Eigenvectors and Eigenvalues Flashcards
Eigenvalue
Definition
-a scalar λ∈F is an eigenvalue for T is there exists a non-zero vector v∈V such that:
T(v) = λv
Eigenvector
Definition
-a vector v∈V is an Eigen vector for T if for some λ∈F we have:
T(v) = λv
Eigenspace
Definition
-if λ∈F the λ-eigenspace of T is;
Vλ = {v | T(v) = λv}
= {v∈V | T(v) = λI(v)}
= {v∈V | (T-λI)v = 0 }
= ker (T - λI)
-hence Vλ is a subspace of V (since it is the kernel of a linear map)Eigenvalue and Eigenspace Lemma
λ is an eigenvalue of T:V->V
<=>
Vλ ≠ {|0}
Eigenvector
Matrix Definition
-let A be an nxn matrix
-a column vector v∈F^n is called an eigenvector of A if for some λ∈F :
Av = λv , i.e. if (A-λI)v=0
Eigenvalue
Matrix Definition
λ∈F is an eigenvalue of A if a non-zero column vector v∈F^n exists with:
Av = λv
Eigenspace
Matrix Definition
-if λ∈F, the eigenspace Vλ of λ is the null space of A-λI i.e. :
Vλ = {v∈F^n | Av = λv } = {v∈F^n | (A-λI)v = 0}
Characteristic Polynomial
Definition
-the characteristic polynomial of an nxn matrix A is;
X(t) = det (A-tI)
= (-1)^n t^n + Cn-1t^(n-1) + … + C1t + Co
-X(t) is a polynomial of degree n
-an equivalent convention is X(t) = det (tI-A), they only differ by a factor of -1 so have the same roots
How to find the characteristic polynomial of a matrix?
1) recall, X(t) = det (A-tI) and sub in
2) find the determinant
3) simplify to a polynomial of degree n (where A is an nxn matrix)
Characteristic Polynomial and Eigenvalues Theorem
-let A be a square matrix with entries in F
-the eigenvalues of F are the roots of its characteristic polynomial i.e.
λ is an eigenvalue of A <=> Xa(λ)=0
Proof:
λ is an eigenvalue of A <=> Av = λv for some non-zero v
<=> (A-λI)v=0
<=> the matrix A-λI is not invertible (i.e. is singular)
<=> det(A-λI) = 0
<=> Xa(λ)=0
Geometric Multiplicity
Definition
- let λ be an Eigen value of nxn matrix A with entries in F, then there exists a column vector v∈F^n such that Av = λv , so (A-λI)v=0
- the geometric multiplicity of λ is the dimension of the λ-eigenspace of A = dim {v∈F^n | (A-λI)v=0}
Algebraic Multiplicity
Definition
- let λ be an Eigen value of nxn matrix A with entries in F, then there exists a column vector v∈F^n such that Av = λv , so (A-λI)v=0
- the algebraic multiplicity of λ is the multiplicity of λ as a root of the characteristic polynomial Xa(t), i.e. the number of times it is repeated as a root of the characteristic polynomial
How to find the eigenvalues and eigenvectors of a matrix A?
1) find the characteristic polynomial using the equation det(A-λI) = 0
2) the roots of this equation are the eigenvalues
3) one at a time substitute each eigenvalue into the equation (A-λI)v=0 to find the corresponding eigenvector for each eigenvalue
How to find algebraic and geometric multiplicity?
1) find the eigenvalues and eigenvectors
2) the number of time each eigenvalue is repeated as a root of the characteristic polynomial is its geometric multiplicity
3) write the eigenvector of each eigenvalue as a span and then find the dimension, this is the algebraic multiplicity
Diagonalisable
Definition
- a matrix A is diagonalisable if it is similar to a diagonal matrix i.e. A is diagonalisable if there exists a non-singular (invertible) matrix P such that P^(-1) A P = Λ with Λ diagonal
- where the columns of P correspond to the eigenvectors of the eigenvalues that form the diagonal of Λ, i.e. the nth column of P is the eigenvector that corresponds to the eigenvalue in the nth column of Λ
Diagonalisability Equivalence Theorem
- let A be an nxn matrix, the following are equivalent:
i) A is diagonalisable
ii) A has n linearly independent eigenvectors - an additional statement of equivalence can be made if A is a matrix over the field of complex numbers, C :
iii) for all eigenvalues, the geometric multiplicity is equal to the geometric multplicity
How to diagonalise a matrix?
1) find n linearly independent eigenvectors (if this is not possible then the matrix is not diagonalisable)
2) columns of P = the eigenvectors of A
3) Λ = nxn diagonal matrix with entries corresponding to the eigenvalues of the eigenvectors in P
4) P^(-1) A P = Λ
Eigenspaces Lemma
-let A be an nxn matrix with entries in F, λ1,…,λm are the distinct eigenvalues of A
-for i = 1 to m consider basis Ri of the eigenspace:
Vλi = { v∈F^n | (A - λi I ) |v = |0 }
-the set R1∩R2∩R3∩…∩Rm obtained by putting together all of the bases Ri of eigenspaces Vλi is linearly independent
Eigenspaces Lemma
Special Case n distinct eigenvalues
- let A be an nxn matrix with entries in F, λ1,…,λm are the distinct eigenvalues of A
- non-zero eigenvectors v1,v2,…,vm associated to different eigenvalues λ1,…,λm of a matrix are linearly independent
- thus if nxn matrix A has n distinct eigenvalues then A is diagonalisable since it will also have n linearly independent eigenvectors
Properties of Matrix Determinants
-if A, B are nxn matrices, then det(AB) = detA detB
-if A is invertible (detA ≠ 0) then det(A^(-1)) = 1/ detA
-similar matrices A and B have the same determinant since B = PAP^(-1)
det B = det (PAP^(-1)) = detP det A det (P^(-1)) = detPdetA/detP = detA
Similar Matrices Characteristic Polynomial Theorem
-similar matrices (A & B) have the same characteristic polynomial:
Χb(t) = det(B - tI) = det(P^(-1)AP - tP^(-1)P) =
det(P^(-1)(A-t)P) = det(P^(-1)) det(A-It) det(P) = det(A-tI) = Xa(t)
-hence if Χb(t) ≠ Xa(t) , then matrices A and B cannot be similar
Cayley-Hamilton Theorem
2x2 Matrices
-let A be a 2x2 matrix
-let p(t) = Xa(t) = t² + mt + n be the characteristic polynomial of A
-then A is a root of it characteristic polynomial:
p(A) = A² + mA + nI = |0
-where I is the 2x2 identity matrix
Cayley Hamilton Theorem
nxn Matrices
-let A be an nxn matrix
-let p(t) = t^n + a_n-1 * t^(n-1) + … + a2t² + a1t + ao be the characteristic polynomial of A
-then p(A) = 0 where 0 is the nxn zero matrix:
A^n + a_n-1A^(n-1) + … + a2A² + a1A + a0I = |0
-where I is the nxn identity matrix
