MATLAB Graphics - A Guide to MATLAB for Beginners and Experienced Users - page 89

appears in the corresponding column of the diagonal matrix R . This says
exactly that AU = UR .
(a) Verify the equality AU = UR for eachof the coefficient matrices in
Problem 10.
(b) In fact, rank( A ) = rank( U ), so when A is nonsingular, then

U − 1 AU = R .

Thus if two diagonalizable matrices A and B have the same set of
eigenvectors, then the fact that diagonal matrices commute implies
the same for A and B . Verify these facts for the two matrices

102
− 104
− 1

52 − 8
36 − 10
33 − 7

, ;

A = B =

− 15

that is, show that the matrices of eigenvectors are the “same” —
that is, the columns are the same up to scalar multiples — and
verify that AB = BA .
13. This problem, having to do with genetic inheritance, is based on Chapter
12 in Applications of Linear Algebra , 3rd ed., by C. Rorres and H. Anton,
John Wiley & Sons, 1984. In a typical inheritance model, a trait in the off-
spring is determined by the passing of a genotype from the parents, where
there are two independent possibilities from each parent, say A and a ,
and eachis equally likely. ( A is the dominant gene, and a is recessive.)
Then we have the following table of probabilities of the possible geno-
types for the offspring for all possible combinations of the genotypes of the
parents:

Genotype of Parents
AA-AA AA-Aa AA-aa Aa-Aa Aa-aa aa-aa

Genotype AA 1 1/2 0 1/4 0 0

of Aa 0 1/2 1 1/2 1/2 0

Offspring aa 0 0 0 1/4 1/2 1

Now suppose one has a population in which mating only occurs with
one's identical genotype. (That's not far-fetched if we are considering con-
trolled plant or vegetable populations.) Next suppose that x 0 , y 0 , and z 0
denote the percentage of the population with genotype AA , Aa , and aa
respectively at the outset of observation. We then denote by x n , y n , and
z n the percentages in the n thgeneration. We are interested in knowing