4.6:3,8,16 4.9:2,3,12,13
DUE: Wednesday, Nov. 1, Beginning of Class
4.6:3) rank A = number of pivot columns.
>From the echelon form B, we see that columns 1,3, and 4 are pivot columns. Thus
rank A = 3 .
Using the Rank Theorem, or that dim(Nul A) = the # of nonpivot columns =
# of free variables for Ax = 0; we get
dim(Nul A) = 2.
A basis for Col A is the pivot columns of A, which { [ 2 ] , [ 6 ] , [ 2 ] }
from the echelon form B are identified as the [-2 ] [-3 ] [-3 ]
1st,3rd, and 4th columns of A. [ 4 ] [ 9 ] [ 5 ]
[-2 ] [ 3 ] [-4 ]
For Bx = 0 we get
x_2, x_5 free (note columns 2 and 5 are nonpivot columns).
To get a basis for Nul A, row reduce B to REDUCED echelon form:
[ 2 -3 6 2 5 ] [ 1 -3/2 0 0 -9/2] [ 1 -3/2 0 0 -9/2]
[ 0 0 3 -1 1 ] [ 0 0 3 0 4 ] [ 0 0 1 0 4/3]
[ 0 0 0 1 3 ] ~ [ 0 0 0 1 3 ] ~ [ 0 0 0 1 3 ]
[ 0 0 0 0 0 ] [ 0 0 0 0 0 ] [ 0 0 0 0 0 ]
Thus [x_1] = [(3/2)x_2 + (9/2)x_5 ] = x_2 [3/2] + x_5 [ 9/2]
[x_2] [ x_2 ] [ 1 ] [ 0 ]
[x_3] [ (-4/3)x_5 ] [ 0 ] [-4/3]
[x_4] [ -3x_5 ] [ 0 ] [ -3 ]
[x_5] [ x_5 ] [ 0 ] [ 1 ]
Thus a basis for Nul A is { (3/2,1,0,0,0) , (9/2,0,-4/3,-3,1) }.
4.6:8) A (a 5x6 matrix) has 4 pivot columns, so
rankA = dim(Col A) = 4.
By the Rank Thm. dim(Nul A) = n - rankA = 6 - 4 = 2.
No, Col A does not equal R^4. Note the pivot columns of A consist of 4
linearly independent column vectors in R^5 which comprise a basis for Col A.
Thus Col A is a 4 dimensional subspace of R^5. However, these pivot column
vectors each have 5 entries (recall A has 5 rows) so are not in R^4.
4.6:16) A is given to be 6x4 (6 rows, 4 columns).
By the Rank Theorem,
dim(Nul A) = 4 - rankA = 4 - (# pivot columns of A).
A could have as many as 4 pivot columns; thus dim(Nul A) could be 0 (smallest
possible).
4.9.2) a)
From
Healthy Ill
P = [ .95 .45 ] Healthy
[ .05 .55 ] Ill To
b) The initial probability state vector (on Monday) is
x_0 = [ proportion healthy ] = [ .80 ]
[ proportion ill ] [ .20 ]
The state vector for Tues. is
x_1 = P x_0 = [ .95 .45 ] [ .80 ] = [ .85 ] Thus 15% are ill
[ .05 .55 ] [ .20 ] [ .15 ] on Tues.
The state vector for Wed. is
x_2 = P x_1 = [ .95 .45 ] [ .85 ] = [ .875 ] Thus 12.5% are ill
[ .05 .55 ] [ .15 ] [ .125 ] on Wed.
c) To find the probability that a student is healthy in two days given that he
is healthy today, find the state vector x_2 given the initial state vector
x_0 = [ probability healthy ] = [ 1 ]
[ probability ill ] [ 0 ]
The state vector one day hence is
x_1 = P x_0 = [ .95 .45 ] [ 1 ] = [ .95 ]
[ .05 .55 ] [ 0 ] [ .05 ]
The state vector two days hence is
x_2 = P x_1 = [ .95 .45 ] [ .95 ] = [ .925 ] The first entry
[ .05 .55 ] [ .05 ] [ .075 ] .925 is answer
4.9:3) a)
From
#1 #2 #3
P = [ .50 .25 .25 ] #1
[ .25 .50 .25 ] #2 To
[ .25 .25 .50 ] #3
The probability that the animal chooses food #2 on second trial given that it
chose food #1 on initial trial is the 2nd entry of x_2 given the initial state
x_0 = (1,0,0).
x_2 = P x_1 = P (P x_0) = P^2 x_0 = P^2 [ 1 ]
[ 0 ]
[ 0 ]
= 1st column of the matrix P^2 = [ .5(.5) + .25(.25) + .25(.25) ] = [.3750]
[.25(.5) + .5(.25) + .25(.25) ] [.3125]
[.25(.5) + .25(.25) + .5(.25) ] [.3125]
Thus the desired probability (2nd entry) is .3125.
Alternatively,
x_2 = P x_1 = P (P x_0) = P [ .50 ] = [.3750]
[ .25 ] [.3125]
[ .25 ] [.3125]
4.9:12) a) Note P is a regular stochastic matrix (here all entries of P are
postive) so a unique steady state probability vector x exists (Theorem 18).
Solve
Px = x
==> (P-I)x = 0
==> [ -.05 .45 ] [ x_1 ] = [ 0 ]
[ .05 -.45 ] [ x_2 ] [ 0 ]
Thus .05 x_1 = .45 x_2 ==> x_1 = 9x_2, x_2 free, so x = x_2 [ 9 ].
[ 1 ]
Now the fact that [ x_1 ] is a probability vector implies x_1 + x_2 = 1
[ x_2 ] ==> x_2 = 1 - x_1.
Thus x_1 = 9 x_2 = 9(1 - x_1) = 9 - 9 x_1 ==> 10 x_1 = 9 ==> [ x_1 ] = [.9]
[ x_2 ] = [.1]
This last step is equivalent to picking x_2 any non-zero # (say 1) and
dividing the resultant vector ( [ 9 ] ) by the sum of the entries (= 10).
[ 1 ]
b) After "many" days, the state probabilities approach those of the steady
state vector, regardless of the initial state of the person (so no, it doesn't
matter whether the person is ill today). Thus the probability that the person
is ill many days in future is the 2nd entry of steady state vector, i.e. .1.
4.9:13) Find the steady state vector from (P-I)x = 0.
[ -.5 .25 .25 0 ] [ -1 1/2 1/2 0 ] [ 1 -1/2 -1/2 0 ]
[ .25 -.5 .25 0 ] ~ [ 1 -2 1 0 ] ~ [ 0 -3/2 3/2 0 ]
[ .25 .25 -.5 0 ] [ 1 1 -2 0 ] [ 0 3/2 -3/2 0 ]
[ 1 0 -1 0 ] Thus x_1 = x_3
~ [ 0 1 -1 0 ] x_2 = x_3
[ 0 0 0 0 ] x_3 free
Since the probabilities sum to 1, x_1 + x_2 + x_3 = 1 ==>
x_3 = 1/3 = x_2 = x_1.
Thus after many days, the animal has no food preference (all foods equally
preferred), i.e. the steady state probability vector is [ 1/3 ].
[ 1/3 ]
[ 1/3 ]