eigensystems and canonical forms Flashcards
eigenvector:
a vector x is an eigenvalue of A (A is square) if x is nonzero and Ax is a multiple of x
i.e. there is some λ such that Ax=λx
eigenvalue:
λ is an eigenvalue of A if Ax=λx with x being an eigenvector
det(λI-A)=0
eigenpair:
the λ and x that are eigenvalue/vector of A
characteristic polynomial:
p(λ)=det(λI-A) - some polynomial equal to 0
spectrum:
the set of all eigenvalues of A
an nxn matrix has n eigenvalues, so Λ(A)={λ1,…,λn}
invariant subspace:
a subspace X is invariant for A if AX in X - x in X implies Ax in X
how to check if the spectrum of B is contained within the spectrum of A:
let the columns of Y in C^(nxp), p<=n, form a basis for a subspace X of C^n. X is invariant iff AY=YB for some B in C^pxp. when the latter holds, the spectrum of B is contained within that of A
similar:
two matrices A and B are similar if there exists a nonsingular matrix P such that B=P^(-1)AP
similarity transformation:
B=P^(-1)AP
transforming matrix:
P in B=P^(-1)AP
similarity and eigenpairs:
if A and B are similar, A and B have the same eigenvalues, and x is an eigenvector of A with associated eigenvalue λ iff P^(-1)x is an eigenvector of B with the same associated eigenvalue
unitarily similar:
two matrices A and B are unitarily similar if there is a unitary matrix U such that B=U*AU
orthogonally similar:
if A and B are real matrices, they are orthogonally similar if there is a real orthogonal matrix U such that B=U^(T)AU
if A is similar to a diagonal matrix:
A is called diagonalizable and/or simple
schur’s theorem:
let A be a square matrix, then there exists a unitary matrix U and an upper triangular matrix T such that T=U^(-1)AU=U*AU
schur decomposition:
A=UTU*, not unique
schur vector:
the columns of U in a schur decomp
normal:
a matrix A is normal if AA=AA
the spectral theorem:
god what a cool name
A is normal iff there is a unitary matrix U and a diagonal matrix Λ such that A=UΛU*
if A in C^nxn has n orthogonal eigenvectors:
A is normal
an nxn matrix A is diagonalizable iff:
A has n linearly independent eigenvectors
a matrix with distinct eigenvalues is:
diagonalizable
the jordan canonical form:
any square matrix can be expressed in the form X^(-1)AX=J=
[J1(λ1)
…
Jp(λp)]
Jk=Jk(λk)=
[λk 1
λk 1
… …
1
Ak] in C^mkxmk
where X is nonsingular and m1+…+mp=n
jordan block:
the mkxmk matrices in the jcf that go along the diagonal
jordan matrix properties:
the number p of jordan block is the number of linearly independent eigenvectors of A, so A is diagonalizable iff n=p
the algebraic multiplicity of an eigenvalue λ is the sum of dimensions of the jordan blocks in which λ appears
the geometric multiplicity of λ is the number of jordan blocks associated with λ - dim(null(A-λI))
defective:
an eigenvalue is defective if it appears in a jordan block of size greater than 1, or equivalently if its algebraic multiplicity exceeds its geometric multiplicity
a matrix is defective if it has a defective eigenvalue or equivalently if it doesn’t have a complete set of linearly independent eigenvectors
how to find the jordan canonical form of a given matrix A:
find all the distinct eigenvalues of A, maybe by finding the roots of the characteristic polynomial
for each distinct eigenvalue λi of A, form (A-λiI), (A-λiI)^2, … and analyse the sequence of ranks as follows:
the smallest value of ki for which rank(A-λiI)^ki attains its minimum value is the order of the largest block corresponding to λi, this is called the index of λi
the no. of blocks of size k in J with eigenvalue λi is rank(A-λiI)^(k-1)+rank(A-λiI)^(k+1)-2rank(A-λiI)^k
generalised eigenvectors:
let X^(-1)AX=J
AX=XJ
the columns of X in positions 1, m1+1, m1+m2+1,… are eigenvectors of A and are linearly independent as X is nonsingular
the other columns are generalised eigenvectors
jordan chain:
equating the first m1 columns of AX=XJ corresponding to the first jordan block J1 gives us Ax1=λ1x1, Axi=λ1xi+x(i-1), i=2,…,m1
the vectors x1,…,xm1 are called a jordan chain
cayley-hamilton theorem:
if p is the characteristic polynomial of a nxn matrix A, then p(A)=0
minimal polynomial:
let A be a nxn matrix with s distinct eigenvalues λ1,…,λs - the minimal polynomial is q(λ)=(s)Π(i=1)(λ-λi)^ni, where ni is the dimension of the largest jordan block in which λi appears
