tma4115 - calculus 3 - week 14 · week 14 louise meier carlsen norwegian university of science and...

TMA4115 - Calculus 3Week 14

Louise Meier CarlsenNorwegian University of Science and TechnologySpring 2013

www.ntnu.no TMA4115 - Calculus 3, Week 14

Review of last week’s lecture

Last week we introduce and studied

the dimension of a vector space,the rank of a matrix,Markov chains.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 2

Review of last week’s lectureLast week we introduce and studied

the dimension of a vector space,

the rank of a matrix,Markov chains.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 2

Review of last week’s lectureLast week we introduce and studied

the dimension of a vector space,the rank of a matrix,

Markov chains.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 2

Review of last week’s lectureLast week we introduce and studied

the dimension of a vector space,the rank of a matrix,Markov chains.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 2

Today’s lecture

Today we shall introduce and study

eigenvectors, eigenvalues and eigenspaces of squarematrices,the characteristic polynomial of a square matrix.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 3

Today’s lectureToday we shall introduce and study

eigenvectors, eigenvalues and eigenspaces of squarematrices,the characteristic polynomial of a square matrix.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 3

Today’s lectureToday we shall introduce and study

eigenvectors, eigenvalues and eigenspaces of squarematrices,

the characteristic polynomial of a square matrix.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 3

Today’s lectureToday we shall introduce and study

eigenvectors, eigenvalues and eigenspaces of squarematrices,the characteristic polynomial of a square matrix.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 3

Eigenvectors and eigenvalues ofsquare matrices

Let A be an n × n matrix.

An eigenvector of A is a nonzero vector x such thatAx = λx for some scalar λ.An eigenvalue of A is a scalar λ such that the equationAx = λx has a nontrivial solution.If λ is an eigenvalue and x is an eigenvector such thatAx = λx, then x is called an eigenvector (of A)corresponding to λ and λ is called the eigenvalue (of A)corresponding to x.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 4

Eigenvectors and eigenvalues ofsquare matrices

Let A be an n × n matrix.

An eigenvector of A is a nonzero vector x such thatAx = λx for some scalar λ.An eigenvalue of A is a scalar λ such that the equationAx = λx has a nontrivial solution.If λ is an eigenvalue and x is an eigenvector such thatAx = λx, then x is called an eigenvector (of A)corresponding to λ and λ is called the eigenvalue (of A)corresponding to x.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 4

Eigenvectors and eigenvalues ofsquare matrices

Let A be an n × n matrix.An eigenvector of A is a nonzero vector x such thatAx = λx for some scalar λ.

An eigenvalue of A is a scalar λ such that the equationAx = λx has a nontrivial solution.If λ is an eigenvalue and x is an eigenvector such thatAx = λx, then x is called an eigenvector (of A)corresponding to λ and λ is called the eigenvalue (of A)corresponding to x.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 4

Eigenvectors and eigenvalues ofsquare matrices

Let A be an n × n matrix.An eigenvector of A is a nonzero vector x such thatAx = λx for some scalar λ.An eigenvalue of A is a scalar λ such that the equationAx = λx has a nontrivial solution.

If λ is an eigenvalue and x is an eigenvector such thatAx = λx, then x is called an eigenvector (of A)corresponding to λ and λ is called the eigenvalue (of A)corresponding to x.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 4

Eigenvectors and eigenvalues ofsquare matrices

Let A be an n × n matrix.An eigenvector of A is a nonzero vector x such thatAx = λx for some scalar λ.An eigenvalue of A is a scalar λ such that the equationAx = λx has a nontrivial solution.If λ is an eigenvalue and x is an eigenvector such thatAx = λx, then x is called an eigenvector (of A)corresponding to λ and λ is called the eigenvalue (of A)corresponding to x.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 4

Eigenspaces

Let A be an n × n matrix and let λ be an eigenvalue of A.

A vector x ∈ Rn is an eigenvector of A corresponding toλ if and only if x is a nontrivial solution of the equation(A− λIn)x = 0.The set of of all solution of the equation (A− λIn)x = 0(i.e., Nul(A− λIn)) is called the eigenspace (of A)corresponding to λ.Notice that an eigenspace of A is a subspace of Rn.Notice also that 0 belongs to any eigenspace, eventhough it is not an eigenvector.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 5

EigenspacesLet A be an n × n matrix and let λ be an eigenvalue of A.

A vector x ∈ Rn is an eigenvector of A corresponding toλ if and only if x is a nontrivial solution of the equation(A− λIn)x = 0.The set of of all solution of the equation (A− λIn)x = 0(i.e., Nul(A− λIn)) is called the eigenspace (of A)corresponding to λ.Notice that an eigenspace of A is a subspace of Rn.Notice also that 0 belongs to any eigenspace, eventhough it is not an eigenvector.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 5

EigenspacesLet A be an n × n matrix and let λ be an eigenvalue of A.

A vector x ∈ Rn is an eigenvector of A corresponding toλ if and only if x is a nontrivial solution of the equation(A− λIn)x = 0.

The set of of all solution of the equation (A− λIn)x = 0(i.e., Nul(A− λIn)) is called the eigenspace (of A)corresponding to λ.Notice that an eigenspace of A is a subspace of Rn.Notice also that 0 belongs to any eigenspace, eventhough it is not an eigenvector.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 5

EigenspacesLet A be an n × n matrix and let λ be an eigenvalue of A.

A vector x ∈ Rn is an eigenvector of A corresponding toλ if and only if x is a nontrivial solution of the equation(A− λIn)x = 0.The set of of all solution of the equation (A− λIn)x = 0(i.e., Nul(A− λIn)) is called the eigenspace (of A)corresponding to λ.

Notice that an eigenspace of A is a subspace of Rn.Notice also that 0 belongs to any eigenspace, eventhough it is not an eigenvector.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 5

EigenspacesLet A be an n × n matrix and let λ be an eigenvalue of A.

A vector x ∈ Rn is an eigenvector of A corresponding toλ if and only if x is a nontrivial solution of the equation(A− λIn)x = 0.The set of of all solution of the equation (A− λIn)x = 0(i.e., Nul(A− λIn)) is called the eigenspace (of A)corresponding to λ.Notice that an eigenspace of A is a subspace of Rn.

Notice also that 0 belongs to any eigenspace, eventhough it is not an eigenvector.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 5

EigenspacesLet A be an n × n matrix and let λ be an eigenvalue of A.

A vector x ∈ Rn is an eigenvector of A corresponding toλ if and only if x is a nontrivial solution of the equation(A− λIn)x = 0.The set of of all solution of the equation (A− λIn)x = 0(i.e., Nul(A− λIn)) is called the eigenspace (of A)corresponding to λ.Notice that an eigenspace of A is a subspace of Rn.Notice also that 0 belongs to any eigenspace, eventhough it is not an eigenvector.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 5

Linearly independent eigenvectors

Theorem 2If v1, . . . ,vr are eigenvectors that correspond to distincteigenvalues λ1, . . . , λr of an n × n matrix A, then the set{v1, . . . ,vr} is linearly independent.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 6

Linearly independent eigenvectors

Theorem 2If v1, . . . ,vr are eigenvectors that correspond to distincteigenvalues λ1, . . . , λr of an n × n matrix A, then the set{v1, . . . ,vr} is linearly independent.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 6

Finding a basis of an eigenspace

Let A be an n× n matrix and let λ be an eigenvalue of A. Theeigenspace of A corresponding to λ is Nul(A− λIn). So wecan find a basis for the eigenspace of A corresponding to λby:

1 row reducing A− λIn to its reduced echelon form,2 write the solutions to the equation (A− λIn)x = 0 (which

is equivalent to the equation Ax = 0) as a linearcombinations of vectors using the free variables asparameters.

3 Then these vectors form a basis of the eigenspace of Acorresponding to λ.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 7

Finding a basis of an eigenspaceLet A be an n× n matrix and let λ be an eigenvalue of A.

Theeigenspace of A corresponding to λ is Nul(A− λIn). So wecan find a basis for the eigenspace of A corresponding to λby:

1 row reducing A− λIn to its reduced echelon form,2 write the solutions to the equation (A− λIn)x = 0 (which

is equivalent to the equation Ax = 0) as a linearcombinations of vectors using the free variables asparameters.

3 Then these vectors form a basis of the eigenspace of Acorresponding to λ.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 7

Finding a basis of an eigenspaceLet A be an n× n matrix and let λ be an eigenvalue of A. Theeigenspace of A corresponding to λ is Nul(A− λIn).

So wecan find a basis for the eigenspace of A corresponding to λby:

1 row reducing A− λIn to its reduced echelon form,2 write the solutions to the equation (A− λIn)x = 0 (which

is equivalent to the equation Ax = 0) as a linearcombinations of vectors using the free variables asparameters.

3 Then these vectors form a basis of the eigenspace of Acorresponding to λ.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 7

Finding a basis of an eigenspaceLet A be an n× n matrix and let λ be an eigenvalue of A. Theeigenspace of A corresponding to λ is Nul(A− λIn). So wecan find a basis for the eigenspace of A corresponding to λby:

1 row reducing A− λIn to its reduced echelon form,2 write the solutions to the equation (A− λIn)x = 0 (which

is equivalent to the equation Ax = 0) as a linearcombinations of vectors using the free variables asparameters.

3 Then these vectors form a basis of the eigenspace of Acorresponding to λ.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 7

Finding a basis of an eigenspaceLet A be an n× n matrix and let λ be an eigenvalue of A. Theeigenspace of A corresponding to λ is Nul(A− λIn). So wecan find a basis for the eigenspace of A corresponding to λby:

1 row reducing A− λIn to its reduced echelon form,

2 write the solutions to the equation (A− λIn)x = 0 (whichis equivalent to the equation Ax = 0) as a linearcombinations of vectors using the free variables asparameters.

3 Then these vectors form a basis of the eigenspace of Acorresponding to λ.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 7

Finding a basis of an eigenspaceLet A be an n× n matrix and let λ be an eigenvalue of A. Theeigenspace of A corresponding to λ is Nul(A− λIn). So wecan find a basis for the eigenspace of A corresponding to λby:

1 row reducing A− λIn to its reduced echelon form,2 write the solutions to the equation (A− λIn)x = 0 (which

is equivalent to the equation Ax = 0) as a linearcombinations of vectors using the free variables asparameters.

3 Then these vectors form a basis of the eigenspace of Acorresponding to λ.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 7

Finding a basis of an eigenspaceLet A be an n× n matrix and let λ be an eigenvalue of A. Theeigenspace of A corresponding to λ is Nul(A− λIn). So wecan find a basis for the eigenspace of A corresponding to λby:

1 row reducing A− λIn to its reduced echelon form,2 write the solutions to the equation (A− λIn)x = 0 (which

is equivalent to the equation Ax = 0) as a linearcombinations of vectors using the free variables asparameters.

3 Then these vectors form a basis of the eigenspace of Acorresponding to λ.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 7

Example

Let A =

4 −1 62 1 62 −1 8

.

Let us determine if 2 is an eigenvalue of A, and let us find abasis for the corresponding eigenspace if it is.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 8

Finding the eigenvalues of a matrix

Let A be an n × n matrix.

A scalar λ is an eigenvalue of A if and only ifdet(A− λIn) = 0.The equation det(A− λIn) = 0 is called the characteristicequation of A.If we regard λ as an independent variable, thendet(A− λIn) is a polynomial of degree n in λ.det(A− λIn) is called the characteristic polynomial of A.The eigenvalues of A are the zeros of the characteristicpolynomial det(A− λIn) of A.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 9

Finding the eigenvalues of a matrixLet A be an n × n matrix.

A scalar λ is an eigenvalue of A if and only ifdet(A− λIn) = 0.The equation det(A− λIn) = 0 is called the characteristicequation of A.If we regard λ as an independent variable, thendet(A− λIn) is a polynomial of degree n in λ.det(A− λIn) is called the characteristic polynomial of A.The eigenvalues of A are the zeros of the characteristicpolynomial det(A− λIn) of A.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 9

Finding the eigenvalues of a matrixLet A be an n × n matrix.

A scalar λ is an eigenvalue of A if and only ifdet(A− λIn) = 0.

The equation det(A− λIn) = 0 is called the characteristicequation of A.If we regard λ as an independent variable, thendet(A− λIn) is a polynomial of degree n in λ.det(A− λIn) is called the characteristic polynomial of A.The eigenvalues of A are the zeros of the characteristicpolynomial det(A− λIn) of A.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 9

Finding the eigenvalues of a matrixLet A be an n × n matrix.

A scalar λ is an eigenvalue of A if and only ifdet(A− λIn) = 0.The equation det(A− λIn) = 0 is called the characteristicequation of A.

If we regard λ as an independent variable, thendet(A− λIn) is a polynomial of degree n in λ.det(A− λIn) is called the characteristic polynomial of A.The eigenvalues of A are the zeros of the characteristicpolynomial det(A− λIn) of A.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 9

Finding the eigenvalues of a matrixLet A be an n × n matrix.

A scalar λ is an eigenvalue of A if and only ifdet(A− λIn) = 0.The equation det(A− λIn) = 0 is called the characteristicequation of A.If we regard λ as an independent variable, thendet(A− λIn) is a polynomial of degree n in λ.

det(A− λIn) is called the characteristic polynomial of A.The eigenvalues of A are the zeros of the characteristicpolynomial det(A− λIn) of A.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 9

Finding the eigenvalues of a matrixLet A be an n × n matrix.

A scalar λ is an eigenvalue of A if and only ifdet(A− λIn) = 0.The equation det(A− λIn) = 0 is called the characteristicequation of A.If we regard λ as an independent variable, thendet(A− λIn) is a polynomial of degree n in λ.det(A− λIn) is called the characteristic polynomial of A.

The eigenvalues of A are the zeros of the characteristicpolynomial det(A− λIn) of A.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 9

Finding the eigenvalues of a matrixLet A be an n × n matrix.

A scalar λ is an eigenvalue of A if and only ifdet(A− λIn) = 0.The equation det(A− λIn) = 0 is called the characteristicequation of A.If we regard λ as an independent variable, thendet(A− λIn) is a polynomial of degree n in λ.det(A− λIn) is called the characteristic polynomial of A.The eigenvalues of A are the zeros of the characteristicpolynomial det(A− λIn) of A.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 9

ExampleLet us find the eigenvalues of the matrix A =

[3 04 −2

].

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 10

ExampleIn a certain region, 5% of a city’s population moves to thesurrounding suburbs each year, and 3% of the suburbanpopulation moves into the city. In 2000, there were 600,000residents in the city and 400,000 in the suburbs.Let us try to find a formula for the number of the people in thecity and the number of people in the suburbs for each year.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 11

Markov chains

A probability vector is a vector with nonnegative entriesthat add up to 1.A stochastic matrix is a square matrix whose columnsare probability vectors.A Markov chain is a sequence x0,x1,x2, . . . of probabilityvectors together with a stochastic matrix P such thatx1 = Px0, x2 = Px1, x3 = Px2, . . . .A steady-state vector (or an equilibrium vector) for astochastic matrix P is a probability vector q such thatPq = q.So a steady-state vector for a stochastic matrix P is aprobability vector which is an eigenvector of Pcorresponding to the eigenvalue 1.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 12

Markov chainsA probability vector is a vector with nonnegative entriesthat add up to 1.

A stochastic matrix is a square matrix whose columnsare probability vectors.A Markov chain is a sequence x0,x1,x2, . . . of probabilityvectors together with a stochastic matrix P such thatx1 = Px0, x2 = Px1, x3 = Px2, . . . .A steady-state vector (or an equilibrium vector) for astochastic matrix P is a probability vector q such thatPq = q.So a steady-state vector for a stochastic matrix P is aprobability vector which is an eigenvector of Pcorresponding to the eigenvalue 1.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 12

Markov chainsA probability vector is a vector with nonnegative entriesthat add up to 1.A stochastic matrix is a square matrix whose columnsare probability vectors.

A Markov chain is a sequence x0,x1,x2, . . . of probabilityvectors together with a stochastic matrix P such thatx1 = Px0, x2 = Px1, x3 = Px2, . . . .A steady-state vector (or an equilibrium vector) for astochastic matrix P is a probability vector q such thatPq = q.So a steady-state vector for a stochastic matrix P is aprobability vector which is an eigenvector of Pcorresponding to the eigenvalue 1.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 12

Markov chainsA probability vector is a vector with nonnegative entriesthat add up to 1.A stochastic matrix is a square matrix whose columnsare probability vectors.A Markov chain is a sequence x0,x1,x2, . . . of probabilityvectors together with a stochastic matrix P such thatx1 = Px0, x2 = Px1, x3 = Px2, . . . .

A steady-state vector (or an equilibrium vector) for astochastic matrix P is a probability vector q such thatPq = q.So a steady-state vector for a stochastic matrix P is aprobability vector which is an eigenvector of Pcorresponding to the eigenvalue 1.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 12

Markov chainsA probability vector is a vector with nonnegative entriesthat add up to 1.A stochastic matrix is a square matrix whose columnsare probability vectors.A Markov chain is a sequence x0,x1,x2, . . . of probabilityvectors together with a stochastic matrix P such thatx1 = Px0, x2 = Px1, x3 = Px2, . . . .A steady-state vector (or an equilibrium vector) for astochastic matrix P is a probability vector q such thatPq = q.

So a steady-state vector for a stochastic matrix P is aprobability vector which is an eigenvector of Pcorresponding to the eigenvalue 1.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 12

Markov chainsA probability vector is a vector with nonnegative entriesthat add up to 1.A stochastic matrix is a square matrix whose columnsare probability vectors.A Markov chain is a sequence x0,x1,x2, . . . of probabilityvectors together with a stochastic matrix P such thatx1 = Px0, x2 = Px1, x3 = Px2, . . . .A steady-state vector (or an equilibrium vector) for astochastic matrix P is a probability vector q such thatPq = q.So a steady-state vector for a stochastic matrix P is aprobability vector which is an eigenvector of Pcorresponding to the eigenvalue 1.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 12

Problem 7 from the exam fromAugust 2010

Let A =

2 0 21 −2 −1−1 6 5

.

1 Solve the equation Ax = 0.2 Find the eigenvalues and eigenvectors of A.

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 13

Problem 6 from August 2012For which numbers a does R2 have a basis of eigenvectors

of the matrix[0 a1 0

]?

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 14

Monday’s lectureTomorrow we shall look at

diagonal matrices,how to diagonalizable a matrix.

Section 5.3 in “Linear Algebras and Its Applications” (pages281—288).

www.ntnu.no TMA4115 - Calculus 3, Week 14, page 15