Krylov subspace methods

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The entries of the matrix ∥bμν∥\left\| b_{\mu \nu }\right\| are determined by the formulas:

bμ+1,ν=bμ1a1ν+bμ2a2ν+…+bμnanν; b_{\mu +1,\nu }=b_{\mu 1}a_{1\nu }+b_{\mu 2}a_{2\nu }+\ldots +b_{\mu n}a_{n\nu }; μ=1,2,…,n, \mu =1,2,\ldots ,n,

where b11,b12,…,b1nb_{11}, b_{12}, \ldots , b_{1n} and aμνa_{\mu \nu } are given numbers. Find the value of the product of the determinants

∣b11b12…b1nb21b22…b2n⋮⋮⋮⋮bn1bn2…bnn∣⋅∣a11−xa12…a1na21a22−x…a2n⋮⋮⋮⋮an1an2…ann−x∣=0. \begin{vmatrix} b_{11} & b_{12} & \ldots & b_{1n} \\ b_{21} & b_{22} & \ldots & b_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ b_{n1} & b_{n2} & \ldots & b_{nn} \end{vmatrix} \cdot \begin{vmatrix} a_{11}-x & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22}-x & \ldots & a_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ a_{n1} & a_{n2} & \ldots & a_{nn}-x \end{vmatrix}=0 .
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Use Krylov's method to determine the characteristic polynomial for the matrix A\mathbf{A} given in Exercise 7.11.1.

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Use the technique for computing a matrix's minimum polynomial via an orthogonalized Krylov sequence (successively Gram--Schmidt-orthogonalizing I,A,A2,…\mathbf{I}, \mathbf{A}, \mathbf{A}^2, \ldots in the Frobenius inner product ⟨X∣Y⟩=trace⁡(XTY)\langle \mathbf{X} \mid \mathbf{Y} \rangle = \operatorname {trace}(\mathbf{X}^T\mathbf{Y}) until a power Ak\mathbf{A}^k is found to be a linear combination of the earlier orthogonalized terms U0,…,Uk−1\mathbf{U}_0, \ldots , \mathbf{U}_{k-1}, with coefficients recovered by solving Rc→=R−1\mathbf{R}\overrightarrow {c} = \mathbf{R}^{-1} triangular system for the αi\alpha_i's) to determine the minimum polynomial for

A=(−7−48−8−4−14−4−16−817−16−6−36−5). \mathbf{A} = \begin{pmatrix} -7 & -4 & 8 & -8 \\ -4 & -1 & 4 & -4 \\ -16 & -8 & 17 & -16 \\ -6 & -3 & 6 & -5 \end{pmatrix}.
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Use the Lanczos algorithm to find an orthogonal matrix P\mathbf{P} such that PTAP=T\mathbf{P}^T\mathbf{A}\mathbf{P} = \mathbf{T} is tridiagonal, where A=(211121112)\mathbf{A} = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}.

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Starting with x→0=0→\overrightarrow {x}_0 = \overrightarrow {0}, apply the conjugate gradient algorithm to solve Ax→=b→\mathbf{A}\overrightarrow {x} = \overrightarrow {b}, where A=(211121112)\mathbf{A} = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix} and b→=(400)\overrightarrow {b} = \begin{pmatrix} 4 \\ 0 \\ 0 \end{pmatrix}.

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Use Arnoldi's algorithm to find an orthogonal matrix Q\mathbf{Q} such that QTAQ=H\mathbf{Q}^T\mathbf{A}\mathbf{Q} = \mathbf{H} is upper Hessenberg, where A=(512−40−2−4−1−1)\mathbf{A} = \begin{pmatrix} 5 & 1 & 2 \\ -4 & 0 & -2 \\ -4 & -1 & -1 \end{pmatrix}.

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Use GMRES to solve Ax→=b→\mathbf{A}\overrightarrow {x} = \overrightarrow {b} for A=(512−40−2−4−1−1)\mathbf{A} = \begin{pmatrix} 5 & 1 & 2 \\ -4 & 0 & -2 \\ -4 & -1 & -1 \end{pmatrix} and b→=(121)\overrightarrow {b} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}.

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References

[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.
[Mey00] Meyer C.D. Matrix Analysis and Applied Linear Algebra. SIAM, 2000.