Karteikarten
Ch. 5: Eigenvalues and Eigenvectors
Karten in diesem Stapel (14)
- 1Define: Eigenvector, EigenvalueAn eigenvector of an n x n matrix A is a nonzero vector x | Ax = (lambda)x
for some scalar lambda. Lambda is an eigenvalue of A if there is a nontrivial solution x of Ax = (lambda)x; such an x is called "an eigenvector corresponding to lambda) - 2Theorem: If {v_1,v_2,...,v_r} are eigenvectors that correspond to distinct eigenvalues of an n x n matrix Athen the set {v_1,v_2,...,v_r} is linearly independent
- 3det A =product of pivots in U (when A is invertible)
0 (when A is not invertible) - 4parts s & t of the IMTLet A be an n x n matrix. Then A is invertible iff:
s. The number 0 is not an eigenvalue of A
t. The determinant A is not zero - 5Properties of DeterminantsLet A and B be n x n matrices:
a. A is invertible iff det(A) != 0
b. det(AB) = det(A) * det(B)
c. det(A^T) = detA
d. If A is triangular, then det(A) is the product of the entries on the main diagonal of A
e. row replacement operations do not change determinant - 6A scalar Lambda is an eigenvalue of an n x n matrix A iff Lambda satisfies the characteristic equation:det(A - Lambda *I) = 0
- 7If n x n matrices of A and B are similar:then they have the same characteristic polynomial and hence the same eigenvalues (with the same multiplicities)
- 8The Diagonalization TheoremAn n x n matrix A is diagonizable iff A has n linearly independent eigenvectors.
In fact, A = PDP^(-1), with D a diagonal matrix, iff the columns of P are n linearly independent eigenvectors of A. In this case, the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors in P. - 9An n x n matrix with n distinct eigenvalues is / isn't diagonalizable (??)An n x n matrix with n distinct eigenvalues is diagonalizable.
- 10Let a be an n x n matrix whose distinct eigenvalues are Lambda_1,... Lambda_p: (part 1)For 1 <= k <= p, the dimension of the eigenspace for Lambda_k is less then or equal to the multiplicity of the eigenvalue Lambda_k
- 11Let a be an n x n matrix whose distinct eigenvalues are Lambda_1,... Lambda_p: (part 2)The Matrix A is diagonalizable iff the sum of the dimensions of the eigenspace equals n, and this happens iff:
1) the characteristic polynomial factors completely factors into linear factors and
2) the dimension of the eigenspace for each Lambda_k equals the multiplicity of Lambda_k - 12Let a be an n x n matrix whose distinct eigenvalues are Lambda_1,... Lambda_p: (part 3)If A is diagonizable and B_k is a basis for the eigenspace corresponding to Lambda_k for each k, then the total collection of vectors in the sets B_1,...B_p forms an eigenvector basis for R^n
- 13Diagonal Matrix RepresentationSuppose A = PDP^(-1), where D is a diagonal n x n matrix. If B is the basis for R^n formed from the columns of P, then D is the B-matrix for the transformation
x -> Ax - 14Let A be a real 2 x 2 matrix with a complex eigenvalue Lambda = a - bi (b != 0) and an associated eigenvector v in C^2. ThenA = PCP^(-1), where P = [Re(v) lmv]
and
C =
[ a -b
b a]
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- Mathematics (Linear Algebra) Karteikarten am Ch. 5: Eigenvalues and Eigenvectors, erstellt von William Hartemink am 12/04/2017.
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