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17 Cards in this Set
- Front
- Back
A matrix is diagonalizable
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if there are enough linearly independent eigenvectors to span the vector space
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Eigenvectors are linearly independent when
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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.
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Characteristic polynomial and diagonalizable operator L
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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.
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degenerate eigenvalue
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if ma(λ0)>1
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deficiency of λ0
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ma(λ0)-mg(λ0)
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operator L is diagonalizable on a complex vector space
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iff all of its eigenvalues have ma=mg
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trace of a matrix A is the
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sum of its diagonal entries
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trace of a linear operator L on a finite dim. vector space V is
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the sume of the diagonal entries of [L]B, where B is any basis of V
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Trick 2
The determinant is |
the product of the eigenvalues
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Trick 1
The trace is |
the sum of the eigenvalues.
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Euler's Formula
e^iϴ |
cosϴ + isinϴ
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Euler's Formula
e^-iϴ |
cosϴ - isinϴ
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Taylor Series
e^x |
∑ (x^n)/n!
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Taylor series
cos x |
∑ (-1ⁿ)(x^2n)/2n!
even powers |
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Taylor series
sin x |
∑ (-1ⁿ)(x^2n+1)/(2n+1)!
even powers |
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if we have diagonalized matrix (A=PDP^-1)
(e^A)= |
(P)(e^D)(P^-1)
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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) |