Eigenvectors and Eigenvalues

Question 1

Eigenvalue

Definition

Answer 1

-a scalar λ∈F is an eigenvalue for T is there exists a non-zero vector v∈V such that:
T(v) = λv

Question 2

Eigenvector

Definition

Answer 2

-a vector v∈V is an Eigen vector for T if for some λ∈F we have:
T(v) = λv

Question 3

Eigenspace

Definition

Answer 3

-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)

Question 4

Eigenvalue and Eigenspace Lemma

Answer 4

λ is an eigenvalue of T:V->V
<=>
Vλ ≠ {|0}

Question 5

Eigenvector

Matrix Definition

Answer 5

-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

Question 6

Eigenvalue

Matrix Definition

Answer 6

λ∈F is an eigenvalue of A if a non-zero column vector v∈F^n exists with:
Av = λv

Question 7

Eigenspace

Matrix Definition

Answer 7

-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}

Question 8

Characteristic Polynomial

Definition

Answer 8

-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

Question 9

How to find the characteristic polynomial of a matrix?

Answer 9

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)

Question 10

Characteristic Polynomial and Eigenvalues Theorem

Answer 10

-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

Question 11

Geometric Multiplicity

Definition

Answer 11

• 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}

Question 12

Algebraic Multiplicity

Definition

Answer 12

• 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

Question 13

How to find the eigenvalues and eigenvectors of a matrix A?

Answer 13

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

Question 14

How to find algebraic and geometric multiplicity?

Answer 14

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

Question 15

Diagonalisable

Definition

Answer 15

• 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

Question 16

Diagonalisability Equivalence Theorem

Answer 16

• 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

Question 17

How to diagonalise a matrix?

Answer 17

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

Question 18

Eigenspaces Lemma

Answer 18

-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

Question 19

Eigenspaces Lemma

Special Case n distinct eigenvalues

Answer 19

• 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

Question 20

Properties of Matrix Determinants

Answer 20

-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

Question 21

Similar Matrices Characteristic Polynomial Theorem

Answer 21

-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

Question 22

Cayley-Hamilton Theorem

2x2 Matrices

Answer 22

-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

Question 23

Cayley Hamilton Theorem

nxn Matrices

Answer 23

-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