# Lecture 10: Eigenvectors and eigenvalues - pha.jhu. neufeld/numerical/ 10: Eigenvectors and...

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### Transcript of Lecture 10: Eigenvectors and eigenvalues - pha.jhu. neufeld/numerical/ 10: Eigenvectors and...

220

Lecture 10:Eigenvectors and eigenvalues

(Numerical Recipes, Chapter 11)

The eigenvalue problem,

A x = x,

occurs in many, many contexts:

classical mechanics, quantum mechanics, optics

221

Eigenvectors and eigenvalues(Numerical Recipes, Chapter 11)

The textbook solution is obtained from

(A I) x = 0, which can only hold for x 0 if

det (A I) = 0.

This is a polynomial equation for of order N

222

Eigenvectors and eigenvalues(Numerical Recipes, Chapter 11)

This method of solution is:

(1) very slow(2) inaccurate, in the presence of roundoff error(3) yields only eigenvalues, not eigenvectors

(4) of no practical value (except, perhaps in the 2 x 2 case where the solution to the resultant quadratic can be written in closed form)

223

Review of some basic definitions in linear algebra

A matrix is said to be:

Symmetric if A = AT (its transpose) i.e. if aij = aji

Hermitian if A = A (AT)* i.e. if aij = aji*(its Hermitian conjugate)

Orthonormal if U1 = UT

Unitary if U1 = U

224

Review of some basic definitions in linear algebra

All 4 types are normal, meaning that they obey the relations

A A = A AU U = U U

225

Eigenvectors and eigenvalues

Hermitian matrices are of particular interest because they have:

real eigenvaluesorthonormal eigenvectors (xi xj = ij )(unless some eigenvalues are degenerate, in which case we can always design an orthonormalset)

226

Diagonalization of Hermitian matrices

Hermitian matrices can be diagonalizedaccording to

A = U D U1

Diagonal matrix consisting of the eigenvalues

Unitary matrix whose column are the eigenvectors

227

Diagonalization of Hermitian matrices

Clearly, if A = U D U1, then A U = U D..which is simply the eigenproblem(written out N times):

The columns of U are the eigenvectors (mutually orthogonal) and the elements of D (non-zero only on the diagonal) are the corresponding eigenvalues.

228

Non-Hermitian matrices

We are often interested in non-symmetric real matrices:

the eigenvalues are real or come in complex-conjugate pairs

the eigenvectors are not orthonormal in general

the eigenvectors may not even span an N-dimensional space

229

Diagonalization of non-Hermitian matrices

For a non-Hermitian matrix, we can identify two (different) types of eigenvectors

Right hand eigenvectors are column vectors which obey:A xR = xR

Left hand eigenvectors are row vectors which obey:xL A = xL

The eigenvalues are the same (the roots of det (A I) = 0 in both cases) but, in general, the eigenvectors are not

230

Diagonalization of non-Hermitian matrices

Note: for a Hermitian matrix, the two types of eigenvectors are equivalent, since

A xR = xR since A = A since is real

xR A = xR xR A = xR

so if any vector xR is a right-hand eigenvector, then xR is a left-hand eigenvector

231

Diagonalization of non-Hermitian matrices

Let D be the diagonal matrix whose elements are the eigenvalues, D = diag (i)

Let XR be the matrix whose columns are the right-hand eigenvectors

Let XL be the matrix whose rows are the left-hand eigenvectors

Then the eigenvalue equations are

A XR = XR D and XL A = D XL

232

Diagonalization of non-Hermitian matrices

The eigenvalue equations

A XR = XR D and XL A = D XL

imply XLA XR = XL XR D and XL A XR = D XL XR

Hence, XL A XR = XL XR D = D XL XRXL XR commutes with DXL XR is also diagonalrows of XL are orthogonal to columns of XR

233

Diagonalization of non-Hermitian matrices

It follows that any matrix can be diagonalizedaccording to

D = XL A XR = XR1 A XR , or equivalently

D = XL A XL1

The special feature of a Hermitian matrix is that the XL and XR are unitary

234

Solving the eigenvalue equation

The problem is entirely equivalent to figuring out how to diagonalize of A according to

A = XR D XR1

All methods proceed by making a series of similarity transformationA = P11 M1 P1 = P11 P21 M2 P2 P1 = = XR D XR1

where M1, M2 are successively more nearly diagonal

235

Solving the eigenvalue equation

There are many routines available:EISPACK + Linpack LAPACK (free)NAG, IMSL (expensive)

Before using a totally general technique, consider what you want:eigenvalues alone, or eigenvectors as well?

..and how special your case isreal symmetric & tridiagonal, real symmetric, real non-symmetric, complex Hermitian, complex non-Hermitian

236

Solution for a real symmetric matrix: the Householder method (Recipes, 11.2)

The Householder method makes use of a similarity transform based upon

P = I 2 w wT,

where w is a column vector for which wT w = 1 and w wT is an N x N symmetric matrixi.e. (w wT)ij = (w wT)ji = wi wj

237

The Householder method

The Householder matrix, P, clearly has the following properties

P is symmetric (being the difference between 2 symmetric matrices, I and 2wwT)

P is orthogonal:PT P = (I 2 w wT)T (I 2 w wT)

= (I 4 w wT + 4 w wT w wT ) = I= 1

238

The Householder method

Let a1 be the first column of A

Let w = a1 |a1| e1| a1 |a1| e1|

where e1 = (1,0,0,0,00)T

239

The Householder method

Lets consider the action of P = I 2 w wT upon the first column of AP a1 ={ I 2 (a1 |a1| e1 ) (a1 |a1| e1 )T } a1

|a1 |a1| e1 |2

= a1 2 (a1 |a1| e1 ) (|a1|2 |a1| e1Ta1)|a1|2 2 |a1| e1. a1 + |a1|2

= |a1| e1

P zeroes out all but the first element of a1

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The Householder methodSuppose A is symmetric

A = P A =

M1 = P A P1 = P A PT =

XXXXXXX

XXXXXXX

XXXXXXX

XXXXXXX

XXXXXXX

XXXXXXX

XXXXXXX

XXXXXX0

XXXXXX0

XXXXXX0

XXXXXX0

XXXXXX0

XXXXXX0

XXXXXXX

XXXXXX0

XXXXXX0

XXXXXX0

XXXXXX0

XXXXXX0

XXXXXX0

000000X

241

The Householder methodSuppose A is symmetric

A = P A =

M1 = P A P1 = P A PT =

XXXXXXX

XXXXXXX

XXXXXXX

XXXXXXX

XXXXXXX

XXXXXXX

XXXXXXX

XXXXXX0

XXXXXX0

XXXXXX0

XXXXXX0

XXXXXX0

XXXXXX0

XXXXXXX

XXXXXX0

XXXXXX0

XXXXXX0

XXXXXX0

XXXXXX0

XXXXXX0

000000X

242

The Householder methodNext step, choose new Householder matrix of the form

P = to operate on M1 =

M2 = P M1 P1 =

XXXXXX0

XXXXXX0

XXXXXX0

XXXXXX0

XXXXXX0

XXXXXX0

000000X

XXXXX00

XXXXX00

XXXXX00

XXXXX00

XXXXX00

00000XX

00000XX

0

0

0

0

0

0

0000001

IN1 2 wN1wN1T

243

The Householder methodAfter N Householder transformations, the matrix is tridiagonal

MN = = PNP2P1 A P11P21 PN1

XX00000

XXX0000

0XXX000

00XXX00

000XXX0

0000XXX

00000XX

244

The Householder method: computational cost

You might think that each Householder transformation would require O(N3) operations (2 matrix multiplications), so the whole series would cost ~ O(N4) operations

However, we can rewrite the product P A P1 asfollows P A P1 = P A P = (I 2 w wT) A (I 2 w wT)

= (I 2 w wT) (A 2 A w wT) = (A2 w wTA 2 A w wT + 4 w wTAw wT)

245

The Householder method: computational cost

Now define an N x 1 vector v = A w

P1A P = (A 2wwTA 2A w wT + 4 w wT A w wT) = (A 2 w vT 2 v wT + 4 w wTv wT )

This calculation involves an N2 operation to compute v, and a series of N2 operations to compute, subtract and add various matrices

The reduction of the matrix to triadiagonal form therefore costs only O(N3) operations, not O(N4)

scalar

246

Diagonalization of a symmetric tridiagonal matrix (Recipes, 11.3)

The diagonalization of a symmetric tridiagonalmatrix can be accomplished by a series of ~N 2 x 2 rotations

Consider the rotation

R =

1000000

0100000

0010000

0001000

0000100

00000C-S

00000SC

(rotation through angle = cos1 C = sin1 S inthe x1 - x2 plane)

247

Diagonalization of a symmetric tridiagonal matrix

Applying this similarity transform to the triadiagonal matrix, we obtain

RT R =

XX00000

XXX0000

0XXX000

00XXX00

000XXX0

0000Xvu

00000ut

XX00000

XXX0000

0XXX000

00XXX00

000XXX0

0000XX0

000000X

248

Diagonalization of a symmetric tridiagonal matrix

To make the off-diagonal elements (RT MN R)12 = (RT MN R)21 zero, we require

tSC + u(C2 S2) vSC = 0

tan = (t v) (t v)2 + 4u2

2u

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