5 Flashcards

Diagonalisation

1
Q

Suppose that A is a square matrix. The number λ is said to be an…

A

eigenvalue of A if for some non-zero vector x, Ax = λx. Any non-zero vector x for which this equation holds is called an eigenvector.

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2
Q

Steps to find eigenvalues and eigenvectors of A

A
  1. The eigenvalues of A are the λ values that satisfy Ax = λx, equivalent to Ax = λIx or (A - λI)x = 0. The eigenvectors of A for eigenvalue λ are any non-zero vector x for which this equation holds. This has a solution other than x=0 when determinant I A - λI I =0, meaning we can solve it to find the eigenvalues of A since eigenvectors x can never be zero vectors.
  2. Find and state characteristic polynomial p(λ).
  3. Factorise p(λ) to find eigenvalues.
  4. To find an eigenvector for λ1 , solve equation (A-λ1I )x = 0.
  5. Any non-zero multiple of this eigenvector is an eigenvector of A for λ1.
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3
Q

Square matrices A and B are similar if

A

there is an invertible matrix P such that P-1AP = B.

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4
Q

The matrix A is diagonalisable if it is

A

similar to a diagonal matrix, i.e. if there is a diagonal matrix D such that P-1AP=D.

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5
Q

A matrix A is diagonalisable if and only if

A

it has n linearly independent igenvectors.

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6
Q

Symmetric: matrices (transpose AT is equal to itself) are always

A

diagonalisable.

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7
Q

A matrix P is orthogonal if

A

PTP = PPT = I, i.e. if P has inverse PT.

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8
Q

A matrix A is orthogonally diagonalisable if

A

there is an orthogonal matrix P such that
PTAP = D where D is a diagonal matrix.

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9
Q

The condition that PTP = I (for matrix P to be orthogonal) means

A
  • any two of the eigenvectors x1,.., xn must
    be orthogonal xiTxj = 0 (i ≠ j)
  • each eigenvector must have length 1, i.e. xiTxj = 0
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10
Q

If the matrix A is symmetric (AT = A) then

A
  • eigenvectors corresponding to different eigenvalues are orthogonal.
  • A is orthogonally diagonalisable
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11
Q

Length of a vector x is

A

IIxII = √(Σxi2) (i=1 to n)

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12
Q

When to normalise vectors of P

A
  • When A is a symmetric matrix so P-1=PT
  • When D = PTAP
  • Find an orthogonal matrix P
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