G. E. Cho and I. C. F. Ipsen. If a matrix has only a single eigenvalue how sensitive is this eigenvalue? Technical report, University of North Carolina, 1997. Home/Search Document Details and Download
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Eigenvalue bounds from the Schur form - Braconnier, Saad (1998) (1 citation) (Correct)
....on the canonical Jordan form of the matrix R and therefore also on the Jordan form of A. Let A be the diagonal matrix A = D I. The following relation holds Taking norm and using Lemma 2.3. 3 from [4] the following error bound can be derived k v q 2 A similar inequality was pointed out in [2] (Corollary 5.1) Note that if IINIl tends to zero, i.e. the matrix R tends to become diagonal then the left hand side is essentially equal to r( 7 4 The Hermitian Case When the matrix is Hermitian, it is fairly easy to derive residual bounds for approximate eigenvaluesor eigenvectors. ....
....that the vector w is zero. This is a strong assumption that has implications on the matrix and not only the approximation. In this case, the relation (8) becomes, and since c = IIPSll2, and = liPvii2 we obtain a remarkably simple inequality, A bound of this type was also pointed out in [2] (Corollary 3.2) Except for the denom inator term this is very similar to the Bauer Fike inequality. In fact the denominator should be close to one if the eigenvector is accurate enough. Another way of rewriting the inequality is: I x 1 cos Z, 11 112 Note that when w = 0, then the left ....
G. E. Cho and I. C. F. Ipsen. If a matrix has only a single eigenvalue how sensitive is this eigenvalue? Technical report, University of North Carolina, 1997.
Eigenvalue bounds from the Schur form - Thierry Braconnier Yousef (1998) (1 citation) (Correct)
....k X i=0 ( Delta Gamma1 N) i Delta Gamma1 Taking norm and using Lemma 2.3. 3 from [4] the following error bound can be derived ( k kN k Gamma1 k 2 ( kN k Gamma2 k 2 : k Gamma1 q 1 fl 2 w kP uk 2 krk 2 A similar inequality was pointed out in [2] (Corollary 5.1) Note that if kNk 2 tends to zero, i.e. the matrix R 1 tends to become diagonal then the left hand side is essentially equal to ( 7 4 The Hermitian Case When the matrix is Hermitian, it is fairly easy to derive residual bounds for approximate eigenvalues or ....
....on the matrix and not only the approximation. In this case, the relation (8) becomes, ff( Gamma ) ffl and since ff = kP uk 2 , and ffl = kP rk 2 we obtain a remarkably simple inequality, j Gamma j kP rk 2 kP uk 2 krk 2 kP uk 2 A bound of this type was also pointed out in [2] (Corollary 3.2) Except for the denominator term this is very similar to the Bauer Fike inequality. In fact the denominator should be close to one if the eigenvector is accurate enough. Another way of rewriting the inequality is: j Gamma j cos 6 (u; u) krk 2 Note that when w = 0, then ....
G. E. Cho and I. C. F. Ipsen. If a matrix has only a single eigenvalue how sensitive is this eigenvalue? Technical report, University of North Carolina, 1997.
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