5 Flashcards
Diagonalisation
Suppose that A is a square matrix. The number λ is said to be an…
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.
Steps to find eigenvalues and eigenvectors of A
- 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.
- Find and state characteristic polynomial p(λ).
- Factorise p(λ) to find eigenvalues.
- To find an eigenvector for λ1 , solve equation (A-λ1I )x = 0.
- Any non-zero multiple of this eigenvector is an eigenvector of A for λ1.
Square matrices A and B are similar if
there is an invertible matrix P such that P-1AP = B.
The matrix A is diagonalisable if it is
similar to a diagonal matrix, i.e. if there is a diagonal matrix D such that P-1AP=D.
A matrix A is diagonalisable if and only if
it has n linearly independent igenvectors.
Symmetric: matrices (transpose AT is equal to itself) are always
diagonalisable.
A matrix P is orthogonal if
PTP = PPT = I, i.e. if P has inverse PT.
A matrix A is orthogonally diagonalisable if
there is an orthogonal matrix P such that
PTAP = D where D is a diagonal matrix.
The condition that PTP = I (for matrix P to be orthogonal) means
- any two of the eigenvectors x1,.., xn must
be orthogonal xiTxj = 0 (i ≠ j) - each eigenvector must have length 1, i.e. xiTxj = 0
If the matrix A is symmetric (AT = A) then
- eigenvectors corresponding to different eigenvalues are orthogonal.
- A is orthogonally diagonalisable
Length of a vector x is
IIxII = √(Σxi2) (i=1 to n)
When to normalise vectors of P
- When A is a symmetric matrix so P-1=PT
- When D = PTAP
- Find an orthogonal matrix P