Continuing in this way we find unitary matrices U 3 U 4 U n 1 so that U n 1 U 2

Continuing in this way we find unitary matrices u 3 u

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Notice that if we take the adjoint of this equation, we getU*A*U=λ0000*λ20. . .0**.........**λnNow lets return to the case whereAis Hermitian. ThenA*=Aso that the matrices appearingin the previous two equations are equal. Thusλ****0λ2*. . .*00.........00λn=λ0000*λ20. . .0**.........**λnThis implies that all the entries denoted*must actually be zero. This also shows thatλi=λiforeveryi. In other words, Hermitian matrices can be diagonalized by a unitary matrix, and all theeigenvalues are real.IV.2.5. Powers and other functions of Hermitian matricesThe diagonalization formula for a Hermitian matrixAwith eigenvaluesλ1, . . . , λnand orthonormalbasis of eigenvectorsv1, . . .vncan be writtenA=UDU*=v1v2. . .vnλ10· · ·00λ2· · ·0............00· · ·λnvT1vT2...vTnThis means that for any vectorwAw=nXi=1λivivTiw.Since this is true for any vectorwit must be thatA=nXi=1λiPiwherePi=vivTi. The matricesPiare the orthogonal projections onto the one dimensional sub-spaces spanned by the eigenvectors. We havePiPi=PiandPiPj= 0 fori6=j.162
Recall that to compute powersAkwe simply need to replaceλiwithλkiin the diagonalizationformula. ThusAk=nXi=1λkiPi.This formula is valid for negative powerskprovided none of theλiare zero. In particular, whenk=-1 we obtain a formula for the inverse.In fact we can definef(A) for any functionfto be the matrixf(A) =nXi=1f(λi)Pi.IV.2.6. Effective resistance revisitedIn this section we will use its eigenvalues and eigenvectors of the Laplacian matrix to expressthe effective resistance between any two nodes of a resistor network.The LaplacianLis a realsymmetric matrix.The basic equation for a resistor network isLv=Jwhere the entries ofv= [v1, v2, . . . , vn]Tare the voltages the nodes, and the entries ofJ=[J1, J2, . . . , Jn] are the currents that are flowing in or out of each node.If we connect abvoltbattery to nodesiandjwe are fixing the voltage difference between these nodes to be the batteryvoltage so thatvi-vj=b.In this situation the there are no currents flowing in or out of the network except at nodesiandj, so except for these two nodesJhas all entries equal to zero. Since the current flowing in mustequal the current flowing out we haveJi=c,Jj=-cfor some value ofc. This can be writtenJ=c(ei-ej)whereeiandejare the standard basis vectors. The effective resistance between theith andjthnodes isR=b/c.This is the quantity that we wish to compute.Letλiandui,i-1. . . , nbe the eigenvalues and eigenvectors forL. SinceLis real symmetric,

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