Spectrum of certain tridiagonal matrices when their dimension goes
... e + 1 non-Toeplitz eigenvalues. Both of these results will be derived by using Sturm sequences (see also [5,6]). We note, however, that there are other methods to derive similar results. In particular, it follows from Theorem 5.23 of [2] that ϕ of the corresponding Toeplitz eigenvalues are uniformly ...
... e + 1 non-Toeplitz eigenvalues. Both of these results will be derived by using Sturm sequences (see also [5,6]). We note, however, that there are other methods to derive similar results. In particular, it follows from Theorem 5.23 of [2] that ϕ of the corresponding Toeplitz eigenvalues are uniformly ...
Math 480 Notes on Orthogonality The word orthogonal is a synonym
... Another Key Example: Given an m × n matrix A ∈ Rm×n , the orthogonal complement of the column space Col(A) is precisely N (AT ). This follows by the same sort of argument as for the first Key Example. The following theorem should seem geometrically obvious, but it is annoyingly difficult to prove di ...
... Another Key Example: Given an m × n matrix A ∈ Rm×n , the orthogonal complement of the column space Col(A) is precisely N (AT ). This follows by the same sort of argument as for the first Key Example. The following theorem should seem geometrically obvious, but it is annoyingly difficult to prove di ...
Calculus II - Basic Matrix Operations
... results as the columns in a new matrix. Since each Aj is an m × 1 column vector, and there are exactly p of them, we find that AB is an m × p matrix. Let’s look at a quick example. Take ...
... results as the columns in a new matrix. Since each Aj is an m × 1 column vector, and there are exactly p of them, we find that AB is an m × p matrix. Let’s look at a quick example. Take ...
1. General Vector Spaces 1.1. Vector space axioms. Definition 1.1
... C is similar to A if there is an invertible matrix P such that C = P −1 AP . Definition 3.13. Properties of similar matrices. (1) Two square matrices are similar if and only if there exist bases with respect to which the matrices represent the same linear operator. (2) Similar matrices have the same ...
... C is similar to A if there is an invertible matrix P such that C = P −1 AP . Definition 3.13. Properties of similar matrices. (1) Two square matrices are similar if and only if there exist bases with respect to which the matrices represent the same linear operator. (2) Similar matrices have the same ...
PDF
... The radical of an associative algebra is the set of elements x ∈ A which are (left) quasi-invertible in every homotope of A. It is this version of an element description of Jacobson radicals which can be generalized to non-associative algebras. First one must define homotopes and isotopes for the g ...
... The radical of an associative algebra is the set of elements x ∈ A which are (left) quasi-invertible in every homotope of A. It is this version of an element description of Jacobson radicals which can be generalized to non-associative algebras. First one must define homotopes and isotopes for the g ...
nae06.pdf
... without row or column exchanges to give its unique solution and the computations are stable with respect to roundo error. De nition of Positive De nite Matrix. A is positive de nite if it is symmetric and if xt Ax > 0 for every x 6= 0. Theorem of Positive De nite Matrix If A is n n positive de ni ...
... without row or column exchanges to give its unique solution and the computations are stable with respect to roundo error. De nition of Positive De nite Matrix. A is positive de nite if it is symmetric and if xt Ax > 0 for every x 6= 0. Theorem of Positive De nite Matrix If A is n n positive de ni ...
aa6.pdf
... 7. Let U+ be an associative C-algebra with two generators E, H, and one defining relation HE − EH = 2E. Let M be an U+ -module. (i) Show that if v ∈ M is a nonzero eigenvector of the operator H : M → M , then E(v) is either zero or is an eigenvector of H again. (ii) Show that if M is a finite dimen ...
... 7. Let U+ be an associative C-algebra with two generators E, H, and one defining relation HE − EH = 2E. Let M be an U+ -module. (i) Show that if v ∈ M is a nonzero eigenvector of the operator H : M → M , then E(v) is either zero or is an eigenvector of H again. (ii) Show that if M is a finite dimen ...
Vector spaces, norms, singular values
... Most of mathematics is best learned by doing. Linear algebra is no exception. You have had a previous class in which you learned the basics of linear algebra, and you will have plenty of practice with these concepts over the semester. This brief refresher lecture is supposed to remind you of what yo ...
... Most of mathematics is best learned by doing. Linear algebra is no exception. You have had a previous class in which you learned the basics of linear algebra, and you will have plenty of practice with these concepts over the semester. This brief refresher lecture is supposed to remind you of what yo ...
Second midterm solutions
... T 2 = 0. Prove that for some basis v1 , . . . , vn of V , the matrix A for T with respect to this basis is in Jordan canonical form. Solution: We have image(T ) ⊂ ker(T ), because w ∈ image(T ) ⇒ w = T (v) ⇒ T (w) = T 2 (v) = 0. Choose a basis w1 , . . . , wj for image(T ), and extend this basis to ...
... T 2 = 0. Prove that for some basis v1 , . . . , vn of V , the matrix A for T with respect to this basis is in Jordan canonical form. Solution: We have image(T ) ⊂ ker(T ), because w ∈ image(T ) ⇒ w = T (v) ⇒ T (w) = T 2 (v) = 0. Choose a basis w1 , . . . , wj for image(T ), and extend this basis to ...
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.