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TMA4115 - Calculus 3Week 14
Louise Meier CarlsenNorwegian University of Science and TechnologySpring 2013
www.ntnu.no TMA4115 - Calculus 3, Week 14
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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
![Page 3: TMA4115 - Calculus 3 - Week 14 · Week 14 Louise Meier Carlsen Norwegian University of Science and Technology Spring 2013 TMA4115 - Calculus 3, Week 14. Review of last week’s lecture](https://reader035.vdocument.in/reader035/viewer/2022062508/60562390e906ce4ee665a767/html5/thumbnails/3.jpg)
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
![Page 4: TMA4115 - Calculus 3 - Week 14 · Week 14 Louise Meier Carlsen Norwegian University of Science and Technology Spring 2013 TMA4115 - Calculus 3, Week 14. Review of last week’s lecture](https://reader035.vdocument.in/reader035/viewer/2022062508/60562390e906ce4ee665a767/html5/thumbnails/4.jpg)
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
![Page 5: TMA4115 - Calculus 3 - Week 14 · Week 14 Louise Meier Carlsen Norwegian University of Science and Technology Spring 2013 TMA4115 - Calculus 3, Week 14. Review of last week’s lecture](https://reader035.vdocument.in/reader035/viewer/2022062508/60562390e906ce4ee665a767/html5/thumbnails/5.jpg)
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
![Page 6: TMA4115 - Calculus 3 - Week 14 · Week 14 Louise Meier Carlsen Norwegian University of Science and Technology Spring 2013 TMA4115 - Calculus 3, Week 14. Review of last week’s lecture](https://reader035.vdocument.in/reader035/viewer/2022062508/60562390e906ce4ee665a767/html5/thumbnails/6.jpg)
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
![Page 7: TMA4115 - Calculus 3 - Week 14 · Week 14 Louise Meier Carlsen Norwegian University of Science and Technology Spring 2013 TMA4115 - Calculus 3, Week 14. Review of last week’s lecture](https://reader035.vdocument.in/reader035/viewer/2022062508/60562390e906ce4ee665a767/html5/thumbnails/7.jpg)
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
![Page 8: TMA4115 - Calculus 3 - Week 14 · Week 14 Louise Meier Carlsen Norwegian University of Science and Technology Spring 2013 TMA4115 - Calculus 3, Week 14. Review of last week’s lecture](https://reader035.vdocument.in/reader035/viewer/2022062508/60562390e906ce4ee665a767/html5/thumbnails/8.jpg)
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
![Page 9: TMA4115 - Calculus 3 - Week 14 · Week 14 Louise Meier Carlsen Norwegian University of Science and Technology Spring 2013 TMA4115 - Calculus 3, Week 14. Review of last week’s lecture](https://reader035.vdocument.in/reader035/viewer/2022062508/60562390e906ce4ee665a767/html5/thumbnails/9.jpg)
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
![Page 10: TMA4115 - Calculus 3 - Week 14 · Week 14 Louise Meier Carlsen Norwegian University of Science and Technology Spring 2013 TMA4115 - Calculus 3, Week 14. Review of last week’s lecture](https://reader035.vdocument.in/reader035/viewer/2022062508/60562390e906ce4ee665a767/html5/thumbnails/10.jpg)
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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
![Page 29: TMA4115 - Calculus 3 - Week 14 · Week 14 Louise Meier Carlsen Norwegian University of Science and Technology Spring 2013 TMA4115 - Calculus 3, Week 14. Review of last week’s lecture](https://reader035.vdocument.in/reader035/viewer/2022062508/60562390e906ce4ee665a767/html5/thumbnails/29.jpg)
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
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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
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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
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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
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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
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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
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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
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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
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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.
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ExampleLet us find the eigenvalues of the matrix A =
[3 04 −2
].
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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Problem 6 from August 2012For which numbers a does R2 have a basis of eigenvectors
of the matrix[0 a1 0
]?
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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).
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