Intermediate Test of Mathematics for Economic Applications 02/12/2016
I M 1) D œ #3 "*† " 3 "' œ # † 3 †"* "* # cos 1 sin 1 "'œ
% 3 %
# † 3 † #"* $ )cos %1 3sin %1 œ # 3#( and D œ # 3#( ; D œ# 3 œ#(
##(#cos 1 sin 1 œ ##(#cos1 sin1 .
1 1
# 3 # % 5 3 % 5 with 5 œ !ß "
We conclude that D œ „ ##(## # œ „ #
# 3 # "$ " 3 .
I M 2) Since the system is an homogeneus one, it has ∞" solutions if and only if 8 Rank ‚ œ " , where 8 is the number of the unknown variables and ‚ is the matrix of the coefficients. To check the rank we reduce the matrix by elementary operations on its rows:
7 8 " " 7 8 " "
7 8 5 " ! ! 5 " !
7 8 5 2 ! ! ! 2 "
Ä
V V V V
# "
$ #
and it is easy to note that Rank ‚ œ $ iff 2 Á " 5 Á ", and at least one from and is different from .7 8 ! I M 3) If is the matrix associated to the linear map , we know that dim ImmJ J is equal to Rank .
Like the previous exercise we can check Rank by elementary operations on the lines of :
7 " ! ! ! " 75 ! !
" 5 ! ! " 5 ! !
! ! 5 " ! ! 5 "
! ! " 7 ! ! " 75 !
Ä Ä
V 7V V V
V 7V G
" # " #
% $
$ G%
Ä
" 5 ! !
! " 75 ! !
! ! " 5
! ! ! " 75
and so we have:
Rank œdim Imm œ %
J # 75 Á "
75 œ "
if
if .
By Sylvester Theorem we know that dim Ker dimImmœdim ‘% œ %; if dimImm is minimumÐ75 œ " Ñ it trivialy follows dimImmœdim Ker œ #. To find a basis for the Kernel remember that belongs to Ker— J if —† œ, that in
system form is: . Choosing
7
Á ! 7 œ "Î5 B B œ !
B 5B œ ! B œ 5B
5B B œ ! B œ 5B
B 7B œ !
Ê 5
" #
" # " #
$ % % $
$ %
and ,
every element of Ker J is of the form:
— œ 5B B B ß 5B #ß #ß $ $œ B# 5 " !ß ! B ! ! "ß 5ß ß $ ß ß . So a basis for Ker J is UKer J œ 5 " !ß ! ß ! ! "ß 5ß ß ß ß .
For a basis of Imm J remember that belongs to the immage of if ˜ J —† œ˜, that in system form is:
7 "Î5
"Î5
B B œ C B B œ C B 5B œ 5C
B 5B œ C B 5B œ C B 5B œ C
5B B œ C 5B B œ C 5B B
B 7B œ C B B œ C
Ê Ê
" # " " # " " # "
" # # " # # " # #
$ % $ $ % $ $ %
$ % % $ % %
œ C C œ 5C
5B B œ 5C
Ê C œ 5C
$ $ %
$ % %
# "
.
Choosing again 5 Á ! and 7 œ "Î5, every element of Imm J is of the form:
˜ œ C C C C "ß #ß $ß % œ C 5C 5C C"ß "ß %ß %œ C " 5 ! ! C ! ! 5 "" ß ß ß % ß ß ß and a basis for Imm J is UImm J œ" 5 ! ! ß ! ! 5 "ß ß ß ß ß ß .
I M 4) Since the matrix # has eigenvectors with two components, thus # is a # ‚ #
square matrix; if # œ B C #
D A
by the conditions the matrix must satisfy:
# † " œ " † " #† " œ " † "
" " " "
and , that in system form is:
B B
B " !
C ! "
œ
C œ " œ ! D A œ " œ "
C œ " D œ "
D A œ " A œ !
Ê Ê # .
I M 5) The characteristic polynomial of is:
: œ œ œ
# " " " " "
" # " ! # "
" " 5 5 " " 5
-
- -
- -
- - -
œ " -# -5 - " 5 - " " - œ
œ " - - # 5 $ - $5 # œ ! , so a5 : " œ !: , and - œ " is an eigen- value of for every 5. Define ; - œ -# 5 $ - $5 #.
If - œ " is a multiple eigenvalue of ; " œ % #5 œ !, and so 5 œ #.
Since -# & % œ- - " % - , t characteristic polynomial is factorized ashe : - œ - " # % - and the eigenvalues of are -" œ-# œ " and -$ œ %. Since is a symmetric matrix, from the Spectral Theorem we know that is a diagona- lizable matrix and so 7 œ # œ 71" +": the geometric multiplicity of - œ " , the dimenti- on of Xf-œ", the eigenspace associated to the eigenvalue - œ ", is equal to , its alge-# braic multiplicity. Solving the homogeneus system " †ˆ† — œ we get:
B B B œ ! B B B œ !
B B B œ !
Ê B œ B B
" # $
" # $
" # $
$ " #. Every element of Xf-œ" is a vector:
— œ B ß B ß B B " # " # but for -" œ-# œ " we must find two orthogonal unit vectors.
For B œ "" and B œ !# we get —" œ "ß !ß " . To get another eigenvector orthogonal
to —" we pose: "ß !ß " † B ß B ß B B" # " #œ ! and we get:
B B B" " # œ !Ê B# œ #B" and so the eigenvector —# œ B ß " #B"ß B". For B œ "" we get —# œ "ß #ß " . From —" and —# we get the unit eigenvectors:
•" œ#Î#ß !ß#Î#, •# œ'Î'ß 'Î$ß 'Î'.
For -$ œ % the homogeneus system % †ˆ† —œ Ê
#
#
#
#
$ $ B B B œ !
B B B œ !
B B B œ !
Ê B œ B B Ê B œ B
B B œ ! B œ B
" # $
" # $
" # $
$ " # $ "
" # # " . Every element of
Xf-œ% is a vector —œ B ß B ß B " " ". For B œ "" we get —$ œ "ß "ß " and the unit eigenvector •# œ$Î$ß$Î$ß $Î$.
The orthogonal requested matrix is ” œ
#Î# 'Î' $Î$
! 'Î$ $Î$
#Î# 'Î' $Î$
.
