
A Backward Stable Hyperbolic QR Factorization Method for Solving
... In Step 1 the factorization can be obtained by applying the bidiagonal factorization methods followed by a block row permutation. The Householder transformation method for computing the bidiagonal factorization can be found in [6, pp. 252]. Since U is only used for computing d1 , we only need to app ...
... In Step 1 the factorization can be obtained by applying the bidiagonal factorization methods followed by a block row permutation. The Householder transformation method for computing the bidiagonal factorization can be found in [6, pp. 252]. Since U is only used for computing d1 , we only need to app ...
Spectrum of certain tridiagonal matrices when their dimension goes
... The sequence {xi (λ), i = 1, 2, . . . N } forms a Sturm sequence if xi (λ) = 0 implies xi−1 (λ)xi+1 (λ) < 0, for i = 2, 3, . . . , N − 1. In Sturm sequences, there is a close relation between the location of zeros and sign variations. For i N a sign variation occurs if xi (λ)xi+1 (λ) < 0 or xi (λ) ...
... The sequence {xi (λ), i = 1, 2, . . . N } forms a Sturm sequence if xi (λ) = 0 implies xi−1 (λ)xi+1 (λ) < 0, for i = 2, 3, . . . , N − 1. In Sturm sequences, there is a close relation between the location of zeros and sign variations. For i N a sign variation occurs if xi (λ)xi+1 (λ) < 0 or xi (λ) ...
1 Matrix Lie Groups
... 1.1 Definition of a Matrix Lie Group We begin with a very important class of groups, the general linear groups. The groups we will study in this book will all be subgroups (of a certain sort) of one of the general linear groups. This chapter makes use of various standard results from linear algebra t ...
... 1.1 Definition of a Matrix Lie Group We begin with a very important class of groups, the general linear groups. The groups we will study in this book will all be subgroups (of a certain sort) of one of the general linear groups. This chapter makes use of various standard results from linear algebra t ...
Lower Bounds on Matrix Rigidity via a Quantum
... most by a constant then the rigidity is Ω(n2 ). For the case θ > r/n, Kashin and Razborov [4] improved the bound to RH (r, θ) = Ω(n3 /rθ2 ). Study of this relaxed notion of rigidity is further motivated by the fact that stronger lower bounds would separate the communication complexity versions of th ...
... most by a constant then the rigidity is Ω(n2 ). For the case θ > r/n, Kashin and Razborov [4] improved the bound to RH (r, θ) = Ω(n3 /rθ2 ). Study of this relaxed notion of rigidity is further motivated by the fact that stronger lower bounds would separate the communication complexity versions of th ...
The alogorithm
... The statement "for j = 2 to n" has no flops. The statement "m = aj1/a11" has one flop and it is done n-1 times so there are n-1 flops altogether The statement "for p = 2 to n" has no flops. The statement "ajp = ajp – ma1p" has two flops. As p runs from 2 to n it is done n-1 times so there are 2(n-1 ...
... The statement "for j = 2 to n" has no flops. The statement "m = aj1/a11" has one flop and it is done n-1 times so there are n-1 flops altogether The statement "for p = 2 to n" has no flops. The statement "ajp = ajp – ma1p" has two flops. As p runs from 2 to n it is done n-1 times so there are 2(n-1 ...
3 Let n 2 Z + be a positive integer and T 2 L(F n, Fn) be defined by T
... 8 Prove or give a counterexample to the following claim: Claim. Let V be a finite-dimensional vector space over F, and let T 2 L(V ) be a linear operator on V . If the matrix for T with respect to some basis on V has all zeros on the diagonal, then T is not invertible. Solution This is false, the ma ...
... 8 Prove or give a counterexample to the following claim: Claim. Let V be a finite-dimensional vector space over F, and let T 2 L(V ) be a linear operator on V . If the matrix for T with respect to some basis on V has all zeros on the diagonal, then T is not invertible. Solution This is false, the ma ...