5.3:10,12,20; 5.6:5,6,11,12;
DUE: Wednesday Nov. 22, Beginning of Class
In the solutions below,"&" replaces "lambda", the Greek symbol conventionally
used to denote an eigenvalue.
5.3:10) The e-values for A = [ 2 3 ] are determined from
[ 4 1 ]
det(A-&I) = 0
==> (2-&)(1-&) - 12 = 0
==> &^2 - 3& - 10 = 0
==> (&-5)(&+2) = 0
==> & = 5, -2.
Since the e-values are distinct, we already know that A is diagonalizable.
For & = 5, to find an e-vector solve the system (A-5I)x = 0:
[ -3 3 0 ] [ 1 -1 0 ] so an e-vector is v_1 = [ 1 ]
[ 4 -4 0 ] ~ [ 0 0 0 ] [ 1 ]
For & = -2, solve
[ 4 3 0 ] [ 4 3 0 ] so x_2 = -4/3 x_1; pick x_1 = 3
[ 4 3 0 ] [ 0 0 0 ] to give e-vector v_2 = [ 3 ]
[-4 ]
A is diagonalizable. A = P D (P^-1) , where P is the 2x2 matrix having
columns equal to the e-vectors:
P = [ v_1 v_2 ] = [ 1 3 ] and D = [ 5 0 ]
[ 1 -4 ] [ 0 -2 ]
5.3:12) The e-values are given to be & = 2 (multiplicity 2) and & = 8.
To find a basis for eigenspace for & = 2, we find a basis for the solution
set of (A-2I)x = 0:
[ 2 2 2 0 ] [ 2 2 2 0 ] [ 1 1 1 0 ]
[ 2 2 2 0 ] ~ [ 0 0 0 0 ] ~ [ 0 0 0 0 ]
[ 2 2 2 0 ] [ 0 0 0 0 ] [ 0 0 0 0 ]
so x_1 = -x_2 - x_3.
There are 2 free variables so the dimension of the e-space is 2, matching the
multiplicity of the e-value, so we can conclude now that A is diagonalizable
(note the dimension of the eigenspace for & = 8 must be 1).
A basis for the eigenspace for & = 2 is { [-1 ] , [-1 ] }
[ 1 ] [ 0 ]
[ 0 ] [ 1 ]
Now solving (A-8I)x = 0 we find an e-vector spanning the 1-dimensional
eigenspace for & = 8:
[ -4 2 2 0 ] [ 2 -1 -1 0 ] [ 2 -1 -1 0 ]
[ 2 -4 2 0 ] ~ [ 0 -3 3 0 ] ~ [ 0 1 -1 0 ]
[ 2 2 -4 0 ] [ 0 3 -3 0 ] [ 0 0 0 0 ]
[ 2 0 -2 0 ] [ 1 0 -1 0 ] [ 1 ]
~ [ 0 1 -1 0 ] ~ [ 0 1 -1 0 ] x_2 = x_3 = x_1 [ 1 ]
[ 0 0 0 0 ] [ 0 0 0 0 ] so an e-vector is [ 1 ]
Hence A = P D (P^-1), where
P = [ -1 -1 1 ] and D = [ 2 0 0 ]
[ 1 0 1 ] [ 0 2 0 ]
[ 0 1 1 ] [ 0 0 8 ]
5.3:20) Notice A is lower triangular matrix so det(A-&I) is the product of the
diagonal entries of A-&I:
[ 4 0 0 0 ] [ 4-& 0 0 0 ]
A = [ 0 4 0 0 ] A-&I = [ 0 4-& 0 0 ]
[ 0 0 2 0 ] [ 0 0 2-& 0 ]
[ 1 0 0 2 ] [ 1 0 0 2-&]
det(A-&I) = (4-&)(4-&)(2-&)(2-&) = 0 ==> & = 4 with multiplicity 2
& = 2 with multiplicity 2
Find a basis for solutions of (A-2I)x = 0:
[ 2 0 0 0 0 ] [ 1 0 0 0 0 ]
[ 0 2 0 0 0 ] ~ [ 0 1 0 0 0 ]
[ 0 0 0 0 0 ] [ 0 0 0 0 0 ]
[ 1 0 0 0 0 ] [ 0 0 0 0 0 ]
Thus x_3, x_4 free, and x_1 = 0 = x_2.
Thus a basis for the e-space for & = 2 is { [ 0 ] , [ 0 ] }
[ 0 ] [ 0 ]
[ 1 ] [ 0 ]
[ 0 ] [ 1 ]
Similarly,
(A-4I)x = 0
[ 0 0 0 0 0 ] [ 1 0 0 -2 0 ]
[ 0 0 0 0 0 ] ~ [ 0 0 1 0 0 ]
[ 0 0 -2 0 0 ] [ 0 0 0 0 0 ]
[ 1 0 0 -2 0 ] [ 0 0 0 0 0 ]
implies x_4, x_2 free
x_3 = 0
x_1 = 2 x_4.
Therefore, any vector in the eigenspace corresponding to & = 4 is of the form
x_2 (0,1,0,0) + x_4 (2,0,0,1).
So a basis for the e-space for & = 4 is { [ 0 ] , [ 2 ] }
[ 1 ] [ 0 ]
[ 0 ] [ 0 ]
[ 0 ] [ 1 ]
A is diagonalizable with A P = P D , where a choice is (note order dependent):
P = [ 0 2 0 0 ] and D = [ 4 0 0 0 ]
[ 1 0 0 0 ] [ 0 4 0 0 ]
[ 0 0 1 0 ] [ 0 0 2 0 ]
[ 0 1 0 1 ] [ 0 0 0 2 ]
5.6:5) det(A-&I) = det [ .4-& .3 ] = (.4-&)(1.2-&) + .0975
[ -.325 1.2-&] = &^2 - 1.6& + .48 + .0975
= &^2 - 1.6& + .5775
is the characteristic polynomial.
To find eigenvalues for A we set the characteristic polynomial to 0 and
solve for & (use quadratic formula):
& = (1/2) [1.6 +- sqrt([1.6]^2 - 4[.5775])]
= (1/2) [1.6 +- sqrt(.25)]
= (1/2) [1.6 +- .5].
Thus &_1 = 1.05 and &_2 = .55 are the eigenvalues.
Solve (A - 1.05I)x = 0 to find a basis for the eigenspace for &_1 = 1.05:
[ (.4 - 1.05) .3 ] [ -.65 .3 0 ] [ -.65 .3 0 ]
[ -.325 (1.2 - 1.05) ] ~ [-.325 .15 0 ] ~ [ 0 0 0 ]
Thus .65x_1 = .3x_2 ==> (13/6)x_1 = x_2, so
an eigenvector is v_1 = [ x_1 ] = [ 6 ].
[ x_2 ] [ 13 ]
The dynamical system x_k+1 = A x_k has solution
x_k = A^k x_0
= c_1 (1.05)^k [ 6 ] + c_2 (.55)^k v_2, where [ c_1 ] is the
[ 13 ] [ c_2 ]
coordinates of x_0 relative to the eigenvector basis { v_1, v_2 }.
If c_1 is not zero, then in the long term (for k large (.55)^k is negligible)
x_k = c_1 (1.05)^k [ 6 ] approximately.
[ 13 ]
So in long term both populations eventually grow at almost 5% rate (per month),
and the eventual ratio of owls to flying squirrels is 6 to 13 (thousand).
5.6:6) det(A-&I) = det [ .4-& .3 ] = (.4-&)(1.2-&) + .15
[ -.5 1.2-& ] = &^2 - 1.6& + .48 + .15
= &^2 - 1.6& + .63
= (& - .9)(& - .7)
Thus &_1 = .9 and &_2 = .7 are the eigenvalues.
Since both eigenvalues are less than 1, the populations will eventually perish
since the dynamical system x_k+1 = A x_k has solution
x_k = A x_0
= c_1 (.9)^k v_1 + c_2 (.7)^k v_2 converges to 0 as k ->infinity.
To find a value of p in the matrix A = [ .4 .3 ]
[ -p 1.2 ]
such that the populations remain constant, we need to
determine what p yields an e-value of &_1 = 1 (and with &_2 <= 1).
Method 1 (for finding p):
If & = 1 is e-value, then there is a (non-zero) e-vector solution of
Ax = x ==> (A-I)x = 0 ==> [ -.6 .3 ] has linearly DEPENDENT rows
has a free [ - p .2 ] ==> p =.4.
variable
Method 2 (alternative):
The characteristic polynomial is
det(A-&I) = &^2 - 1.6& + .48 + .3p
= &^2 - 1.6& + .3(1.6 + p)
The roots are (using quadratic formula)
& = (1/2) [ 1.6 +- sqrt{(2.56) - (1.2)(1.6+p)} ]
Thus we want sqrt(discriminant) to equal .4 (then &_1 = 1, &_2 = .6);
that is we want discriminant to equal .16.
Set .16 = 2.56 - (1.2)(1.6+p) and solve for p.
(1.2)(1.6+p) = 2.4
==> 1.2p + 1.92 = 2.4 ==> p = .48/1.2 = .4.
An eigenvector for & = 1 is found by solving (A-I)x = 0 to get v_1 = [ 1 ].
[ 2 ]
For large k, x_k = c_1 (1)^k [ 1 ] = c_1 [ 1 ] assuming c_1 non-zero.
[ 2 ] [ 2 ]
Thus the eventual relative population sizes are 1 owl per 2(thousand)squirrels.
5.6:11)
det(A-&I) = det [ .4-& .5 ] = (.4-&)(1.3-&) + .2 = &^2 - 1.7& + .72
[ -.4 1.3-&]
= (&-.9)(&-.8)
Thus the e-values are &_1 = .9 and &_2 = .8.
Since both of these are < 1, x_k --> 0 vector as k goes to infinity; the origin
is an attractor.
To find an e-vector for &_2 = .8 solve the system (A-.8I)x = 0 to get
v = [ 5 ] (or any non-zero multiple), which is direction of greatest
[ 4 ] attraction since .8 smallest.
5.6:12)
det(A-&I) = det [ .5-& .6 ] = (.5-&)(1.4-&) + .18 = &^2 - 1.7& + .88
[ -.3 1.4-&]
= (&-1.1)(&-.8)
Thus the e-values are &_1 = 1.1 and &_2 = .8.
Since 1.1 > 1 > .8, the origin is a saddle.
The direction of greatest attraction corresponds to any e-vector for
&_2 = .8 (the e-value less than 1 in magnitude), such as [ 2 ].
[ 1 ]
The direction of greatest repulsion corresponds to any e-vector for
&_1 = 1.1, such as (line through origin and) [ 1 ].
[ 1 ]