MfEA 1 TASK MATHEMATICS for ECONOMIC APPLICATIONS 04/07/2015
I M 1) The complex number 3 " has argument αœ $ Î%1 and modulus 3 œ# thus $ 3 " œ $ # $ 3 $ œ% %
cos 1 sen 1
œ # $ † " #5 3 $ † " #5 œ
% $ $ % $ $
' cos 1 1 sen 1 1
œ # 5 3 5 5 œ !ß "ß #
% $ % $
# #
' cos1 sen1
1 1 with . The three roots are:
D œ # 3 œ " 3
% % #
%
!
'
$
cos 1 sen1
;
D œ # "" 3 "" œ # # $ " #3 $ " œ
"# "# % %
"
' '
cos 1 sen 1
œ % $ " 3 $ "
%
$ ;
D œ # "* 3 "* œ # # $ " 3 # $ " œ
"# "# % %
#
' '
cos 1 sen 1
œ % $ " 3 $ "
%
$ .
I M 2) The characteristic polynomial of is -
-
-
-
: œ œ
" # "
# "
% % 5
œ " " # # " # œ
% 5 % 5 % %
- - -
- -
œ " - - # 5 % # #5 # % ) %- - -œ œ -$ " 5 -# % 5 - % & 5 ;
if - œ ! is an eigenvalue of matrix , : - must satisfy condition : ! œ ! or
% & 5 œ ! and so 5 œ &. For 5 œ & :, - becomes:
: - œ -$ '-# * œ - - - $# and thus the three eigenvalues of are
-" œ !, -# œ-$ œ $. To check for diagonalizability of the matrix we must find the
geometric multiplicity of multiple eigenvalue -$ œ $; to do so, we consider the system:
$ˆ† @ œ ! Ê Ê
@ @ @
@ @ @
@ @ @
@ @ @
@ @ @
# # œ !
# $ œ !
% % # œ !
# # œ !
# $ œ !
" # $
" # $
" # $
" # $
" # $
(the third equation is a multipler of the first), with solution:
@ œ @ ß ß #@ " ! "œ @ "ß ß #" ! .
As we can simply note, the eigenvectors associated with eigenvalue -$ œ $ belong to a line, thus geometric multiplicity of eigenvalue -$ œ $ is , less than its algebraic multi-"
plicity, so the matrix is not a digonalizable one.
A basis for the eigenspace associated to -$ œ $ is UIW $ œ"ß ß #! .
I M 3) For a linear map 0 À‘% Ä‘$, 0 — œ —† , Kernel's dimension is given by dim ‘% dim Imm 0 œ % Rank , so Kernel's dimension is maximum when Rank is minimum. To calculate the rank of the matrix we reduce it by elementary operations on its lines:
MfEA 2
" " " " " " " "
" " 7 " ! ! 7 " !
" " " 5 ! ! ! 5 "
Ê
V Ã V V V Ã V V
# # "
$ $ "
and from the last matrix we get:
Rank .
if and
if and or and
if and
œ
" œ " œ "
# œ " Á " Á " œ "
$ Á " Á "
7 5
7 5 7 5
7 5
Kernel's dimension is maximum when 7 œ 5 œ " and in this case dimKer 0 œ $. The matrix becomes œ
" " " "
" " " "
" " " "
.
To find bases for Kernel and Image of the linear map we begin determining the elements of Kernel: —† œÊB B " # B$B% œ !Ê B% œ B B " # B$. The elements of Kernel have the form: — œB ß B ß" # B ß$ B B " # B$œ
œB"†"ß !ß!ß "B#†!ß "ß!ß " B †$ !ß !ß"ß " ; so a basis may be:
UKer 0 œ"ß !ß!ß " ß !ß "ß !ß " ß !ß !ß "ß ". The elements of the Image have the form:
˜œ —† œB B " # B$B%ß B B " # B$ B%ß B B " # B$B%œ œCß Cß CœC†"ß "ß "; so UImm 0 œ"ß "ß ".
I M 4) Remember that square matrices and of the same order are similar if a non singular matrix exists such that † œ † , or œ"† † . So:
œ# ! †" # † # !œ † # ! † % %
" # # " " # " # $ # Ê
"
"
%
Ê # #
œ #
"Î
# .
II M 1) Since 0 T œ " , the condition is satisfied then à f0 Bß C œ #Bß #C, f0 T œ #ß # and ‡ 0 œ‡0 T œ# !
! # . Since 0 T Á !Cw an implicit function B Ä C B exists, B œ! "
#. We calculate Cw B and Cww B and we get: Cw B œ 0 T œ "
! ! ! 0 TBw
Cw
and
Cww Cw Cw # #
B œ 0 T # 0 T † B 0 T † B œ œ #
0 T
# ! #
#
! BBww BCww CCww
! !
Cw
.
II M 2) From W@0 " ß " œ f0 " ß " †@ we have f0 œ/BC B/BCß B/BC, f0 " ß " œ #ß " and W@0 " ß " œ #ß " † cosαßsenαœ#cosαsenα. From condition W@0 " ß " œ ! it follows:
#cosαsenα α α 1 α 1
œ ! Ê œ # Ê œ #
# #
tg arctg , .
II M 3) We construct the Lagrangian function:
ABß Cß Dß-œ B C D -B C D $# # # . First Order Conditions:
f œ " A # Bß - " # Cß- " #-D B C D $ß # # # , from which:
MfEA 3
" œ ! œ "Î
" œ ! œ "Î
" D œ ! D œ "Î
B C D œ $ $Î œ $
Ê Ê
œ „"
œ …"
D œ „"
œ „"Î
# B B #
# C C #
# #
%
B C
#
- -
- -
- -
- -
# # # #
.
We get two constrained critical points: T" œ " ß ß" " and T# œ " " ß ß ". Second Order Conditions:
‡ A -
-
-
œ
D
D
!
! !
! !
! !
#B #C #
#B #
#C #
# #
.
The problem has three variables and one constraint, so we must consider two leading minors, ‡$and :‡%
‡ -
- -
-
$ œ œ # œ
!
!
!
!
!
#B #C
#B #
#C #
#B #B C #B #
#C # #C
œ ) B ) C œ )- # - # -B C# # ;
‡ -
-
-
% œ œ
D
D
!
! !
! !
! !
#B #C #
#B #
#C #
# #
œ #D
#B #
# #B
# #
#C #D #C
! # !
# !
!
! ! C
# œ
! B
- -
- -
-
œ %D#%-# #- #B % B #C % C œ - -
œ "' B "' C "' D œ "'-# # -# # -# # -#B C D# # # .
It's easy to note that ‡% ! at T" and T#, ‡$ ! at T" while ‡$ ! at T#.
It follows that T" is a maximum point with 0 T" œ $ while T# is a minimum point with .0 T# œ $
II M 4) The feasible region is a square represented in red colour in the left figure inX the next page. Note that the function is simmetric respect to the origin 0 Sœ ! ! ß
0 Bß C œ 0 Bß C and 0 Bß C œ 0 Bß C thus we can study the optimiza- tion problem only on the right part of the square .X
First case (free optimization)
First Order Conditions: f0 œ . œ ! Ê œ !
œ ! œ !
#B %Cß %B #C #B %C B
%B #C C
.
Free optimization presents only one critical point S that belongs to the interior of .X Second Order Conditions: ‡ 0 œ ‡# S œ # %œ " ; so
% # # ! S is a
saddle point.
Second case (optimization along the border of )X
0 Bß " B œ B # %B" B " B # œ 'B# 'B ", that is a parabola with mi- nimum equal "Î# at point B œ "· #.
0 Bß B " œ B # %BB " B " # œ #B # #B ", that is a parabola with maximum equal $Î# at point B œ "· #.
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We conclude that Max 0 œ $Î# at points "Î#ß "Î# and "Î "Î#ß # (blue points on the left figure below), min 0 œ "Î# at points "Î "Î#ß # and "Î#ß "Î# (pink points).On the right figure below there have been drawn negative level curves (pink), zero level curves (yellow) and positive level curves (blue).
