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A skew-symmetric matrix is a matrix \(S\) that satisfies $$ s^{T}=-S $$ (a) Show that for any general matrix \(A\), the matrix \(\left(A-A^{T}\right) / 2\) is skew-symmetric. (This matrix is in fact referred to as the "skew- symmetric part" of \(A\).) (b) Show that the diagonal of a skew-symmetric matrix \(S\) must be zero component-wise. (c) Show that the eigenvalues of \(S\) must be purely imaginary. (d) If \(S\) is \(n \times n\) with \(n\) an odd number, then it is necessarily singular. Why? (e) Suppose that the skew-symmetric \(S\) is nonsingular and sparse. In the process of solving a linear system associated with \(S\), a procedure equivalent to Arnoldi or Lanczos is applied to form an orthogonal basis for the corresponding Krylov subspace. Suppose the resulting matrices satisfy the relation $$ S Q_{k}=Q_{k+1} U_{k+1, k} $$ where \(Q_{k}\) is an \(n \times k\) matrix whose orthonormal columns form the basis for the Krylov subspace, \(Q_{k+1}\) is the matrix of basis vectors containing also the \((k+1)\) st basis vector, and \(U_{k+1, k}\) is a \((k+1) \times k\) matrix. i. Determine the nonzero structure of \(U_{k+1, k} .\) Specifically, state whether it is tridiagonal or upper Hessenberg, and explain what can be said about the values along the main diagonal. ii. Preconditioners for systems with a dominant skew-symmetric part often deal with the possibility of singularity by solving a shifted skew-symmetric system, where instead of solving for \(S\) one solves for \(S+\beta_{k} I\) with \(\beta_{k}\) a scalar. Suppose we have the same right-hand-side, but we need to solve the system for several values of \(\beta_{k}\). Can the Arnoldi or Lanczos type procedure outlined above be applied once and for all and then be easily adapted? iii. Describe the main steps of a MINRES-like method for solving a skew- symmetric linear system.

Short Answer

Expert verified
Given a skew-symmetric matrix, we've shown that: (a) A matrix computed as \(\left(A-A^{T}\right) / 2\) is skew-symmetric for any general matrix \(A\). (b) The diagonal of a skew-symmetric matrix must be zero component-wise. (c) The eigenvalues of a skew-symmetric matrix must be purely imaginary. (d) An odd-order skew-symmetric matrix is singular. For a nonlinear Arnoldi or Lanczos type procedure applied to a skew-symmetric linear system: i. The matrix \(U_{k+1, k}\) will also have a tridiagonal nonzero pattern since it is generated from the product of two tridiagonal matrices, but its main diagonal can't be specified a priori. ii. The procedure may not be easily adapted for solving shifted skew-symmetric systems with the same right-hand-side as the orthogonal bases formed from the shifted systems might not be similar. iii. A MINRES-like method consists of initializing an initial guess, computing the residual vector, generating an orthonormal basis, forming the matrix \(U_{k+1, k}\), solving the linear system, updating the solution, and iterating until the desired accuracy is reached.

Step by step solution

01

(a) Proving skew-symmetry of (A-AT)/2

Given a general matrix \(A\), consider the matrix \(\left(A-A^{T}\right) / 2\). Let's denote this matrix as \(B\). To prove that \(B\) is skew-symmetric, we need to show that \(B^T=-B\). Now, let's compute the transpose of \(B\): $$ B^T = \left(\frac{A-A^T}{2}\right)^T = \frac{1}{2}(A^T - A), $$ where we used the property that the transpose of the difference between two matrices is equal to the difference of their transposes. Notice that \(B^T = -B\). Therefore, \(B\) is skew-symmetric.
02

(b) Diagonal of skew-symmetric matrix must be zero component-wise

Let \(S\) be a skew-symmetric matrix. Recall that a matrix is skew-symmetric if \(S^T = -S\). Let's consider the diagonal of \(S\). If the diagonal of \(S\) contains any non-zero components, then the transposition will not change the value of these components. However, since \(S^T = -S\), this implies that the diagonal of \(S\) must be zero component-wise.
03

(c) Eigenvalues of skew-symmetric matrix must be purely imaginary

Let \(S\) be a skew-symmetric matrix and \(\lambda\) be an eigenvalue of \(S\). Then, we have \(Sv = \lambda v\) for some eigenvector \(v\). Taking the transpose of the equation, we get \((Sv)^T = (\lambda v)^T\). Using properties of transposes, this yields \(v^TS^T = \lambda^* v^T\), where \(\lambda^*\) is the complex conjugate of \(\lambda\). Since \(S^T = -S\), we can rewrite the equation as \(v^T(-S) = \lambda^* v^T\) or \(-v^TS = \lambda^* v^T\). Multiplying both sides of the eigenvalue equation \(Sv = \lambda v\) by \(v^T\) from the left, we get \(v^TSv = \lambda v^Tv\). On the other hand, multiplying the equation \(-v^TS = \lambda^* v^T\) by \(v\) from the right, we get \(-v^TSv = \lambda^* v^Tv\). Comparing these two results, we have \(\lambda v^Tv = -\lambda^* v^Tv\). Since \(v^Tv\) is nonzero, we can conclude \(\lambda = -\lambda^*\). Thus, the eigenvalues of a skew-symmetric matrix must be purely imaginary.
04

(d) Odd-order skew-symmetric matrix is singular

Let \(S\) be an \(n \times n\) skew-symmetric matrix where \(n\) is an odd number. Denote \(S^T\) by \(T\), and then we have \(T = -S\). We can use the determinant property that \(\det(T) = \det(-S)\). Also, we know that \(\det(T) = \det(S^T) = \det(S)\). Therefore, we have \(\det(S) = \det(-S)\). For a square matrix with odd order, the determinant of its negative is equal to the determinant of itself multiplied by \((-1)^n\), so we have \(\det(S) = (-1)^n \det(S)\). Since \(n\) is odd, this implies that \(\det(S) = -\det(S)\). The only way this equation can be satisfied is if \(\det(S) = 0\). Thus, an odd-order skew-symmetric matrix is necessarily singular.
05

(e) Arnoldi or Lanczos type procedure applied to a skew-symmetric linear system

i. The matrix \(S\) is skew-symmetric, and \(U_{k+1, k}\) is a \((k+1)\times k\) matrix. From the given relation, \(SQ_{k} = Q_{k+1}U_{k+1, k}\). Since \(S\) is skew-symmetric, the product \(SQ_{k}\) will have a specific pattern along its diagonal: the two adjacent diagonals above and below the main diagonal will have opposite signs, meaning the matrix will be tridiagonal. Since \(U_{k+1, k}\) is obtained from the product of two tridiagonal matrices, \(U_{k+1, k}\) itself will also be tridiagonal. The main diagonal of \(U_{k+1, k}\) can't be said a priori. ii. If we are solving for several values of \(\beta_{k}\), we need to find if the procedure outlined above can be applied once and easily adapted. Given \(S+\beta_kI\), the eigenvalues of the resulting matrix will change. However, the matrix \(Q_k\) is not dependent on \(\beta_k\), so we can find it once and use it for all systems having the same right-hand-side. However, since \(S+\beta_kI\) has different eigenvalues, and we are trying to form orthogonal bases, the procedure may not be easily adapted unless the orthogonal bases formed from the shifted systems are similar. iii. Steps of a MINRES-like method for solving a skew-symmetric linear system: 1. Given the skew-symmetric matrix \(S\) and the right-hand-side vector \(b\), initialize an initial guess vector \(x_0\). 2. Compute the residual vector \(r_0 = b - Sx_0\). 3. From the initial residual \(r_0\), generate an orthonormal basis \(Q_k\) for the Krylov subspace using an Arnoldi, Lanczos, or any other block orthonormalization procedure. 4. Form the matrix \(U_{k+1, k}\) using the given relation, \(U_{k+1, k} = Q_{k+1}^TSQ_{k}\). 5. Solve the linear system \(U_{k+1, k}z = \|r_0\|_2e_1\) for the required weights \(z\), where \(e_1\) is the first canonical basis vector. 6. Update the solution with \(x_{k+1} = x_0 + Q_kz\). 7. Compute the new residual vector \(r_{k+1} = b - Sx_{k+1}\). If the desired accuracy is not reached, return to step 3.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Skew-symmetric Matrix
A skew-symmetric matrix is a special kind of square matrix. Its defining property is that its transpose is equal to its negative. This means for a matrix \( S \), if you take the transpose, you will get \( S^T = -S \). This implies that if you reflect the matrix over its main diagonal, all the values will have flipped signs.

This property leads to interesting characteristics:
  • All diagonal elements of a skew-symmetric matrix must be zero. If a nonzero element were on the diagonal, it would imply that the element is equal to its own negative, which is only possible if the element is zero.
  • The skew-symmetric property \( S^T = -S \) is preserved in linear combinations, making these matrices useful in various applications like simplifying certain problems or representing rotational dynamics in physics.
Eigenvalues
Eigenvalues are central to understanding the behavior of matrices in linear algebra. For skew-symmetric matrices, these eigenvalues have a particular characteristic: they must be purely imaginary, or zero. Purely imaginary numbers are numbers of the form \( bi \), where \( b \) is a real number and \( i \) is the square root of -1.

Why is this important?
  • Purely imaginary eigenvalues imply rotation, rather than scaling, in transformations represented by skew-symmetric matrices.
  • This property is crucial for understanding the behavior of differential equations and stability analysis in dynamic systems.
This unique property comes from the fact that, for an eigenvalue \( \lambda \) of a skew-symmetric matrix \( S \), the equation \( \lambda = -\lambda^* \) (where \( * \) signifies the complex conjugate) holds, leading to \( \lambda \) being purely imaginary.
Krylov Subspace
Krylov subspaces are powerful mathematical constructs used in the analysis of matrices, especially for solving linear systems and eigenvalue problems. These are generated by applying a matrix to a vector repeatedly.

In practical terms, if you have a skew-symmetric matrix and a starting vector, you can generate a sequence of vectors that form the Krylov subspace. This sequence consists of the starting vector, the matrix applied once to the vector, the matrix applied twice, and so on. This is especially useful:
  • When an orthonormal basis for these vectors is needed, procedures like Arnoldi or Lanczos are applied, which are computationally efficient.
  • Krylov subspaces help in computing matrix functions, including finding eigenvectors and eigenvalues for large-scale problems.
Singular Matrix
A singular matrix is a matrix that does not have an inverse. For skew-symmetric matrices, there is a particular interest when the matrix order is odd. In such cases, the matrix is always singular.

The intuition behind this is tied to the determinant of the matrix. The determinant of a skew-symmetric matrix of odd order is zero, which is also the condition of singularity. This occurs because:
  • The eigenvalues of an odd dimension skew-symmetric matrix include zero due to the pairing nature of non-zero purely imaginary eigenvalues. This directly implies a zero determinant.
  • It signifies that such matrices cannot represent a full-rank transformation, i.e., their transformations result in a loss of dimensionality.
Understanding singular matrices is crucial for scenarios where you need to invert a matrix to solve linear systems. In contexts like these, if a matrix is singular, alternative methods like pseudo-inversion or regularization techniques might be necessary.

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Most popular questions from this chapter

Suppose CG is applied to a symmetric positive definite linear system \(A \mathbf{x}=\mathbf{b}\) where the righthand-side vector \(\mathbf{b}\) happens to be an eigenvector of the matrix \(A\). How many iterations will it take to converge to the solution? Does your answer change if \(A\) is not SPD and instead of CG we apply GMRES?

Show that the energy norm is indeed a norm when the associated matrix is symmetric positive definite.

Let \(\alpha\) be a scalar, and consider the iterative scheme $$ \mathbf{x}_{k+1}=\mathbf{x}_{k}+\alpha\left(\mathbf{b}-A \mathbf{x}_{k}\right) $$ This is the gradient descent method with a fixed step size \(\alpha\). (a) If \(A=M-N\) is the splitting associated with this method, state what \(M\) and the iteration matrix \(T\) are. (b) Suppose \(A\) is symmetric positive definite and its eigenvalues are \(\lambda_{1}>\lambda_{2}>\cdots>\lambda_{n}>0 .\) i. Derive a condition on \(\alpha\) that guarantees convergence of the scheme to the solution \(\mathbf{x}\) for any initial guess. ii. Express the condition on \(\alpha\) in terms of the spectral condition number of \(A, \kappa(A)\). iii. Show the best value for the step size in terms of maximizing the speed of convergence is $$ \alpha=\frac{2}{\lambda_{1}+\lambda_{n}} $$ Find the spectral radius of the iteration matrix in this case, and express it in terms of the condition number of \(A\). (c) Determine whether the following statement is true or false. Justify your answer. "If \(A\) is strictly diagonally dominant and \(\alpha=1\), then the iterative scheme converges to the solution for any initial guess \(\mathrm{x}_{0} .\)

The linear system $$ \begin{aligned} &10 x_{1}+x_{2}+x_{3}=12 \\ &x_{1}+10 x_{2}+x_{3}=12 \\ &x_{1}+x_{2}+10 x_{3}=12 \end{aligned} $$ has the unique solution \(x_{1}=x_{2}=x_{3}=1\). Starting from \(\mathbf{x}_{0}=(0,0,0)^{T}\) perform \- two iterations using Jacobi; \- one iteration using Gauss-Seidel. Calculate error norms for the three iterations in \(\ell_{1}\). Which method seems to converge faster?

\(x\) $$ A=\left(\begin{array}{cc} 2 & -1 \\ -1 & 2 \end{array}\right) $$ and suppose we are required to solve \(A \mathbf{x}=\mathbf{b}\). (a) Write down explicitly the iteration matrices corresponding to the Jacobi, Gauss-Seidel, and SOR schemes. (b) Find the spectral radius of the Jacobi and Gauss-Seidel iteration matrices and the asymptotic rates of convergence for these two schemes. (c) Plot a graph of the spectral radius of the SOR iteration matrix vs. the relaxation parameter \(\omega\) for \(0 \leq \omega \leq 2\). (d) Find the optimal SOR parameter, \(\omega^{*}\). What is the spectral radius of the corresponding iteration matrix? Approximately how much faster would SOR with \(\omega^{*}\) converge compared to Jacobi?

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