TASK MATHEMATICS for ECONOMIC APPLICATIONS 19/03/2016 I M 1) Putting the complex number in trigonometric form we get:
D œ3 $3 3 $3"* ' ") $ œ3 $3 3 $3$ # # $ œ
œ % œ % % œ % # # # % # ( (
# # % %
3 %3$ # 3 3 œ 3
cos 1 sen 1.
The fourth roots of D are D œ % # cos( sen( œ
% %
% % 1 3 1
œ % # ( Î% #5 ( Î% #5 œ
% %
% cos 1 1 sen 1 1
3
œ % # ( 5 ( 5 œ !ß "ß #ß $
"' # "' #
% cos 1 1 sen 1 1
3 , .5
The imaginary part of % D is positive iff sen ( 5 ! and this is true iff
"'1 1#
! ( 5 Ê ( 5 * Ê œ !ß "
"'1 1# 1 ) )
5 . The requested roots are:
D œ % # ( ( œ
"' "'
! % cos 1 3sen 1
œ % # # # # # # #
#
% 3 ;
D œ % # "& "& œ
"' "'
" % cos 1 3sen 1
œ % # # # # # # #
#
% 3 .
I M 2) By the Sylvester Theorem Dim Ker Dim Imm œDim ‘% œ %. If Dim Ker œ # trivialy follows Dim Imm œRank œ #.
To find and such that Rank7 5 is we reduce matrix by elementary operations# on the lines:
" " # # " " # #
" # " " ! " " $
$ # 7 5 ! " 7 ' 5 '
Ê Ê
V V
V $V V V
# "
$ " $ #
Ê œ #
" " # #
! " " $
! ! 7 ( 5 *
and from the last matrix it follows that Rank iff
7 ( œ ! and 5 * œ ! Ê 7 œ ( and 5 œ *.
So the matrix is equal to Kernel
" " # #
" # " "
$ # ( *
. To find a basis for the remember that belongs to the Ker— 0 if —† œ, that in system form is:
B B #B #B œ ! B B #B #B œ !
B #B B B œ ! B B $B œ !
$B #B (B *B œ ! B B $B œ !
Ê Ê
" # $ % " # $ %
" # $ % # $ %
" # $ % # $ %
Ê B œ $B &B 0 B œ B $B
" $ %
# $ % , and so a generic element of Ker is:
— œ $B &B B $B B ß B $ %ß $ %ß $ %œ B$ $ " "ß ! B & $ !ß "ß ß % ß ß and a basis for Ker 0 is UKer 0 œ $ " "ß ! ß & $ !ß "ß ß ß ß .
To find a basis for the Immage remember that belongs to Imm˜ 0 if —† œ˜, that in system form is:
B B #B #B œ C B B #B #B œ C
B #B B B œ C B B $B œ C C
$B #B (B *B œ C B B $B œ C $C
Ê Ê
" # $ % " " # $ % "
" # $ % # # $ % # "
" # $ % $ # $ % $ "
Ê
B B #B #B œ C B B $B œ C C
! œ C C %C
" # $ % "
# $ % # "
$ # "
.
So a generic element of Imm 0 is a vector ˜œ C C C "ß #ß $ with C C %C œ !$ # "
or C œ %C C$ " #; ˜œ C C %C C "ß #ß " #œ C " ! % C ! "" ß ß # ß ß " and a basis for Imm 0 is UImm 0 œ" ! % ß ! "ß ß ß ß ".
I M 3) To find the characteristic polynomial of we calculate:
: œ œ œ
$ # "
" % 5
" # $
- -ˆ
-
-
-
œ œ # % # % œ
# ! %
" % 5
" # $
% 5
# $
- -
-
-
- - - -
- œ # - - #- "# #5 % -# -œ
œ # - - # # #5 "'- .
Note that a5ß : # œ ! , this implies that - œ # is an eigenvalue of the matrix ß a5. If ; - œ-# # #5 "'- , - œ # is a multiple eingenvalue for iff:
; # œ % % #5 "' œ ! Ê 5 œ ) . In this case the three eingevalues for are
-" œ ! and -#ß$ œ #. To study the diagonalizability of we must check the dimension
of the eingespace associated to - œ #, that is:
$ # œ $ œ
" # "
" # )
" # &
Rank Rank
ˆ
œ $ œ $ # œ "
" # "
! ! *
! ! '
Rank .
Matrix isn't diagonalizable since the algebraic multiplicity of - œ # is greater than its geometric multiplicity.
I M 4) Two similar matrices have equal characteristic polynomials, and so we get:
: œ œ " # œ % "
" $
- -ˆ - - -
-
# and
: œ œ 7 œ 5 7
" 5
- -ˆ - - -
-
# .
The two polynomials are equal iff 5 œ % and 7 œ ".
To find a matrix œ : : we must satisfy † œ † Ê
: :
""#" "###
Ê " # † : : œ : : † ! " Ê
" $ : : : : " %
#""" ##"# ""#" ##"#
Ê : #: : #: œ : %: :
: $: : $: : %: :
"""" #"#" "#"# #### ##"# "### ""#". In system form we have:
: #: œ : : #: œ :
: #: œ %: : : #: œ $: : œ #
: $: œ : : $: œ :
: $: œ %: : : : œ :
Ê Ê
"" #" "# "" #" "#
"# ## "# "" "" ## "# ""
"" #" ## "" #" ##
"# ## ## #" "# #" ##
: : : œ : :
#" "#
## "# #"
.
Choosing, for instance, : œ # and : œ ", one possible matrix is ! #
" $
"# #" .
II M 1) For any pair of unit vectors @ß A, H@ßA# 0 Bß C œ @ †‡ 0 †AX where ‡ 0 is the Hessian matrix of . From this it follows easily that 0 H 0 Bß C#@ß@ œH#@ß@0 Bß C . So H 0 Bß C#@ß@ H#@ß@0 Bß C œ ! iff H 0 Bß C œ !#@ß@ . Since:
f0 œ œ
#B ß #C # #B " %BC
%BC # #C "
/ / 0 / /
/ /
B C B C B C B C
B C B C
# # # #
# # # #
# # # #
and ‡ # #
we have:
H 0 Bß C œ œ
# #
/ /
/ /
@ß@#
B C B C
B C B C
#
#
# #
ß # #B " %BC
%BC # #C "
#
#
#
#
# # # #
# # # #
œ#B "# /B C# #%BC/B C# ##C "# /B C# # œ #B C#/B C# #. So we have H 0 Bß C œ !#@ß@ iff B C œ !.
II M 2) The equation is satisfied at point T œ !ß " , since 0 Bß C œ logB C# # BC Êlog " ! œ !. Then
f0 œ #B #C f0 œ
ß !ß " ß #
B C# # C B C# # B with " , and so
C !ß "
!ß "
w w
B Cw
! œ 0 œ "
0 #.
For the second order derivative we have: Cww BBww # BCww † C w CCww † Cw # ,
Cw
œ 0 0 0
0
‡ 0 œ ‡0 œ #
# C B %BC
%BC # B C
# #
# #
B C B C
B C B C
# ## # ##
# ## # ##
"
" !ß " "
" #
, and so
Cww # " † # † #
! œ # œ Þ
# %
"
"# "#
II M 3) From C œ0 B ß B " # and B ß B" #œ 1> ß > ß >" # $, by the chain rule (derivative of a composite function) we get: ` ` ` B ß B
` > ß > ß > ` B ß B ` > ß > ß >
C C
œ †
" # $ " # " # $
" #
or
" $ %œ † " " #
! # "
` `
`B `B
C C
" # that in system form gives:
`
`B `
` `
`B `B `
` `
`B `B
`B
`B " #
C
C C
C C
C C
"
" #
" #
"
#
œ "
# œ $ Ê Ê œ
# œ %
œ "
œ #
` C
` B ß B " #
.
II M 4) The triangle of vertexes X !ß ! , "ß ! and !ß " is red-drawed in the figure:
X is a compact set and the function 0 is continuous so it admits absolute maximum and minimum on ; X a Bß C − X B , and are both not negative and trivially follows thatC min 0 œ ! at point !ß ! . For the maximum we observe that when at least one bet- ween B and increases, then increases, thus MaxC 0 0 must be searched on the upper border of having equation X C œ " B. If we define:
1 B œ0B " B œß B $ " B , we get:$
1w B œ $B $# " B# œ $B # " B# #B œ $ #B " and 1w B ! iff $ #B " ! Ê B "Î#.
Function increases from point 1 "Î#ß "Î# to the vertexes "ß ! and !ß " of X (grey arrows in the figure): by the simmetry of we conclude that Max0 0 œ " at points
"ß ! and !ß " . Point "Î#ß "Î# is a relative minimum point.
In the figure there are drawn zero level curve (yellow) and positive level curves (blue).
