A proof of the Jordan normal form theorem
... Indeed, assume there is a vector v ∈ Ker(A − λ Id)k ∩ Im(A − λ Id)k . This means that (A − λ Id)k (v) = 0 and that there exists a vector w such that v = (A − λ Id)k (w). It follows that (A − λ Id)2k (w) = 0, so w ∈ Ker(A − λ Id)2k = N2k (λ). But from the previous lemma we know that N2k (λ) = Nk (λ) ...
... Indeed, assume there is a vector v ∈ Ker(A − λ Id)k ∩ Im(A − λ Id)k . This means that (A − λ Id)k (v) = 0 and that there exists a vector w such that v = (A − λ Id)k (w). It follows that (A − λ Id)2k (w) = 0, so w ∈ Ker(A − λ Id)2k = N2k (λ). But from the previous lemma we know that N2k (λ) = Nk (λ) ...
Eigenvalues - University of Hawaii Mathematics
... (3) In the case of a symmetric matrix, the n different eigenvectors will not necessarily all correspond to different eigenvalues, so they may not automatically be orthogonal to each other. However (if the entries in A are all real numbers, as is always the case in this course), it’s always possible ...
... (3) In the case of a symmetric matrix, the n different eigenvectors will not necessarily all correspond to different eigenvalues, so they may not automatically be orthogonal to each other. However (if the entries in A are all real numbers, as is always the case in this course), it’s always possible ...
2 - UCSD Math Department
... We skipped the example below and the equations of a plane. Equations of planes. a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0, where (x0 , y0 , z0 ) is a point in the plane and n = (a, b, c) is normal to the plane. In fact, if (x, y, z) is a point in the plane then the vector (x − x0 , y − y0 , z − z0 ) ...
... We skipped the example below and the equations of a plane. Equations of planes. a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0, where (x0 , y0 , z0 ) is a point in the plane and n = (a, b, c) is normal to the plane. In fact, if (x, y, z) is a point in the plane then the vector (x − x0 , y − y0 , z − z0 ) ...
MODULE 11 Topics: Hermitian and symmetric matrices Setting: A is
... (meaning that det(A − λI) = (λ − µ)k g(λ), where g(µ) 6= 0), then dim N (A − µI) = k. We simply find an orthogonal basis of this null space. The process is, as usual, the application of Gaussian elimination to (A − µI)x = 0 and finding k linearly independent vectors in the null space which can then ...
... (meaning that det(A − λI) = (λ − µ)k g(λ), where g(µ) 6= 0), then dim N (A − µI) = k. We simply find an orthogonal basis of this null space. The process is, as usual, the application of Gaussian elimination to (A − µI)x = 0 and finding k linearly independent vectors in the null space which can then ...
Summary of week 8 (Lectures 22, 23 and 24) This week we
... It is easy to generalize the above discussion to the complex case: it is simply necessary to take complex conjugates in a few places. Definition. If A is a matrix over the complex field then we define the conjugate of A to be the matrix A whose (i, j)-entry is the complex conjugate of the (i, j)-en ...
... It is easy to generalize the above discussion to the complex case: it is simply necessary to take complex conjugates in a few places. Definition. If A is a matrix over the complex field then we define the conjugate of A to be the matrix A whose (i, j)-entry is the complex conjugate of the (i, j)-en ...
5QF
... The sum of its diagonal elements equals its rank A diagonal matrix has its eigenvalues equal to the diagonal elements of the matrix The identity matrix has all its eigenvalues equal to 1 Any set of mutually orthonormal vectors can be used as eigenvectors ...
... The sum of its diagonal elements equals its rank A diagonal matrix has its eigenvalues equal to the diagonal elements of the matrix The identity matrix has all its eigenvalues equal to 1 Any set of mutually orthonormal vectors can be used as eigenvectors ...
4 LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
... 13. If V is a vector space of finite dimension and W is a subspace of V, then prove that dim V / W = dim V – dim W. 14. Prove that T A(V ) is invertible if and only if the constant term of the minimal polynomial for T is non-zero. 15. Prove that T A(V ) is invertible if and only if whenever v1, ...
... 13. If V is a vector space of finite dimension and W is a subspace of V, then prove that dim V / W = dim V – dim W. 14. Prove that T A(V ) is invertible if and only if the constant term of the minimal polynomial for T is non-zero. 15. Prove that T A(V ) is invertible if and only if whenever v1, ...
notes
... • We say f (n) = o(g(n)) if for every C > 0, there is an N such that f (n) ≤ Cg(n) for all n ≥ N . This is almost the same as saying f (n) ...
... • We say f (n) = o(g(n)) if for every C > 0, there is an N such that f (n) ≤ Cg(n) for all n ≥ N . This is almost the same as saying f (n) ...
1 The Covariance Matrix
... any symmetric positive semi-definite real-valued matrix Σ has real-valued orthogonal eigenvectors with real eigenvalues. We can see that two eigenvectors with different eigenvalues must be orthogonal by observing the following for any two eigenvectors x and y. yΣx = λx (y · x) ...
... any symmetric positive semi-definite real-valued matrix Σ has real-valued orthogonal eigenvectors with real eigenvalues. We can see that two eigenvectors with different eigenvalues must be orthogonal by observing the following for any two eigenvectors x and y. yΣx = λx (y · x) ...
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.