Exam 3
... (b) Construct a particular solution to the nonhomogeneous equation L(y) = f , where f (t) = t3 . (c) Write down the general solution to the equation L(y) = f . ...
... (b) Construct a particular solution to the nonhomogeneous equation L(y) = f , where f (t) = t3 . (c) Write down the general solution to the equation L(y) = f . ...
(pdf).
... (b) Let a = 0 in C. Assume that such a matrix is an echelon form of some matrix A. What value of c and d so that rank(A) = 2? (c) Let d = 1 and c = 1 and a = 0 in the matrix C. Assume that the matrix you obtain is the reduced echelon form of some matrix A. Write the last column of A as linear combin ...
... (b) Let a = 0 in C. Assume that such a matrix is an echelon form of some matrix A. What value of c and d so that rank(A) = 2? (c) Let d = 1 and c = 1 and a = 0 in the matrix C. Assume that the matrix you obtain is the reduced echelon form of some matrix A. Write the last column of A as linear combin ...
2
... name: Mathematics 220 second quiz Thursday, July 9, 2015 please show your work to get full credit ...
... name: Mathematics 220 second quiz Thursday, July 9, 2015 please show your work to get full credit ...
Section 7.2
... Thus the columns u1, ..., un are orthogonal eigenvectors of A; and they form a basis for V . For a Hermitian matrix A, the eigenvalues are all real; and there is an orthogonal basis for the associated vector space V consisting of eigenvectors of A. In dealing with such a matrix A in a problem, the b ...
... Thus the columns u1, ..., un are orthogonal eigenvectors of A; and they form a basis for V . For a Hermitian matrix A, the eigenvalues are all real; and there is an orthogonal basis for the associated vector space V consisting of eigenvectors of A. In dealing with such a matrix A in a problem, the b ...
Physics 3730/6720 – Maple 1b – 1 Linear algebra, Eigenvalues and Eigenvectors
... their elements. To do matrix and vector products and see the result you need evalm. (Maple does it, otherwise, but silently.) Multiplication by scalars works with *, but multiplication of matrices with matrices and matrices with vectors requires the special multiplication operator &*. > eigenvals(A) ...
... their elements. To do matrix and vector products and see the result you need evalm. (Maple does it, otherwise, but silently.) Multiplication by scalars works with *, but multiplication of matrices with matrices and matrices with vectors requires the special multiplication operator &*. > eigenvals(A) ...
Solution of 2x2
... Eigensolution of a 2x2 Matrix The eigenvectors and eigenvalues of a 2x2 matrix can be easily calculated in closed form. Using the standard eigensolution formula: ...
... Eigensolution of a 2x2 Matrix The eigenvectors and eigenvalues of a 2x2 matrix can be easily calculated in closed form. Using the standard eigensolution formula: ...
the jordan normal form
... Recall that by linearizing around hyperbolic fixed points and periodic orbits of nonlinear systems, one can obtain a locally accurate picture of the dynamics. Hence, it is important to understand the different kinds of evolution that can occur in a linear system. The JNF allows a complete representa ...
... Recall that by linearizing around hyperbolic fixed points and periodic orbits of nonlinear systems, one can obtain a locally accurate picture of the dynamics. Hence, it is important to understand the different kinds of evolution that can occur in a linear system. The JNF allows a complete representa ...
Applying transformations in succession Suppose that A and B are 2
... The matrix for a rotation or a reflection is invertible: their determinants are non-zero. Suppose that A is the 2 × 2 matrix representing TA, and it has inverse A−1. Then TA−1 (TA (x)) = A−1(Ax) = (A−1A)x = I2x = x, and similarly, TA(TA−1 (x)) = x. So to get the matrix for the inverse, we just take ...
... The matrix for a rotation or a reflection is invertible: their determinants are non-zero. Suppose that A is the 2 × 2 matrix representing TA, and it has inverse A−1. Then TA−1 (TA (x)) = A−1(Ax) = (A−1A)x = I2x = x, and similarly, TA(TA−1 (x)) = x. So to get the matrix for the inverse, we just take ...
Jordan normal form
In linear algebra, a Jordan normal form (often called Jordan canonical form)of a linear operator on a finite-dimensional vector space is an upper triangular matrix of a particular form called a Jordan matrix, representing the operator with respect to some basis. Such matrix has each non-zero off-diagonal entry equal to 1, immediately above the main diagonal (on the superdiagonal), and with identical diagonal entries to the left and below them. If the vector space is over a field K, then a basis with respect to which the matrix has the required form exists if and only if all eigenvalues of the matrix lie in K, or equivalently if the characteristic polynomial of the operator splits into linear factors over K. This condition is always satisfied if K is the field of complex numbers. The diagonal entries of the normal form are the eigenvalues of the operator, with the number of times each one occurs being given by its algebraic multiplicity.If the operator is originally given by a square matrix M, then its Jordan normal form is also called the Jordan normal form of M. Any square matrix has a Jordan normal form if the field of coefficients is extended to one containing all the eigenvalues of the matrix. In spite of its name, the normal form for a given M is not entirely unique, as it is a block diagonal matrix formed of Jordan blocks, the order of which is not fixed; it is conventional to group blocks for the same eigenvalue together, but no ordering is imposed among the eigenvalues, nor among the blocks for a given eigenvalue, although the latter could for instance be ordered by weakly decreasing size.The Jordan–Chevalley decomposition is particularly simple with respect to a basis for which the operator takes its Jordan normal form. The diagonal form for diagonalizable matrices, for instance normal matrices, is a special case of the Jordan normal form.The Jordan normal form is named after Camille Jordan.