Eigenvectors, Eigenvalues, and Finite Strain · 2017. 8. 17. · 8/17/17 2 9. EIGENVECTORS,...

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Eigenvectors,Eigenvalues,andFiniteStrain

GG303,2016

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StrainedConglomerateSierraNevada,California

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9.EIGENVECTORS,EIGENVALUES,ANDFINITESTRAIN

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HomogenousdeformaMondeformsaunitcircletoa“strainellipse”

ObjecMve:Toquan%fythesize,shape,andorientaMonofstrainellipseusingitsaxes


I MainTopicsAEquaMonsforellipses

BRotaMonsinhomogeneousdeformaMon

CEigenvectorsandeigenvalues

DSoluMonsforgeneralhomogeneousdeformaMonmatrices

E Keyresults

F Appendices(1,2,3,4)

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II EquaMonsofellipsesA EquaMonofaunitcircle

centeredattheorigin1

2

3

4

x2 + y2 = 1

x y[ ] 1 00 1⎡

⎣⎢

⎤

⎦⎥

xy

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 1

X[ ]T F[ ] X[ ] = 1

x y[ ] xy⎡

⎣⎢⎢

⎤

⎦⎥⎥= x y[ ] 1x + 0y0x +1y

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 1

Symmetric

Here,[F]istheidenMtymatrix[I].SoposiMonvectorsthatdefineaunitcircletransformtothosesameposiMonvectorsbecause[X’]=[F][X].

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II EquaMonsofellipsesB EquaMonofanellipse

centeredattheoriginwithitsaxesalongthex-andy-axes

1

2

3

4

ax2 + 0xy + dy2 = 1

x y[ ] a 00 d⎡

⎣⎢

⎤

⎦⎥

xy

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 1

X[ ]T F[ ] X[ ] = 1

x y[ ] ax + 0y0x + dy⎡

⎣⎢⎢

⎤

⎦⎥⎥= 1

SymmetricPosiMonvectorsthatdefineaunitcircletransformtoposiMonvectorsthatdefineanellipsebecause[X’]=[F][X].

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II EquaMonsofellipsesC “Symmetric”equaMon

ofanellipsecenteredattheorigin

1

2

3

4

ax2 + 2bxy + dy2 = 1

x y[ ] a bb d

⎡

⎣⎢

⎤

⎦⎥

xy

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 1

X[ ]T F[ ] X[ ] = 1

x y[ ] ax + bybx + dy

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 1

Symmetric

Displacementvectorsareinblack.Bluenumbersarefinalaxiallengths.RednumbersareiniMalradii.Displacementvectorsaresymmetricaboutaxesofellipse.

Example : F = 2 11 2

⎡

⎣⎢

⎤

⎦⎥

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II EquaMonsofellipsesD GeneralequaMonofan

ellipsecenteredattheorigin

1

2

3

4

ax2 + b + c( )xy + dy2 = 1

x y[ ] a bc d

⎡

⎣⎢

⎤

⎦⎥

xy

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 1

X[ ]T F[ ] X[ ] = 1

x y[ ] ax + bycx + dy

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 1

Notsymmetricifb≠c

Example : F = 2 10 2

⎡

⎣⎢

⎤

⎦⎥

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Vectorsalongaxesofellipsetransformbacktoperpendicularvectorsalongaxesofunitcircle

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A Let[X]bethesetofallposiMonvectorsthatdefineaunitcircle

B Let[X’]bethesetofallposiMonvectorsthatdefineanellipsedescribedbyahomogenousdeformaMonatapoint

C [X’]=[F][X] (Forwarddef.)D [X]=[F-1][X’](Reversedef.)E Thematrices[F]and[F-1]

containconstants

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IIIRotaMonsinhomogenousdeformaMon


F ThedifferenMaltangentvectors[dX’]and[dX]comefromdifferenMaMng[X’]=[F][X]and[X]=[F-1][X’],respecMvely.

G [dX’]=[F][dX](Forwarddef.)H [dX]=[F-1][dX’](Reversedef.)I [F]transforms[X]to[X’],and

[dX]to[dX’]J [F-1]transforms[X’]to[X],and

[dX’]to[dX]K PosiMonvectorsarepairedto

correspondingtangents

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IIIRotaMonsinhomogenousdeformaMon(cont.)

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L Whereanon-zeroposiMonvectoranditstangentareperpendicular,theposiMonvectorachievesitsgreatestandsmallest(squared)lengths,asshownbelow

MN Maximaandminimaof

(squared)lengthsoccurwheredQ’=0

OP

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′Q = ′!X • ′!X = ′X[ ]T ′X[ ]

d ′Q = d ′

!X • ′!X( ) = ′!X • d ′!X + d ′!X • ′!X = 0

2 ′!X • d ′

!X( ) = 0⇒ ′!X • d ′!X( ) = 0


Q ThetangentvectorperpendiculartothelongestposiMonvectorparallelstheshortestposiMonvector(whichliesalongthesemi-minoraxis),andvice-versa.

R Similarreasoningappliestothecorrespondingunitcircle.

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S Fortheunitcircle,alliniMalposiMonvectorsareradialvectors,andeachiniMaltangentvectorisperpendiculartotheassociatedradialposiMonvector.TherediniMalvectorpair[X*,dX*]andtheblueiniMalvectorpair[X*,dX*]bothshowthis.

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T AllthefinalposiMon-tangentvectorpairsfortheellipsehavecorrespondinginiMalposiMon-tangentvectorpairsfortheunitcircle(andvice-versa).

U EveryposiMon-tangentvectorpairfortheunitcirclecontainsperpendicularvectors.

V OnlytheposiMon-tangentvectorpairfortheellipsethatparallelthemajorandminoraxes(i.e.,theredpair[X*’,dX*’])areperpendicular.

W “Retro-transforming”[X*’,dX*’]by[F-1]yieldstheiniMalredpairofperpendicularvectors[X*,dX*].

X Conversely,theforwardtransformaMonoftheredpairofiniMalperpendicularvectors[X*,dX*]using[F]yieldsthefinalperpendicularvectorspair[X*’,dX*’].

Y ThetransformaMonfrom[X*,dX*]to[X*’,dX*’]involvesarotaMon,andthatishowtherotaMonisdefined.

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• Thelongest(X1’)andshortest(X2’)posiMonvectorsoftheellipseareperpendicular,alongtheredaxesoftheellipse,andparallelthetangents.

• Thecorrespondingretro-transformedvectors([X1]=[F]-1[X1’],and[X2]=[F]-1[X2’])(alongtheblackaxes)areperpendicularunitvectorsthatmaintainthe90°anglebetweentheprincipaldirecMons.

• TheangleofrotaMonisdefinedastheanglebetweentheperpendicularpair{X1andX2}alongtheblackaxesoftheunitcircleandtheperpendicularprincipalpair{X1’,X2’}alongtheredaxesoftheellipse.

• TheseresultsextendtothreedimensionsifallthreesecMonsalongtheprincipalaxesofthe“strain”(stretch)ellipsoidareconsidered.

• SeeAppendix4formoreexamples.

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!′X

!X1

!X2

!X1′

!X2′

!X


IVEigenvectorsandeigenvalues(usedtoobtainstretchesandrotaMons)

ATheeigenvaluematrixequaMon[A][X]=λ[X]

1  [A]isa(known)squarematrix(nxn)2  [X]isanon-zerodirecMonaleigenvector(nx1)3  λisanumber,aneigenvalue4  λ[X]isavector(nx1)parallelto[X]5  [A][X]isavector(nx1)parallelto[X]

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ATheeigenvaluematrixequaMon[A][X]=λ[X](cont.)6 Thevectors[[A][X]],λ[X],and[X]sharethesamedirecMonif[X]isaneigenvector

7  If[X]isaunitvector,λisthelengthof[A][X]8Eigenvectors[Xi]havecorrespondingeigenvalues[λi],andvice-versa

9 InMatlab,[vec,val]=eig(A),findseigenvectors(vec)andeigenvalues(val)

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IVEigenvectorsandeigenvaluesBExample:MathemaMcalmeaningof[A][X]=λ[X]

A = 2 11 2

⎡

⎣⎢

⎤

⎦⎥

A − 22

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 2 1

1 2⎡

⎣⎢

⎤

⎦⎥

− 22

⎡

⎣⎢⎢

⎤

⎦⎥⎥= − 2

2

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 1 − 2

2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

Twoeigenvectors

Twoeigenvalues

A 22

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 2 1

1 2⎡

⎣⎢

⎤

⎦⎥

22

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 3 2

3 2

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 3 2

2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

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X ' = FX

F = 2 11 2

⎡

⎣⎢

⎤

⎦⎥

• EigenvectorsofsymmetricFgivedirecMonsoftheprincipalstretches

• EigenvaluesofsymmetricF(i.e.,λ1,λ2)aremagnitudesoftheprincipalstretchesS1andS2

AfA0

= πλ1λ2πr2

= λ1rλ2r= S1S2

Δ =Af − A0A0

=AfA0

− A0A0

= S1S2 −1

r = 1λ1 = 3λ2 = 1

IVEigenvectorsandeigenvaluesC Example:Geometricmeaning

of[A][X]=λ[X]

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Unitcircle

9.EIGENVECTORS,EIGENVALUES,ANDFINITESTRAINIVEigenvectorsandeigenvalues

D Example:MatlabsoluMonof[A][X]=λ[X]

A = 2 11 2

⎡

⎣⎢

⎤

⎦⎥

Eigenvalues(λ)

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Anglebetweenx-axisandlargesteigenvector

Anglebetweenx-axisAndsmallesteigenvector

*Matlabin2016doesnotordereigenvaluesfromlargesttosmallest

>>A=[21;12]A=2112>>[vec,val]=eig(A)vec=-0.70710.70710.70710.7071val=1003>>theta1=atan2(vec(2,2),vec(2,1))*180/pitheta1=45>>theta2=atan2(vec(1,2),vec(1,1))*180/pitheta2=135

Δ = det A[ ]−1Here,Δ = 3−1= 2

λ1 = 3λ2 = 1

Eigenvectors[X]givenbytheirdirecMoncosines

Eigenvector/eigenvaluepairs

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9.EIGENVECTORS,EIGENVALUES,ANDFINITESTRAINIVEigenvectorsandeigenvalues(cont.)

E GeometricmeaningsoftherealmatrixequaMon[A][X]=[B]=01 |A|≠0;

a [A]-1existsb Describestwolines(or3

planes)thatintersectattheorigin

c XhasauniquesoluMon2 |A|=0;

a [A]-1doesnotexistb Describestwoco-linearlines

thatthatpassthroughtheorigin(orthreeplanesthatintersectinalineorinaplanethroughtheorigin)

c [X]hasnouniquesoluMon;canhavemulMplesoluMons

Parallel lines have parallel normals

nx(1) ny

(1) x d1=0nx

(2) ny(2) y d2=0

AX = B = 0

=

|A| = nx(1) * ny

(2) - ny(1) * nx

(2) = 0 n1 x n2 = 0

Intersecting lines have non-parallel normals

nx(1) ny

(1) x d1=0nx

(2) ny(2) y d2=0

AX = B = 0

=

|A| = nx(1) * ny

(2) - ny(1) * nx

(2) ≠ 0 n1 x n2 ≠ 0n1 n2

n1

n2Det[A]=area(volume)definedbyparallelogram(parallelepiped)basedonunitnormals

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IVEigenvectorsandeigenvalues(cont.)DAlternaMveformofaneigenvalueequaMon

1[A][X]=λ[X]SubtracMngIλ[X]=λ[IX]=λ[X]frombothsidesyields:

2[A-Iλ][X]=0(sameformas[A][X]=0)E SoluMoncondiMonsandconnecMonswithdeterminants

1UniquetrivialsoluMonof[X]=0ifandonlyif|A-Iλ|≠02MulMpleeigenvectorsoluMons([X]≠0)

ifandonlyif|A-Iλ|=0* Seepreviousslide

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IVEigenvectorsandeigenvalues(cont.)

FCharacterisMcequaMon:|A-Iλ|=01TherootsofthecharacterisMcequaMonaretheeigenvalues(λ)

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FCharacterisMcequaMon:|A-Iλ|=0(cont.)2Eigenvaluesofageneral2x2matrix

a

b

c

d

A − Iλ = a − I bc d − λ

= 0

a − λ( ) d − λ( ) − bc = 0

λ2 − a + d( )λ + ad − bc( ) = 0

λ1,λ2 =a + d( ) ± a + d( )2 − 4 ad − bc( )

2

(a+d)=tr(A)(ad-bc)=|A|

A = a bc d

⎡

⎣⎢

⎤

⎦⎥

λ1 + λ2 = tr A( )λ1λ2 = A

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GTosolveforeigenvectors,subsMtuteeigenvaluesbackintoAX=lXandsolveforX(seeAppendix1)

HEigenvectorsofrealsymmetricmatricesareperpendicular(fordisMncteigenvalues);seeAppendix3

*Allthesepointsareimportant

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IVSoluMonsforgeneralhomogeneousdeformaMonmatricesA Eigenvalues

1  StartwiththedefiniMonofquadra%celongaMonQ,whichisascalar

2  Expressusingdotproducts

3  Clearthedenominator.DotproductsandQarescalars.

!′X •!′X!

X •!X

=Q

!′X •!′X =!X •!X( )Q

Lf2

L02 =Q

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9.EIGENVECTORS,EIGENVALUES,ANDFINITESTRAINIV SoluMonsforgeneralhomogeneous

deformaMonmatricesA Eigenvalues

4  ReplaceX’with[FX]5  Re-arrangebothsides6  BothsidesofthisequaMonlead

offwith[X]T,whichcannotbeazerovector,soitcanbedroppedfrombothsidestoyieldaneigenvectorequaMon

7 [FTF]issymmetric:[FTF]T=[FTF]8  Theeigenvaluesof[FTF]arethe

principalquadraMcelongaMonsQ=(Lf/L0)2

9  Theeigenvaluesof[FTF]1/2aretheprincipalstretchesS=(Lf/L0)

F[ ]nxn

X[ ]nx1

⎡⎣⎢

⎤⎦⎥

T

F[ ]nxn

X[ ]nx1

⎡⎣⎢

⎤⎦⎥= X

nx1⎡⎣

⎤⎦TX[ ]nx1Q1x1

Xnx1⎡⎣

⎤⎦TFnxn⎡⎣

⎤⎦TFnxnXnx1

⎡⎣

⎤⎦ = Xnx1

⎡⎣

⎤⎦TQ1x1

Xnx1⎡⎣

⎤⎦

Fnxn

T Fnxn

⎡⎣

⎤⎦ Xnx1⎡⎣

⎤⎦ =Q Xnx1

⎡⎣

⎤⎦

!′X •!′X =!X •!X( )Q

" A[ ] X[ ] = λ X[ ]"

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IV SoluMonsforgeneralhomogeneousdeformaMonmatricesB SpecialCase:[F]issymmetric

1  [FTF]=[F2]becauseF=FT2  Theprincipalstretches(S)againare

thesquarerootsoftheprincipalquadraMcelongaMons(Q)(i.e.,thesquarerootsoftheeigenvaluesof[F2])

3  Theprincipalstretches(S)alsoaretheeigenvaluesof[F],directly

4  ThedirecMonsoftheprincipalstretches(S)aretheeigenvectorsof[F],andof[FTF]=[F2]!

5  Theaxesoftheprincipal(greatestandleast)straindonotrotate

FTF⎡⎣ ⎤⎦ X[ ] =Q X[ ]

Q =Lf2

L02 ; S =

LfL0

⇒ Q = S

F2⎡⎣ ⎤⎦ X[ ] =Q X[ ]

F[ ] X[ ] = S X[ ]

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F = 2 20.5 1

⎡

⎣⎢

⎤

⎦⎥

R = 0.89 0.45−0.45 0.89

⎡

⎣⎢

⎤

⎦⎥

′X[ ] = F[ ] X[ ]; F[ ] = R[ ] U[ ]

F[ ] = 2 20.5 1⎡

⎣⎢

⎤

⎦⎥; F[ ]T = 2 0.52 1

⎡

⎣⎢

⎤

⎦⎥

U[ ] = F[ ]T F[ ]⎡⎣ ⎤⎦1/2

= 4.25 4.54.5 5

⎡

⎣⎢

⎤

⎦⎥

1/2

= 1.56 1.341.34 1.79

⎡

⎣⎢

⎤

⎦⎥

R[ ] = F[ ] U[ ]−1 = 2 20.5 1⎡

⎣⎢

⎤

⎦⎥

1.79 −1.34−1.34 1.56

⎡

⎣⎢

⎤

⎦⎥ =

0.89 0.45−0.45 0.89

⎡

⎣⎢

⎤

⎦⎥

U[ ] = 1.56 1.341.34 1.79⎡

⎣⎢

⎤

⎦⎥

First,symmetricallystretchtheunitcircleusing[U]

Second,rotatetheellipse(notthereferenceframe)using[R]

[F]=[R][U]

F[ ] X[ ]

U[ ] X[ ]

X[ ]

X[ ]U[ ] X[ ]

R[ ] U[ ] X[ ]

Example1 Eigenvaluesof[U]giveprincipalstretchmagnitudes

Eigenvectorsof[U]arealongaxesofblueellipses.Rotatedeigenvectorsof[U]giveprincipalstretchdirecMons

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F = 2 20.5 1

⎡

⎣⎢

⎤

⎦⎥

R = 0.89 0.45−0.45 0.89

⎡

⎣⎢

⎤

⎦⎥

′X[ ] = F[ ] X[ ]; F[ ] = V[ ] R[ ]

F[ ] = 2 20.5 1⎡

⎣⎢

⎤

⎦⎥; F[ ]T = 2 0.52 1

⎡

⎣⎢

⎤

⎦⎥

V[ ] = F[ ] F[ ]T⎡⎣ ⎤⎦1/2

= 8 33 1.5

⎡

⎣⎢

⎤

⎦⎥

1/2

= 2.68 0.890.89 0.67

⎡

⎣⎢

⎤

⎦⎥

R[ ] = V[ ]−1 F[ ] = 0.67 −0.89−0.89 2.68

⎡

⎣⎢

⎤

⎦⎥

2 20.5 1

⎡

⎣⎢

⎤

⎦⎥ =

0.89 0.45−0.45 0.89

⎡

⎣⎢

⎤

⎦⎥

First,rotatetheunitcircleusing[R]

Second,stretchtherotatedunitcirclesymmetricallyusing[V]

[F]=[V][R]

R = 0.89 0.45−0.45 0.89

⎡

⎣⎢

⎤

⎦⎥

F = 2 20.5 1

⎡

⎣⎢

⎤

⎦⎥

V[ ] = 2.68 0.890.89 0.67⎡

⎣⎢

⎤

⎦⎥

F[ ] X[ ] X[ ]

X[ ]R[ ] X[ ] R[ ] X[ ]V[ ] R[ ] X[ ]

Example2 Eigenvaluesof[V]alsogiveprincipalstretchmagnitudes

Unrotatedeigenvectorsof[V]giveprincipalstretchdirecMonsdirectly

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Example

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Example

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VI KeyresultsA ForsymmetricFmatrices(F=FT)

1  EigenvectorsofFgivedirecMonsofprincipalstretches2  EigenvectorsofFareperpendicular3 EigenvaluesofFgivemagnitudesofprincipalstretches4 Eigenvectorsdonotrotate

B Fornon-symmetricFmatrices(F≠FT)1  ThedirecMonsoftheprincipalstretchesaregivenbyrotatedeigenvectorsof[FTF]2  Eigenvectorsof[FTF]areperpendicular;eigenvectorsofFarenot3  Eigenvaluesof[FTF]givemagnitudesofprincipalquadraMcelongaMons4  FcanbedecomposedintoasymmetricstretchandrotaMon(orvice-versa)

a ThestretchmatrixU=[FTF]1/2b ThestretchmatrixV=[FFT]1/2

5 TherotaMonmatrixR=F[FTF]1/2=[FFT]1/2F

C NeedtoknowiniMallocaMonsandfinallocaMons,orF,tocalculatestrainsD TheF-matrixdoesnotuniquelydeterminethedisplacementhistory:e.g.,F=RU=VR

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Appendix1

Examplesoflong-handsoluMonsforeigenvaluesandeigenvectors


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CharacterisMcequaMon:|A-Iλ|=0

Eigenvaluesforsymmetric[A]

a

bc

d

e

A − Iλ = a − λ bc d − λ

= 2 − λ 11 2 − λ

= 0

a − λ( ) d − λ( )− bc = 2 − λ( ) 2 − λ( )− 1( ) 1( ) = 0λ2 − a + d( )λ + ad − bc( ) = 0

λ1,λ2 =a + d( ) ± a + d( )2 − 4 ad − bc( )

2

=2 + 2( ) ± 2 + 2( )2 − 4 2 × 2 −1×1( )

2= 2 ±1

tr(A)=(a+d)=4|A|=(ad-bc)=3

A = a bc d

⎡

⎣⎢

⎤

⎦⎥ =

2 11 2

⎡

⎣⎢

⎤

⎦⎥

tr A( ) = λ1 + λ2 = 4A = λ1λ2 = 3

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λ1 = 3, λ2 = 1


EigenvalueequaMon:AX=λX

Eigenvectorsforsymmetric[A]

a bc d

⎡

⎣⎢

⎤

⎦⎥

α1β1

⎡

⎣⎢⎢

⎤

⎦⎥⎥= λ1

α1β1

⎡

⎣⎢⎢

⎤

⎦⎥⎥⇒ 2 1

1 2⎡

⎣⎢

⎤

⎦⎥

α1β1

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

2α1 + β1α1 + 2β1

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 3

α1β1

⎡

⎣⎢⎢

⎤

⎦⎥⎥⇒ β1 =α1

a bc d

⎡

⎣⎢

⎤

⎦⎥

α 2β2

⎡

⎣⎢⎢

⎤

⎦⎥⎥= λ2

α 2β2

⎡

⎣⎢⎢

⎤

⎦⎥⎥⇒ 2 1

1 2⎡

⎣⎢

⎤

⎦⎥

α 2β2

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

2α 2 + β2α 2 + 2β2

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 1

α 2β2

⎡

⎣⎢⎢

⎤

⎦⎥⎥⇒ β2 = −α 2

A = a bc d

⎡

⎣⎢

⎤

⎦⎥ =

2 11 2

⎡

⎣⎢

⎤

⎦⎥

36GG3038/17/17

θ1 = tan−1 β1α1

= tan−1α1α1

= tan−1 11= 45!

θ2 = tan−1 β2α 2

= tan−1 −α 2α 2

= tan−1 −11

= −45!

Angleforeigenvector1


λ1

λ2

8/17/17

19


CharacterisMcequaMon:|A-Iλ|=0

Eigenvaluesfornon-symmetric[A]

a

bc

d

e

A − Iλ = a − λ bc d − λ

= 2 − λ 01 2 − λ

= 0

a − λ( ) d − λ( )− bc = 2 − λ( ) 2 − λ( )− 0( ) 1( ) = 0λ2 − a + d( )λ + ad − bc( ) = 0

λ1,λ2 =a + d( ) ± a + d( )2 − 4 ad − bc( )

2

=2 + 2( ) ± 2 + 2( )2 − 4 2 × 2 − 0 ×1( )

2= 2 ± 0

tr(A)=(a+d)=4|A|=(ad-bc)=4

A = a bc d

⎡

⎣⎢

⎤

⎦⎥ =

2 01 2

⎡

⎣⎢

⎤

⎦⎥

tr A( ) = λ1 + λ2 = 4A = λ1λ2 = 4

37GG3038/17/17

λ1 = 2, λ2 = 0


EigenvalueequaMon:AX=λX

Eigenvectorsfornon-symmetric[A]

a bc d

⎡

⎣⎢

⎤

⎦⎥

α1β1

⎡

⎣⎢⎢

⎤

⎦⎥⎥= λ1

α1β1

⎡

⎣⎢⎢

⎤

⎦⎥⎥⇒ 2 0

1 2⎡

⎣⎢

⎤

⎦⎥

α1β1

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

2α1α1 + 2β1

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 2

α1β1

⎡

⎣⎢⎢

⎤

⎦⎥⎥⇒α1 = 0

a bc d

⎡

⎣⎢

⎤

⎦⎥

α 2β2

⎡

⎣⎢⎢

⎤

⎦⎥⎥= λ2

α 2β2

⎡

⎣⎢⎢

⎤

⎦⎥⎥⇒ 2 0

1 2⎡

⎣⎢

⎤

⎦⎥

α 2β2

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

2α 2α 2 + 2β2

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 2

α 2β2

⎡

⎣⎢⎢

⎤

⎦⎥⎥⇒α 2 = 0

A = a bc d

⎡

⎣⎢

⎤

⎦⎥ =

2 01 2

⎡

⎣⎢

⎤

⎦⎥

38GG3038/17/17

θ1 = tan−1 β1α1

= tan−1 β10= tan−1∞ = ±90!

θ2 = tan−1 β2α 2

= tan−1 β20

= tan−1∞ = ±90!



λ1

λ2

8/17/17

20

Appendix2

ProofthatthevectorsλXarethelongestandshortestvectorsfromthecenterofanellipsetoitsperimeter


39GG3038/17/17

9.EIGENVECTORS,EIGENVALUES,ANDFINITESTRAINVI Eigenvectorsofasymmetricmatrix

A MaximumandminimumsquaredlengthsSetderivaMveofsquaredlengthstozerotofindorientaMonofmaximaandminimumdistancefromorigintoellipse

B PosiMonvectors(X’)withmaximumandminimum(squared)lengthsarethosethatareperpendiculartotangentvectors(dX’)alongellipse

!′X •!′X = Lf

2

d!′X •!′X( )

dθ=!′X • d!′X

dθ+ d!′X

dθ•!′X = 0

2!′X • d!′X

dθ⎛⎝⎜

⎞⎠⎟= 0

!′X • d!′X

dθ⎛⎝⎜

⎞⎠⎟= 0

40GG3038/17/17

8/17/17

21

9.EIGENVECTORS,EIGENVALUES,ANDFINITESTRAINVIEigenvectorsofa

symmetricmatrixCAX=λXD SinceeigenvectorsXof

symmetricmatricesaremutuallyperpendicular,sotooarethetransformedvectorsλX

E AtthepointidenMfiedbythetransformedvectorλX,theperpendiculareigenvector(s)mustparalleldX’andbetangenttotheellipse

41GG3038/17/17

9.EIGENVECTORS,EIGENVALUES,ANDFINITESTRAINVI Eigenvectorsofasymmetric

matrixF RecallthatposiMonvectors

(X’)withmaximumandminimum(squared)lengthsarethosethatareperpendiculartotangentvectors(dX’)alongellipse.Hence,thesmallestandlargesttransformedvectorsλXgivetheminimumandmaximumdistancestoanellipsefromitscenter.

G Theλvaluesaretheprincipalstretches

H Theseconclusionsextendtothreedimensionsandellipsoids

42GG3038/17/17

8/17/17

22

Appendix3

ProofthatdisMncteigenvectorsofarealsymmetricmatrixA=ATare

perpendicular


43GG3038/17/17


1a AX1=λ1X1 1b AX2=λ2X2EigenvectorsX1andX2parallelAX1andAX2,respecMvelyDo{ngAX1byX2andAX2byX1cantestwhetherX1andX2areorthogonal.

2a X2•AX1=X2•λ1X1=λ1(X2•X1)2b X1•AX2=X1•λ2X2=λ2(X1•X2)IfA=AT,thenthele|sidesof(2a)and(2b)areequal:3X2•AX1=AX1•X2=[AX1]T[X2]=[[X1]T[A]T][X2] =[X1]T[A][X2]=[X1]T[[A][X2]]=X1•AX2

8/17/17 GG303 44

8/17/17

23


Sincethele|sidesof(2a)and(2b)areequal,therightsidesmustbeequaltoo.Hence,

4  λ1(X2•X1)=λ2(X1•X2)Nowsubtracttherightsideof(4)fromthele|5  (λ1–λ2)(X2•X1)=0• Theeigenvaluesgenerallyaredifferent,soλ1–λ2≠0.• For(5)tohold,thenX2•X1=0.• Therefore,theeigenvectors(X1,X2)ofarealsymmetric2x2

matrixareperpendicularwhereeigenvaluesaredisMnct• Theeigenvectorscanbechosentobeperpendicularifthe

eignevactorsarethesame

8/17/17 GG303 45

Appendix4

RotaMonsinhomogenousdeformaMon:AnalgebraicperspecMve


46GG3038/17/17

8/17/17

24

9.EIGENVECTORS,EIGENVALUES,ANDFINITESTRAINVI RotaMonsinhomogeneousdeformaMon

A Justge{ngthesizeandshapeofthe“strain”(stretch)ellipseisnotenoughif[F]isnotsymmetric.Needtoconsiderhowpointsontheellipsetransform

B CandothisthroughacombinaMonofstretchesandrotaMons1 F=VR(which“R”?)

a V=symmetricstretchmatrixb R=rotaMonmatrix

2 F=RU(which“U”?“R”?)a R=rotaMonmatrixb U= symmetricstretchmatrix

3 Thechoicesbecomeuniqueforsymmetricstretchmatrices

8/17/17 GG303 47

9.EIGENVECTORS,EIGENVALUES,ANDFINITESTRAINVI RotaMonsinhomogeneous

deformaMonC Ifanellipseistransformedto

aunitcircle,theaxesoftheellipsearetransformedtoo.

D Ingeneral,theaxesoftheellipsesdonotmaintaintheirorientaMonwhentheellipseistransformedbacktoaunitcircle

E IfFisnotsymmetric,theaxesoftheredellipseandtheretro-deformed(black)axeswillhaveadifferentabsoluteorientaMon

F ThetransformaMonfromthetheretro-deformed(black)axestothetheorientaMonoftheprincipalaxesgivestherotaMonoftheaxes.

8/17/17 GG303 48

8/17/17

25


deformaMonG Weknowhowtofindthe

principalstretchmagnitudes:theyarethesquarerootsoftheeigenvaluesofthesymmetricmatrix[[FT][F]]

H Theeigenvectorsof[[FT][F]]givesomeoftheinformaMonneededtofindthedirecMonoftheprincipalstretchaxes.TherotaMondescribestheorientaMondifferencebetweenthe(red)principalstrain(stretch)axesandtheir(black)retro-deformedcounterparts

8/17/17 GG303 49


deformaMonI TofindtherotaMonofthe

principalaxes,startwiththeparametricequaMonforanellipseanditstangent,andtherequirementthattheposiMonvectorsforthesemi-axesoftheellipseareperpendiculartothetangent

LetθgivetheorientaMonofX,whereXtransformstoX’.

8/17/17 GG303 50

!′X = acosθ + bsinθ( )

!i + ccosθ + d sinθ( )

!j

d ′!Xdθ

= −asinθ + bcosθ( )!i + −csinθ + d cosθ( )

!j

′!X • d ′

!Xdθ

= 0WhatvalueofθwillyieldavectorXsuchthatX’willbeperpendiculartothetangentoftheellipse?

XisaposiMonvectorforaunitcircle.[X’]=[F][X].

8/17/17

26


deformaMonNowsolveforθsaMsfying

X’•dX’/dθ=0

8/17/17 GG303 51

!′X = acosθ + bsinθ( )

!i + ccosθ + d sinθ( )

!j

d ′!Xdθ

= −asinθ + bcosθ( )!i + −csinθ + d cosθ( )

!j

′!X • d ′

!Xdθ

= 0

= −a2 sinθ cosθ + abcos2θ − absin2θ + b2 sinθ cosθ− c2 sinθ cosθ + cd cos2θ − cd sin2θ + d 2 sinθ cosθ

= − a2 − b2 + c2 − d 2( )sinθ cosθ + ab + cd( )cos2θ − ab + cd( )sin2θ= − a2 − b2 + c2 − d 2( )sinθ cosθ + ab + cd( ) cos2θ − sin2θ( )=− a2 − b2 + c2 − d 2( )

2sin2θ + ab + cd( )cos2θ

=a2 − b2 + c2 − d 2( )

2sin −2θ( ) + ab + cd( )cos −2θ( ) = 0


deformaMon(Cont.)

8/17/17 GG303 52

a2 − b2 + c2 − d 2( )2

sin −2θ( ) + ab + cd( )cos −2θ( ) = 0

tan −2θ( ) = −2 ab + cd( )a2 − b2 + c2 − d 2

θ1 =12tan−1 2 ab + cd( )

a2 − b2 + c2 − d 2⎛⎝⎜

⎞⎠⎟, θ2 =

12tan−1 2 ab + cd( )

a2 − b2 + c2 − d 2⎛⎝⎜

⎞⎠⎟± 90!

Recallthattwoanglesthatdifferby180°havethesametangent

Soθ1andθ2are90°apart.SoX1andX2thattransformtoX1’andX2’areperpendicular.

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Eigenvectors, Eigenvalues, and Finite Strain · 2017. 8. 17. · 8/17/17 2 9. EIGENVECTORS,...

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