5.1:2,6; 5.2:6,13,25
DUE: Wednesday Nov. 15, Beginning of Class
In the solutions below,"&" replaces "lambda", the Greek symbol conventionally
used to denote an eigenvalue.
5.1:2) Is -2 an eigenvalue of A? Equivalently, does the system
[ A - (-2I) ] x = 0 have a non-zero solution?
<==> [ 9 3 ] [ x_1 ] = [0] has a nontrivial solution.
[ 3 1 ] [ x_2 ] [0]
Now [ 9 3 0 ] ~ [ 3 1 0 ]
[ 3 1 0 ] [ 0 0 0 ]
Thus there is a free variable; hence nontrivial solutions above, so -2 is
an eigenvalue of A. Any nonzero multiple of [ 1 ] is an eigenvector; that
[-3 ]
is, { [ 1 ] } is a basis for the eigenspace (equals the space Nul(A+2I) ).
[-3 ]
5.1:6) To determine if v is an e-vector for A, we do NOT need to find all
e-vectors of A. Just check if Av is a multiple of v:
Av = [ 3 6 7 ] [ 1 ] = [ (3 - 12 + 7) ] = [-2 ] = -2 [ 1 ] = -2v
[ 3 3 7 ] [-2 ] [ (3 - 6 + 7) ] [ 4 ] [-2 ]
[ 5 6 5 ] [ 1 ] [ (5 - 12 + 5) ] [-2 ] [ 1 ]
Thus v is an eigenvector with eigenvalue -2.
5.2:6) det(A-&I) = det [ 3-& -4 ] = (3-&)(8-&) + 16 = &^2 - 11& + 40
[ 4 8-&]
is the characteristic polynomial.
To find eigenvalues for A we set the characteristic polynomial to 0 and
solve for & (equivalent to finding the roots of a quadratic polynomial):
& = (1/2) [11 +- sqrt(121 - 160)] = (1/2) [11 +- sqrt(-39)]
so the e-values are not real.
5.2:13) det(A-&I) = det [ 6-& -2 0 ] (expand on column 3)
[ -2 9-& 0 ]
[ 5 8 3-&]
= (3-&) det [ 6-& -2 ] = (3-&)[(6-&)(9-&) - 4]
[ -2 9-&]
= (3-&)[&^2 - 15& + 50] = (3-&)(&-10)(&-5)
= -&^3 + 18(&^2) - 95& + 150
is the characteristic polynomial.
5.2:25)
a) det(A-&I) = det [ .6-& .3 ] = (.6-&)(.7-&) - .12 = &^2 - 1.3& + .30
[ .4 .7-&]
= (&-.3)(&-1) = 0
Thus the e-values are .3 and 1.
Note v_1 = [ 3/7 ] ,given in text, is an eigenvector for & = 1 (v_1 is the
[ 4/7 ]
steady state probability vector for the regular stochastic matrix A).
To find an e-vector for & = .3 solve the system (A-.3I)x = 0.
[ .3 .3 0 ] [ 1 1 0 ]
[ .4 .4 0 ] ~ [ 0 0 0 ]
==> x_1 = - x_2 , so an e-vector is x_2 [ -1 ] for any non-zero x_2.
[ 1 ]
To comprise a basis for R^2 consisting of v_1 and another e-vector, we must
choose a non-zero multiple of [-1 ] for v_2.
[ 1 ]
Recall that e-vectors corresponding to distinct e-values are linearly
independent. A basis for R^2 is thus
{ v_1 = [ 3/7 ] , v_2 = [-1 ] }
[ 4/7 ] [ 1 ]
b) Solve the system c_1*v_1 + c_2*v_2 = x_0:
(3/7)c_1 - c_2 = .5
(4/7)c_1 + c_2 = .5
[ 3/7 -1 .5 ] [ 6 -14 7 ] [ 0 -14 1 ] c_2 = -1/14
[ 4/7 1 .5 ] ~ [ 1 0 1 ] ~ [ 1 0 1 ] c_1 = 1
Thus x = v_1 - (1/14)v_2.
c) There exists a steady state probability vector for this regular stochastic
matrix, which is unique from our knowledge of Markov chains from Section 4.9,
and thus as k goes to infinity
x_k = A^k x_0 converges to the probability vector x where (A-I)x = 0.
Thus x is an eigenvector for e-value 1. So x must be the multiple
of v_1 that gives a probability vector (whose entries sum to 1). Since in fact
v_1's entries sum to 1, x = v_1 = [ 3/7 ].
[ 4/7 ]
Alternatively
x_k = A^k x_0
= A^k { v_1 + (1/14)v_2 }
= A^k v_1 + (1/14)A^k v_2
= (1)^k v_1 + (1/14)(.3)^k v_2
which clearly converges to v_1 as k goes to infinity since (.3)^k goes to 0.