Flashcards in Eigenvectors and Eigenvalues Deck (23):

1

##
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

2

##
Eigenvector

Definition

###
-a vector v∈V is an Eigen vector for T if for some λ∈F we have:

T(v) = λv

3

##
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)

4

## Eigenvalue and Eigenspace Lemma

###
λ is an eigenvalue of T:V->V

<=>

Vλ ≠ {|0}

5

##
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

6

##
Eigenvalue

Matrix Definition

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

Av = λv

7

##
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}

8

##
Characteristic Polynomial

Definition

###
-the characteristic polynomial of an nxn matrix A is;

X(t) = det (A-tI)

= (-1)^n t^n + Cn-1*t^(n-1) + ... + C1*t + 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

9

## 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)

10

## 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

11

##
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}

12

##
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

13

## 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

14

## 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

15

##
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

16

## 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

17

## 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

18

## 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

19

##
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

20

## 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

21

## 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

22

##
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

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