GEOG4110/5100AdvancedRemoteSensing

Lecture16

1GEOG4110/5100

• Review:PrincipalComponentAnalysis• Displacementfromfeaturetracking

*Formoreinformationonworkingwithmatrices,refertoRichards,AppendixA)

MultispectralTransformationsofImageData

• Itispossibletotransformbrightnessdatathroughlinearoperationsonthesetofspectralbands– Canmakeimagefeaturesvisiblethatarenotdiscernableinthe

originaldata– Canpreserveimagequalityareducednumberoftransformed

dimensions• E.g.fordisplayonacolormonitor

• PrincipalComponenttransformation– Seekstominimizecorrelationinordertominimizeredundancyof

spectralbands

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http://www.ce.yildiz.edu.tr/personal/songul/file/1097/principal_components.pdf

forafairlysimpleexplanation:

PrincipalComponents• Seeknewcoordinatesysteminvectorspaceinwhichdatacan

berepresentedwithoutcorrelation– Covariancematrixisdiagonal

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y = Gx = Dtx

MeanVectorandCovariance

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Themeanvector(m)isthevectoraverageoftheindividualcomponentsofavector

Cov(X,Y ) =1

n 1(Xi x )(Yi y )

i=1

n

Thecovariancebetweentworeal-valuedrandomdescribeshowonevariablevariesinrelationtoanother.

C = 1n−1

(Xi − x )(Xi − x )i=1

n

∑xT

∑- -

RelationshipBetweenx andy CovarianceMatrices

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y = Gx = Dtx

Sy = ξ[(y-my)(y-my)t]

my = ξ[y] = ξ[Dtx] = Dtξ [x] = Dtmx

Sy = ξ[(Dtx-Dtmx)(Dtx-Dtmx)t]

Sy = Dtξ[(x-mx)(x-mx)t]D

Sy = DtSxDSx isthecovarianceofthepixeldatainxspace(Sy in y)

- mx andmy arethemeanvectorsinxandyrespectively

- ξ istheExpectedvalue(heretakenasthemeanform)

Each component of y is a linear combination of all of the elements of x; the weighting coefficients are the elements of the matrix G (or DT)

IdentifyingaycoordinatespaceinwhichthepixeldataexhibitsnocorrelationrequiresSy tobeadiagonalmatrix

EigenvaluesandEigenvectors

GEOG4110/5100 6

http://math.mit.edu/linearalgebra/ila0601.pdf (morecomplexexplanation)http://www.ce.yildiz.edu.tr/personal/songul/file/1097/principal_components.pdf (simpleexplanation)

Eigenvector

Eigenvalue

Whenwehaveatransformationmatrixoperatingonavector,anewvectorisproduced:

Sometimesthatnewvectorissimplytheproductofascalarandtheoriginalvector

Whenthisisthecase,thescalarisreferredtoastheEigenvalue,andthevectorisreferredtoastheEigenvector

Eigenvalues andEigenvectors• Eigenvalues (l)andeigenvectors(x)ofaMatrix(M)arescalar

andvectortermssuchthatthemultiplicationofx byl hasthesameresultasthematrixtransformationofx bymatrixM

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Mx = lx (i.e. y = lx is equivalent to y = Mx)or

Mx - lx = 0 à (M-lI)x =0; where I is the identity matrix(x is a vector with n elements, where n = number of bands)

For the above to be true, then either x = 0 or

|M-lI | = 0

This is the “characteristic equation” from which the eigenvalues(l) can be determinedWhen plugged into the equation: (M-lI)x =0, the eigenvectors(x) can be determined

CalculatingDeterminants

From:http://www.mathsisfun.com/algebra/matrix-determinant.html

PrincipalComponentTransformation

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Sy = DtSxD

- Sx isthecovarianceofthepixeldatainx space- DisamatrixofEigenvectorsderivedfromSx- Thecovariancematrixiny-spaceisgivenby:

Thenth component(n =1…N)representsz percentofthevariancewhere

- Sy is by definition a diagonal covariance matrix with its elements representing the variance in the transformed coordinates

- The greatest variance occurs in the first dimension of the transformed coordinate system, the next greatest in the 2nd, and so-on such that the least variance is found in the nth dimension

€

=y∑

λ1 0 00 λ2 0 0 0 λN

$

%

& & & &

'

(

) ) ) )

WhereNisthedimensionality,andlirepresentstheeigenvalues indescendingorder

€

ζ n =λn

λ1 + λ2...+ λn

PrincipalComponentTransformation

• Theeigenvectorsdeterminethetransformationmatrixthatproduceseachprincipalcomponent

• Theeigenvaluedescribesthepercentageofthevariancethatiscontainedwithineachprincipalcomponent– Thehighertheeigenvalueasafractionofthesumoftheeigenvalues,

themorerelativeinformationiscontainedinthecorrespondingprincipalcomponent

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PrincipalComponentsTransformationExamplein2dimensions

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€

=x∑1.9 1.11.1 1.1#

$ %

&

' (

€

=x∑2.40 00 1.87

#

$ %

&

' (

PrincipalComponentsTransformationExamplein2dimensions

GEOG4110/5100 12

€

=x∑1.9 1.11.1 1.1#

$ %

&

' (

Firstweneedtofindtheeigenvalues |Sx –lI| = 0

l2- 3.0l + 0.88 = 0 à l =2.67and0.33

€

1.9 − λ 1.11.1 1.1− λ

= 0

PrincipalComponentsTransformationExamplein2dimensions

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=y∑

2.67 00 0.33

"

#$

%

&'

Firstcomponentcontains2.67/(2.67+0.33)=89%ofthevarianceinthisexample(usuallyweordertheeigenvaluesindescendingorder)

NowweseektofindtheprincipalcomponentstransformationmatrixG=DT

WhereDT isthetransposedmatrixofeigenvectors.

Thefirsteigenvector(g1)correspondstothefirsteigenvalue l1

[Sx –lI]g1 = 0 with for the two dimensional case

Substituting Sx and l1 (2.67) gives the pair of equations:-0.77g11 + 1.10g21 = 01.10g11 – 1.57g21 = 0

yieldsg11=1.43g21

€

g1 =g11g21

"

# $

%

& ' = d1

t

PrincipalComponentsTransformationExamplein2dimensions

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Wehavetheaddedconstraintthattheeigenvectorsmustbenormalized(i.e.theGmatrixmustbeorthogonalsuchthatGt =G-1)

(g11)2 + (g21)2 =1

Thisproducesthefollowingeigenvectors

Whichinturnproducethefollowingtransformationmatrix€

g1 =0.820.57"

# $

%

& '

€

g2 =−0.570.82#

$ %

&

' (

G = Dt = 0.82 −0.570.57 0.82

"

#$

%

&'

t

= 0.82 0.57−0.57 0.82

"

#$

%

&'

Remember,Disthematrixofeigenvectors

g2 isthe2nd eigenvectorderivedfromthe2nd eigenvalue(replace2.67onpreviouspagewith0.33)

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PrincipalComponentsTransformation

Examplein2dimensions

(a)FourLandat MSSbansfortheregionofAndamooka inCentralAustralia;(b)Thefourprincipalcomponentsoftheimagesegment;(c)comparisonofstandardfalsecolorcomposite(R=band7;G=band5;B=band4)withaprincipalcomponentcomposite(R,G,Bare1st,2nd,and3rd componentsrespectively)

ba

c

a

b c d

e

17

Highlycorrelatedbands1,2,and3

a

b c d

e

18

Bands4,3,2 PC3,PC2,PC1 PC4,PC3,PC2

a

b c d

e

19

PrincipalComponentTransformationSteps1. Computethecovariancematrixofthedatasetinvectorspace2. Calculatetheeigenvaluesofthecovariancematrix3. Thediagonalmatrixwiththeeigenvaluesalongthediagonalwillbethe

covariancematrixofthetransformedaxes(principalcomponentaxes)4. Findthematrixofeigenvectors(Di)foreach individual l of interest by

solvingfor[Sx –liI]gi = 0. for that l.5. Transpose the Matrix D to produce principal component transformation

matrix (g). The number of rows in g will equal the number of spectral dimensions from which the eigenvalues and eigenvectors were calculated

6. For each g matrix (derived from a given l) the original data values (in original x coordinate system) are multiplied by the rows in g (g1, g2, … gnwhere n is the number of dimensions in vector space), to produce coordinates in the transformed dimension (new y coordinate system). Each axis in the original spectral space will be multiplied by its corresponding row in the g matrix to produce the transformed coordinate system (principal component)

7. Steps 4 – 6 are repeated until the desired number of principal component transformations have been executed.

• PCAisatechniquethattransformstheoriginalvectorimagedataintosmallersetofuncorrelatedvariables.

• Thevariablesrepresentmostoftheimageinformationandeasiertointerpret.

• PrincipalcomponentsarederivedsuchthatthefirstPCaccountsformuchofthevariationoftheoriginaldata.Thesecond(vertical)accountsformostoftheremainingvariation.

• PCAisusefulinreducingthedimensionality(numberofbands)thatusedforanalysis.Minimumnoisefraction(MNF)methodcanbeusedwithhyperspectral datafornoisereduction.

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PrincipalComponentsAnalysis(PCA)

TMExampleforPCTransformation

• Computethen-dimensionalcovariancematrix(7x7forLandsat TM).

• Thevariancesoftheprincipalcomponents(eigenvalues)containusefulinformation(e.g.determinethe%oftotalvarianceexplainedbyeachoftheprincipalcomponents)

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BandNumber

1 2 3 4 5 7 6

Variance 100.93 34.14 68.83 248.40 568.84 154.92 17.78

TableshowsthevarianceofdifferentbandsofTMscene.AdaptedfromJensen,2005.

eigenvalueofthepth component

sumoftheeigenvalueofallcomponents

Componentp(eigenvalues)

1 2 3 4 5 6 7

eigenvalue 1010.92 131.20 37.60 6.73 3.95 2.17 1.24

23

TMExampleforPCTransformation

Sumofeigenvaluesofallcomponents=1193.81%ofvarianceexplainedbyPC1=(1010.92/1193.81)*100=84.68%%ofvarianceexplainedbyPC2=(131.2/1193.81)*100=10.99%

Tableshowsthevarianceofdifferentprincipalcomponents.AdaptedfromJensen,2005.

Band Componentp

1 2 3 4 5 6 7

1 0.205 0.637 0.327 -0.054 0.249 -0.611 -0.079

2 0.127 0.342 0.169 -0.077 0.012 0.396 0.821

3 0.204 0.428 0.159 -0.076 -0.075 0.649 -0.562

4 0.443 -0.471 0.739 0.107 -0.153 -0.019 -0.004

5 0.742 -0.177 -0.437 -0.300 0.370 0.007 0.011

7 0.376 0.197 -0.309 -0.312 -0.769 -0.181 0.051

6 0.106 0.033 -0.080 0.887 0.424 0.122 0.005

Tableshowstheeigenvectors(coefficients)foreachprincipalcomponentineachcolumn.AdaptedfromJensen,2005.

• Thecorrelationofeachbandwitheachcomponentpiscalculatedtodeterminewhichbandisassociatedwitheachprincipalcomponent.Thishelpsinunderstandingtheinformationcontainsbyeachcomponent.

Howprincipalcomponentimagesarecreated?• Identifytheoriginalbrightnessvaluesofagivenpixel(e.g.the

firstpixelatcolumn1androw1).• Obtainthenewpixelvaluebysummationofthe

multiplicationoftheeigenvectorofthecomponentofeachbandbytheoriginalvalue

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TMExampleforPCTransformation