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17 Cards in this Set

  • Front
  • Back
A matrix is diagonalizable
if there are enough linearly independent eigenvectors to span the vector space
Eigenvectors are linearly independent when
B={b₁,b₂,..bm} is a collection of eigenvectors of L, corresponding to eigenvalues λ₁,λ₂,..λm, all of which are different. Then B is linearly independent.
Characteristic polynomial and diagonalizable operator L
Let L be an operator on an n-dim vec. sp. If the characteristic polynomial pL(λ) has n distinct roots, the L is diagonalizable.
degenerate eigenvalue
if ma(λ0)>1
deficiency of λ0
ma(λ0)-mg(λ0)
operator L is diagonalizable on a complex vector space
iff all of its eigenvalues have ma=mg
trace of a matrix A is the
sum of its diagonal entries
trace of a linear operator L on a finite dim. vector space V is
the sume of the diagonal entries of [L]B, where B is any basis of V
Trick 2
The determinant is
the product of the eigenvalues
Trick 1
The trace is
the sum of the eigenvalues.
Euler's Formula
e^iϴ
cosϴ + isinϴ
Euler's Formula
e^-iϴ
cosϴ - isinϴ
Taylor Series
e^x
∑ (x^n)/n!
Taylor series
cos x
∑ (-1ⁿ)(x^2n)/2n!

even powers
Taylor series
sin x
∑ (-1ⁿ)(x^2n+1)/(2n+1)!

even powers
if we have diagonalized matrix (A=PDP^-1)

(e^A)=
(P)(e^D)(P^-1)
Discrete-Time Evolution

Find x(n) given A and x(0)
1)Find eigenvalues, eigenvectors, and diagonalize A
2)decompose x(0) by multiplying {P^-1}*{x(0)}
3)solve x(n)=∑ (λⁿ)(aₓ(o))(bₓ)