Numerical Methods for Solving Large Scale Eigenvalue Problems

Peter Arbenz Computer Science Department, ETH Zürich E-mail: arbenz@inf.ethz.ch arge scale eigenvalue problems, Lecture 2, February 28, 2018 1/46 Numerical Methods for Solving Large Scale Eigenvalue Problems Lecture 2, February 28, 2018: Numerical linear algebra basics http://people.inf.ethz.ch/arbenz/ewp/

arge scale eigenvalue problems, Lecture 2, February 28, 2018 2/46 Survey on lecture 1. Introduction 2. Numerical linear algebra basics Definitions Similarity transformations Schur decompositions SVD 3. Newtons method for linear and nonlinear eigenvalue problems 4. The QR Algorithm for dense eigenvalue problems 5. Vector iteration (power method) and subspace iterations 6. Krylov subspaces methods Arnoldi and Lanczos algorithms Krylov-Schur methods 7. Davidson/Jacobi-Davidson methods 8. Rayleigh quotient minimization for symmetric systems 9. Locally-optimal block preconditioned conjugate gradient (LOBPCG) method

arge scale eigenvalue problems, Lecture 2, February 28, 2018 3/46 Survey on lecture Notation Statement of the problem Similarity transformations Schur decomposition The real Schur decomposition Hermitian matrices Jordan normal form Projections The singular value decomposition (SVD)

Large scale eigenvalue problems, Lecture 2, February 28, 2018 4/46 Survey on lecture Literature G. H. Golub and C. F. van Loan. Matrix Computations, 4th edition. Johns Hopkins University Press. Baltimore, 2012. R. A. Horn and C. R. Johnson, Matrix Analysis, Cambridge University Press, Cambridge, 1985. Y. Saad, Numerical Methods for Large Eigenvalue Problems, SIAM, Philadelphia, PA, 2011. E. Anderson et al. LAPACK Users Guide, 3rd edition. SIAM, Philadelphia, 1999. http://www.netlib.org/lapack/

Large scale eigenvalue problems, Lecture 2, February 28, 2018 5/46 Notation Notations R: The field of real numbers C: The field of complex numbers R n : The space of vectors of n real components C n : The space of vectors of n complex components Scalars : lowercase letters, a, b, c..., and α, β, γ.... Vectors : boldface lowercase letters, a, b, c,.... x 1 x R n x 2 x =., x i R. x n We often make statements that hold for real or complex vectors. x F n.

Large scale eigenvalue problems, Lecture 2, February 28, 2018 6/46 Notation The inner product of two n-vectors in C: (x, y) = n x i ȳ i = y x, i=1 y = (ȳ 1, ȳ 2,..., ȳ n ): conjugate transposition of complex vectors. x and y are orthogonal, x y, if x y = 0. Norm in F, (Euclidean norm or 2-norm) x = ( n ) 1/2 (x, x) = x i 2. i=1

arge scale eigenvalue problems, Lecture 2, February 28, 2018 7/46 Notation a 11 a 12... a 1n A F m n a 21 a 22... a 2n A =..., a ij F. a m1 a m2... a mn ā 11 ā 21... ā m1 A F n m A ā 12 ā 22... ā m2 =... ā 1n ā 2n... ā nm is the Hermitian transpose of A. For square matrices: A F n n is called Hermitian A = A. Real Hermitian matrix is called symmetric. U F n n is called unitary U 1 = U. Real unitary matrices are called orthogonal. A F n n is called normal A A = AA. Both, Hermitian and unitary matrices are normal.

Large scale eigenvalue problems, Lecture 2, February 28, 2018 8/46 Notation Norm of a matrix (matrix norm induced by vector norm): A := max x 0 Ax x = max x =1 Ax. The condition number of a nonsingular matrix: κ(a) = A A 1. U unitary = Ux = x for all x = κ(u) = 1.

arge scale eigenvalue problems, Lecture 2, February 28, 2018 9/46 Statement of the problem The (standard) eigenvalue problem: Given a square matrix A F n n. Find scalars λ C and vectors x C n, x 0, such that Ax = λx, (1) i.e., such that (A λi )x = 0 (2) has a nontrivial (nonzero) solution. We are looking for numbers λ such that A λi is singular. The pair (λ, x) be a solution of (1) or (2). λ is called an eigenvalue of A, x is called an eigenvector corresponding to λ

Statement of the problem (λ, x) is called eigenpair of A. The set σ(a) of all eigenvalues of A is called spectrum of A. The set of all eigenvectors corresponding to an eigenvalue λ together with the vector 0 form a linear subspace of C n called the eigenspace of λ. The eigenspace of λ is the null space of λi A: N (λi A). The dimension of N (λi A) is called geometric multiplicity g(λ) of λ. An eigenvalue λ is a root of the characteristic polynomial χ(λ) := det(λi A) = λ n + a n 1 λ n 1 + + a 0. The multiplicity of λ as a root of χ is called the algebraic multiplicity m(λ) of λ. 1 g(λ) m(λ) n, λ σ(a), A F n n. arge scale eigenvalue problems, Lecture 2, February 28, 2018 10/46

arge scale eigenvalue problems, Lecture 2, February 28, 2018 11/46 Statement of the problem y is called left eigenvector corresponding to λ y A = λy Left eigenvector of A is a right eigenvector of A, corresponding to the eigenvalue λ, A y = λy. A is an upper triangular matrix, a 11 a 12... a 1n a 22... a 2n A =...., a ik = 0 for i > k. det(λi A) = a nn n (λ a ii ). i=1

Large scale eigenvalue problems, Lecture 2, February 28, 2018 12/46 Statement of the problem (Generalized) eigenvalue problem Given two square matrices A, B F n n. Find scalars λ C and vectors x C, x 0, such that or, equivalently, such that has a nontrivial solution. The pair (λ, x) is a solution of (3) or (4). λ is called an eigenvalue of A relative to B, Ax = λbx, (3) (A λb)x = 0 (4) x is called an eigenvector of A relative to B corresponding to λ. (λ, x) is called an eigenpair of A relative to B, The set σ(a; B) of all eigenvalues of (3) is called the spectrum of A relative to B.

Large scale eigenvalue problems, Lecture 2, February 28, 2018 13/46 Similarity transformations Similarity transformations Matrix A is similar to a matrix C, A C, there is a nonsingular matrix S such that S 1 AS = C. (5) The mapping A S 1 AS is called a similarity transformation. Theorem Similar matrices have equal eigenvalues with equal multiplicities. If (λ, x) is an eigenpair of A and C = S 1 AS then (λ, S 1 x) is an eigenpair of C.

arge scale eigenvalue problems, Lecture 2, February 28, 2018 14/46 Similarity transformations Similarity transformations (cont.) Proof: Ax = λx and C = S 1 AS = CS 1 x = S 1 ASS 1 x = λs 1 x Hence A and C have equal eigenvalues and their geometric multiplicity is not changed by the similarity transformation. det(λi C) = det(λs 1 S S 1 AS) = det(s 1 (λi A)S) = det(s 1 ) det(λi A) det(s) = det(λi A) the characteristic polynomials of A and C are equal and hence also the algebraic eigenvalue multiplicities are equal.

Large scale eigenvalue problems, Lecture 2, February 28, 2018 15/46 Similarity transformations Unitary similarity transformations Two matrices A and C are called unitarily similar (orthogonally similar) if S (C = S 1 AS = S AS) is unitary (orthogonal). Reasons for the importance of unitary similarity transformations: 1. U is unitary U = U 1 = 1 κ(u) = 1. Hence, if C = U 1 AU C = U AU and C = A. If A is disturbed by δa ( roundoff errors introduced when storing the entries of A in finite-precision arithmetic) U (A + δa)u = C + δc, δc = δa. Hence, errors (perturbations) in A are not amplified by a unitary similarity transformation. This is in contrast to arbitrary similarity transformations.

arge scale eigenvalue problems, Lecture 2, February 28, 2018 16/46 Similarity transformations Unitary similarity transformations (cont.) 2. Preservation of symmetry: If A is symmetric A = A, U 1 = U : C = U 1 AU = U AU = C 3. For generalized eigenvalue problems, similarity transformations are not so crucial since we can operate with different matrices from both sides. If S and T are nonsingular Ax = λbx TAS 1 Sx = λtbs 1 Sx. This is called equivalence transformation of A, B. σ(a; B) = σ(tas 1, TBS 1 ). Special Case: B is invertible & B = LU is LU-factorization of B. Set S = U and T = L 1 TBU 1 = L 1 LUU 1 = I σ(a; B) = σ(l 1 AU 1, I ) = σ(l 1 AU 1 ).

Large scale eigenvalue problems, Lecture 2, February 28, 2018 17/46 Schur decomposition Schur decomposition Theorem (Schur decomposition) If A C n n then there is a unitary matrix U C n n such that U AU = T (6) is upper triangular. The diagonal elements of T are the eigenvalues of A. Proof: By induction: 1. For n = 1, the theorem is obviously true. 2. Assume that the theorem holds for matrices of order n 1.

Large scale eigenvalue problems, Lecture 2, February 28, 2018 18/46 Schur decomposition Schur decomposition (cont.) 3. Let (λ, x), x = 1, be an eigenpair of A, Ax = λx. Construct a unitary matrix U 1 with first column x (e.g. the Householder reflector U 1 with U 1 x = e 1 ). Partition U 1 = [x, U]. Then [ x U1 Ax x ] [ ] AU λ AU 1 = U Ax U = AU 0 Â as Ax = λx and U x = 0 by construction of U 1. By assumption, there exists a unitary matrix Û C (n 1) (n 1) such that Û ÂÛ = ˆT is upper triangular. Setting U := U 1 (1 Û), we obtain (6).

Schur decomposition Schur vectors U = [u 1, u 2,..., u n ] U AU = T is a Schur decomposition of A AU = UT. The k-th column of this equation is k 1 Au k = λu k + t ik u i, λ k = t kk. (7) i=1 = Au k span{u 1,..., u k }, k. The first k Schur vectors u 1,..., u k form an invariant subspace for A. (A subspace V F n is called invariant for A if AV V.) From (7): the first Schur vector is an eigenvector of A. The other columns of U, are in general not eigenvectors of A. The Schur decomposition is not unique.the eigenvalues can be arranged in any order in the diagonal of T. Large scale eigenvalue problems, Lecture 2, February 28, 2018 19/46

Large scale eigenvalue problems, Lecture 2, February 28, 2018 20/46 The real Schur decomposition The real Schur decomposition * Real matrices can have complex eigenvalues. If complex eigenvalues exist, then they occur in complex conjugate pairs! Theorem (Real Schur decomposition) If A R n n then there is an orthogonal matrix Q R n n such that R 11 R 12 R 1m Q T R 22 R 2m AQ =...... (8) R mm is upper quasi-triangular. The diagonal blocks R ii are either 1 1 or 2 2 matrices. A 1 1 block corresponds to a real eigenvalue, a 2 2 block corresponds to a pair of complex conjugate eigenvalues.

Large scale eigenvalue problems, Lecture 2, February 28, 2018 21/46 The real Schur decomposition The real Schur decomposition (cont.) Remark: The matrix [ ] α β, α, β R, β α has the eigenvalues α + iβ and α iβ. Let λ = α + iβ, β 0, be an eigenvalue of A with eigenvector x = u + iv. Then λ = α iβ is an eigenvalue corresponding to x = u iv. Ax = A(u + iv) = Au + iav, λx = (α + iβ)(u + iv) = (αu βv) + i(βu + αv). A x = A(u iv) = Au iav, = (αu βv) i(βu + αv) = (α iβ)u i(α iβ)v = (α iβ)(u iv) = λ x.

Large scale eigenvalue problems, Lecture 2, February 28, 2018 22/46 The real Schur decomposition The real Schur decomposition (cont.) k: the number of complex conjugate pairs. Now, let s prove the theorem by induction on k. Proof: First k = 0. In this case, A has real eigenvalues and eigenvectors. We can repeat the proof of the Schur decomposition in real arithmetic to get the decomposition (U AU = T ) with U R n n and T R n n. So, there are n diagonal blocks R jj all of which are 1 1. R 11 R 12 R 1m R 22 R 2m...... R mm

arge scale eigenvalue problems, Lecture 2, February 28, 2018 23/46 The real Schur decomposition The real Schur decomposition (cont.) Assume that the theorem is true for all matrices with fewer than k complex conjugate pairs. Then, with λ = α + iβ, β 0 and x = u + iv, [ ] α β A[u, v] = [u, v]. β α Let {x 1, x 2 } be an orthonormal basis of span{[u, v]}. Then, since u and v are linearly independent (If u and v were linearly dependent then it follows that β must be zero.), there is a nonsingular 2 2 real square matrix C with [x 1, x 2 ] = [u, v]c.

The real Schur decomposition The real Schur decomposition (cont.) [ ] α β A[x 1, x 2 ] = A[u, v]c = [u, v] C β α [ ] = [x 1, x 2 ]C 1 α β C =: [x β α 1, x 2 ]S. [ ] α β S and are similar and therefore have equal β α eigenvalues. Now, construct an orthogonal matrix [x 1, x 2, x 3,..., x n ] =: [x 1, x 2, W ]. [ [x1, x 2 ], W ] T [ A [x1, x 2 ], W ] x T 1 = x T [ 2 [x 1, x 2 ]S, AW ] W T [ S [x1, x = 2 ] T ] AW O W T. AW Large scale eigenvalue problems, Lecture 2, February 28, 2018 24/46

Large scale eigenvalue problems, Lecture 2, February 28, 2018 25/46 The real Schur decomposition The real Schur decomposition (cont.) The matrix W T AW has less than k complex-conjugate eigenvalue pairs. Therefore, by the induction assumption, there is an orthogonal Q 2 R (n 2) (n 2) such that the matrix Q T 2 (W T AW )Q 2 is quasi-triangular. Thus, the orthogonal matrix ( ) I2 O Q = [x 1, x 2, x 3,..., x n ] O Q 2 transforms A similarly to quasi-triangular form.

Large scale eigenvalue problems, Lecture 2, February 28, 2018 26/46 Hermitian matrices Hermitian matrices Matrix A F n n is Hermitian if A = A. In the Schur decomposition A = UΛU for Hermitian matrices the upper triangular Λ is Hermitian and therefore diagonal. Λ = Λ = (U AU) = U A U = U AU = Λ, each diagonal element λ i of Λ satisfies λ i = λ i = Λ must be real. Hermitian/symmetric matrix is called positive definite (positive semi-definite) if all its eigenvalues are positive (nonnegative). HPD or SPD = Cholesky factorization exists.

Hermitian matrices Spectral decomposition Theorem (Spectral theorem for Hermitian matrices) Let A be Hermitian. Then there is a unitary matrix U and a real diagonal matrix Λ such that A = UΛU = n λ i u i u i. (9) i=1 The columns u 1,..., u n of U are eigenvectors corresponding to the eigenvalues λ 1,..., λ n. They form an orthonormal basis for F n. The decomposition (9) is called a spectral decomposition of A. As the eigenvalues are real we can sort them with respect to their magnitude. We can, e.g., arrange them in ascending order such that λ 1 λ 2 λ n. Large scale eigenvalue problems, Lecture 2, February 28, 2018 27/46

Large scale eigenvalue problems, Lecture 2, February 28, 2018 28/46 Hermitian matrices If λ i = λ j, then any nonzero linear combination of u i and u j is an eigenvector corresponding to λ i, A(u i α + u j β) = u i λ i α + u j λ j β = (u i α + u j β)λ i. Eigenvectors corresponding to different eigenvalues are orthogonal. Au = uλ and Av = vµ, λ µ. and thus λu v = (u A)v = u (Av) = u vµ, (λ µ)u v = 0, from which we deduce u v = 0 as λ µ.

Hermitian matrices Eigenspace The eigenvectors corresponding to a particular eigenvalue λ form a subspace, the eigenspace {x F n, Ax = λx} = N (A λi ). They are perpendicular to the eigenvectors corresponding to all the other eigenvalues. Therefore, the spectral decomposition is unique up to ± signs if all the eigenvalues of A are distinct. In case of multiple eigenvalues, we are free to choose any orthonormal basis for the corresponding eigenspace. Remark: The notion of Hermitian or symmetric has a wider background. Let x, y be an inner product on F n. Then a matrix A is symmetric with respect to this inner product if Ax, y = x, Ay for all vectors x and y. All the properties of Hermitian matrices hold similarly for matrices symmetric with respect to a certain inner product. Large scale eigenvalue problems, Lecture 2, February 28, 2018 29/46

Hermitian matrices Matrix polynomials p(λ): polynomial of degree d, p(λ) = α 0 + α 1 λ + α 2 λ 2 + + α d λ d. A j = (UΛU ) j = UΛ j U Matrix polynomial: d p(a) = α j A j = j=0 d d α j UΛ j U = U α j Λ j U. j=0 j=0 This equation shows that p(a) has the same eigenvectors as the original matrix A. The eigenvalues are modified though, λ k becomes p(λ k ). More complicated functions of A can be computed if the function is defined on the spectrum of A. Large scale eigenvalue problems, Lecture 2, February 28, 2018 30/46

Large scale eigenvalue problems, Lecture 2, February 28, 2018 31/46 Jordan normal form Theorem (Jordan normal form) For every A F n n there is a nonsingular matrix X F n n such that X 1 AX = J = diag(j 1, J 2,..., J p ), where λ k 1. J k = J mk (λ k ) = λ.. k... 1 Fm k m k λk are called Jordan blocks and m 1 + + m p = n. The values λ k need not be distinct. The Jordan matrix J is unique up to the ordering of the blocks. The transformation matrix X is not unique.

arge scale eigenvalue problems, Lecture 2, February 28, 2018 32/46 Jordan normal form Jordan normal form Matrix diagonalizable all Jordan blocks are 1 1 (trivial). In this case the columns of X are eigenvectors of A. One eigenvector associated with each Jordan block [ ] [ ] λ 1 1 J 2 (λ)e 1 = = λ e 0 λ 0 1. Nontrivial blocks give rise to so-called generalized eigenvectors e 2,..., e mk since (J k (λ) λi )e j+1 = e j, j = 1,..., m k 1. Computation of Jordan blocks is unstable.

arge scale eigenvalue problems, Lecture 2, February 28, 2018 33/46 Jordan normal form Jordan normal form (cont.) Let Y := X and let X = [X 1, X 2,..., X p ] and Y = [Y 1, Y 2,..., Y p ] be partitioned according to J. Then, A = XJY = = p X k J k Yk = k=1 p (λ k P k + D k ), k=1 p (λ k X k Yk + X kn k Yk ) k=1 where N k = J mk (0), P k := X k Y k, D k := X k N k Y k. Since P 2 k = P k, P k is a projector on R(P k ) = R(X k ). It is called a spectral projector.

Large scale eigenvalue problems, Lecture 2, February 28, 2018 34/46 Projections Projections A matrix P that satisfies P 2 = P is called a projection. A projection is a square matrix. If P is a projection then Px = x for all x in the range R(P) of P. In fact, if x R(P) then x = Py for some y F n and Px = P(Py) = P 2 y = Py = x. x 2 x 1

Large scale eigenvalue problems, Lecture 2, February 28, 2018 35/46 Projections Projections (cont.) Example: Let P = ( 1 2 0 0 The range of P is R(P) = span{e 1 }. The effect of P is depicted in the figure of the previous page: All points x that lie on a line parallel to span{(2, 1) } are mapped on the same point on the x 1 axis. So, the projection is along span{(2, 1) } which is the null space N (P) of P. ). If P is a projection then also I P is a projection. If Px = 0 then (I P)x = x. = range of I P equals the null space of P: R(I P) = N (P). It can be shown that R(P) = N (P ).

Large scale eigenvalue problems, Lecture 2, February 28, 2018 36/46 Projections Projections (cont.) Notice that R(P) R(I P) = N (I P) N (P) = {0}. So, any vector x can be uniquely decomposed into x = x 1 + x 2, x 1 R(P), x 2 R(I P) = N (P). The most interesting situation occurs if the decomposition is orthogonal, i.e., if x 1 x 2 = 0 for all x. A matrix P is called an orthogonal projection if (i) P 2 = P (ii) P = P.

arge scale eigenvalue problems, Lecture 2, February 28, 2018 37/46 Projections Projections (cont.) Example: Let q be an arbitrary vector of norm 1, q = q q = 1. Then P = qq is the orthogonal projection onto span{q}. Example: Let Q F n p with Q Q = I p. Then QQ is the orthogonal projector onto R(Q), which is the space spanned by the columns of Q.

Rayleigh quotient Rayleigh quotient The Rayleigh quotient of A at x is defined as ρ(x) := x Ax x x, x 0 If x is an approximate eigenvector, then ρ(x) is a reasonable choice for the corresponding eigenvalue. Using the spectral decomposition A = UΛU, x Ax = x UΛU x = n λ i u i x 2. Similarly, x x = n i=1 u i x 2. With λ 1 λ 2 λ n, we have λ 1 n u i x 2 i=1 i=1 n λ i u i x 2 λ n i=1 n u i x 2. Large scale eigenvalue problems, Lecture 2, February 28, 2018 38/46 i=1

Large scale eigenvalue problems, Lecture 2, February 28, 2018 39/46 Rayleigh quotient Rayleigh quotient (cont.) = λ 1 ρ(x) λ n, for all x 0. ρ(u k ) = λ k, the extremal values λ 1 and λ n are attained for x = u 1 and x = u n. Theorem Let A be Hermitian. Then the Rayleigh quotient satisfies λ 1 = min ρ(x), λ n = max ρ(x). (10) As the Rayleigh quotient is a continuous function it attains all values in the closed interval [λ 1, λ n ].

Large scale eigenvalue problems, Lecture 2, February 28, 2018 40/46 Rayleigh quotient Theorem (Minimum-maximum principle) Let A be Hermitian. Then λ p = min X F n p rank(x )=p max ρ(x x) x 0 Proof: Let U p 1 = [u 1,..., u p 1 ]. For every X F n p with full rank we can choose x 0 such that Up 1 X x = 0. Then 0 z := X x = n i=p z iu i and ρ(z) λ p. For equality choose X = [u 1,..., u p ].

Monotonicity principle Theorem (Monotonicity principle) Let A be Hermitian and let Q := [q 1,..., q p ] with Q Q = I p. Let A := Q AQ with eigenvalues λ 1 λ p. Then λ k λ k, 1 k p. Proof: Let w 1,..., w p F p, w i w j = δ ij, be the eigenvectors of A, A w i = λ iw i, 1 i p. Vectors Qw 1,..., Qw p are normalized and mutually orthogonal. Construct normalized vector x 0 = Q(a 1 w 1 + + a kw k ) Qa that is orthogonal to the first k 1 eigenvectors of A, x 0 u i = 0, 1 i k 1. Minimum-maximum principle: = λ k R(x 0 ) = a Q AQa = k i=1 a 2 i λ i λ k. Large scale eigenvalue problems, Lecture 2, February 28, 2018 41/46

Large scale eigenvalue problems, Lecture 2, February 28, 2018 42/46 Trace of a matrix Trace of a matrix The trace of a matrix A F n n is defined to be the sum of the diagonal elements of a matrix. Matrices that are similar have equal trace. Hence, by the spectral theorem, trace(a) = n a ii = i=1 n λ i. i=1 Theorem (Trace theorem) λ 1 + λ 2 + + λ p = min trace(x AX ) X F n p,x X =I p

arge scale eigenvalue problems, Lecture 2, February 28, 2018 43/46 The singular value decomposition (SVD) The singular value decomposition (SVD) Theorem (Singular value decomposition) If A C m n, m n, then there exist unitary matrices U C m m and V C n n such that σ 1 [ ] U Σ1... AV = Σ = = O, where σ 1 σ 2 σ p 0. O m n n Hence, Av j = u j σ j and A u j = v j σ j for j = 1,..., n. σ n

Large scale eigenvalue problems, Lecture 2, February 28, 2018 44/46 The singular value decomposition (SVD) The singular value decomposition (SVD) (cont.) A 2 = max Ax 2 = max UΣ V x x 2 =1 x 2 =1 }{{} 2 = max Σy 2 = σ 1 y y 2 =1 2 because UΣV x 2 2 = x V Σ U UΣV x = y Σ Σy = y Σ 2 1y = Σ 1 y 2 The maximum is assumed for y = e 1, i.e., x = v 1. If A C n n is nonsingular then σ n > 0 and A 1 2 = 1. σ n By consequence, κ 2 (A) = σ 1 /σ n.

Large scale eigenvalue problems, Lecture 2, February 28, 2018 45/46 The singular value decomposition (SVD) The singular value decomposition (SVD) (cont.) The SVD A = UΣV is related to various symmetric eigenvalue problems A A = V Σ 2 V AA = UΣ 2 U [ ] [ ] [ ] [ ] O A U O O Σ U O A = O O V Σ T O O V [ 1 ] 2 1 U 1 2 U 1 U Σ 1 O O 1 2 = 1 2 V 1 O Σ 2 V O 1 O 2 U1 1 2 U1 1 O O O O where U 1 = [u 1,..., u n ]. U 2 1 2 V 2 V

Large scale eigenvalue problems, Lecture 2, February 28, 2018 46/46 Exercise Exercise 2 (Variations on the Schur decomposition) http://people.inf.ethz.ch/arbenz/ewp/exercises/exercise02.pdf