• Cyberian's Gold

    @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

    Two matrices with the same characteristic polynomial need not be similar.

    Two similar matrices have the same characteristic polynomial. The converse however is not true in general: two matrices with the same characteristic polynomial need not be similar. The matrix A and its transpose have the same characteristic polynomial. … In this case A is similar to a matrix in Jordan normal form.

  • Cyberian's Gold

    Two matrices with the same characteristic polynomial need not be similar.

  • Cyberian's Gold

    @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

    If A is a nxn triangular matrix (upper triangular, lower triangular) or diagonal matrix , the eigenvalues of A are the diagonal entries of A.

    The eigenvalues of B are 1,4,6 since B is an upper triangular matrix and eigenvalues of an upper triangular matrix are diagonal entries. We claim that the eigenvalues of A and B are the same.

  • Cyberian's Gold

    If A is a nxn triangular matrix (upper triangular, lower triangular) or diagonal matrix , the eigenvalues of A are the diagonal entries of A.

  • Cyberian's Gold

    @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

    @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

    If n x n matrices A and B are similar, then they have the same eigenvalues (with the same multiplicities).

    Since similar matrices A and B have the same characteristic polynomial, they also have the same eigenvalues. If B = PAP−1 and v = 0 is an eigenvector of A (say Av = λv) then B(Pv) = PAP−1(Pv) = PA(P−1P)v = PAv = λPv. Thus Pv (which is non-zero since P is invertible) is an eigenvector for B with eigenvalue λ.

    The matrix B has the same A as an eigenvalue. M−1x is the eigenvector. If two matrices are similar, they have the same eigenvalues and the same number of independent eigenvectors (but probably not the same eigenvectors).

  • Cyberian's Gold

    @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

    If n x n matrices A and B are similar, then they have the same eigenvalues (with the same multiplicities).

    Since similar matrices A and B have the same characteristic polynomial, they also have the same eigenvalues. If B = PAP−1 and v = 0 is an eigenvector of A (say Av = λv) then B(Pv) = PAP−1(Pv) = PA(P−1P)v = PAv = λPv. Thus Pv (which is non-zero since P is invertible) is an eigenvector for B with eigenvalue λ.

  • Cyberian's Gold

    If n x n matrices A and B are similar, then they have the same eigenvalues (with the same multiplicities).

  • Cyberian's Gold

    @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

    The Jacobi’s method is a method of solving a matrix equation on a matrix that has ____ zeros along its main diagonal.

    The Jacobi’s method is a method of solving a matrix equation on a matrix that has no zeros along its main diagonal. Power method is applicable if the eigen vectors corresponding to eigen values are linearly independent.

  • Cyberian's Gold

    The Jacobi’s method is a method of solving a matrix equation on a matrix that has ____ zeros along its main diagonal.

  • Cyberian's Gold

    @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

    The Gauss-Seidel method is applicable to strictly diagonally dominant or symmetric positive definite matrices A.

    Explanation: Gauss-Seidel method is applicable to strictly diagonally dominant or symmetric positive definite matrices because only in this case convergence is possible. … This is the modification made to Jacobi’s method, which is now called as Gauss-seidal method.

  • Cyberian's Gold

    The Gauss-Seidel method is applicable to strictly diagonally dominant or symmetric positive definite matrices A.

  • Cyberian's Gold

    @zaasmi
    If λk is the eigenvalue that is closest to the number q, then µk is the dominant eigenvalue for B and so it can be determined using the power method. Moreover, to find the eigenvalue of A that is smallest in magnitude is equivalent to find the dominant eigenvalue of the matrix B = A−1.

  • Cyberian's Gold

    @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

    Power method is applicable if the eigen values are ______________.

    The Jacobi’s method is a method of solving a matrix equation on a matrix that has no zeros along its main diagonal. Power method is applicable if the eigen vectors corresponding to eigen values are linearly independent.

    The Power Method is used to find a dominant eigenvalue (one having the largest absolute value), if one exists, and a corresponding eigenvector. To apply the Power Method to a square matrix A, begin with an initial guess u0 for the eigenvector of the dominant eigenvalue.

  • Cyberian's Gold

    Power method is applicable if the eigen values are ______________.

  • Cyberian's Gold

    @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

    While solving a system of linear equations by Gauss Jordon Method, after all the elementary row operations if there lefts also zeros on the main diagonal then which of the is true about the system?

    Answer
    Answer

  • Cyberian's Gold

    While solving a system of linear equations by Gauss Jordon Method, after all the elementary row operations if there lefts also zeros on the main diagonal then which of the is true about the system?

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