -y and z = 0 so that v = (-y,y,0)T = y(-1,1,0)T. Hence the eigenvector of A associated with the eigenvalue 2= -2 is v2 = (-1,1,0)T.
The RREF of the matrix A – 3I3 is
1
0
1
0
1
1
0
0
0
If v = (x,y,z)T, then the equation (A – 3I3)v = 0 is equivalent tox+z = 0 or, x = -z and y+z = 0 or, y = -z so that v = (-z,-z,z)T = z(-1,-1,1)T. Hence the eigenvector of A associated with the eigenvalue 3= 2 is v3 = (-1,-1,1)T.
(e ). E1 = span{(0,0,1)T }, E2 = span{(-1,1,0)T } and E3 = span{(-1,-1,1)T },
-
2
0
2
-
0
1
1
4-
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