Completeness of eigenvectors of Hermitian Matrix.Eigenvectors of a normal matrixOrthonormalization of non-hermitian matrix eigenvectorsHermitian Matrix Unitarily DiagonalizableEigenvectors of non-singular matrixDiagonalizable Matrices vs Hermitian matricesHow can eigenvectors of a Hermitian matrix be entangled?Eigenvectors of Hermitian Toeplitz matrixDoes an n by n Hermitian matrix always has n independent eigenvectors?Eigenvectors of a Hermitian matrixOrthogonality of eigenvectors of a Hermitian matrix
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Completeness of eigenvectors of Hermitian Matrix.
Eigenvectors of a normal matrixOrthonormalization of non-hermitian matrix eigenvectorsHermitian Matrix Unitarily DiagonalizableEigenvectors of non-singular matrixDiagonalizable Matrices vs Hermitian matricesHow can eigenvectors of a Hermitian matrix be entangled?Eigenvectors of Hermitian Toeplitz matrixDoes an n by n Hermitian matrix always has n independent eigenvectors?Eigenvectors of a Hermitian matrixOrthogonality of eigenvectors of a Hermitian matrix
$begingroup$
How do you show that eigenvectors of a Hermitian matrix form a complete set of basis?
linear-algebra matrices eigenvalues-eigenvectors numerical-linear-algebra eigenfunctions
asked Jan 28 '14 at 13:34
Saurabh Uday ShringarpureSaurabh Uday Shringarpure
173137
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bumped to the homepage by Community♦ 17 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
$begingroup$
How do you show that eigenvectors of a Hermitian matrix form a complete set of basis?
linear-algebra matrices eigenvalues-eigenvectors numerical-linear-algebra eigenfunctions
asked Jan 28 '14 at 13:34
Saurabh Uday ShringarpureSaurabh Uday Shringarpure
173137
$endgroup$
bumped to the homepage by Community♦ 17 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
$begingroup$
How do you show that eigenvectors of a Hermitian matrix form a complete set of basis?
linear-algebra matrices eigenvalues-eigenvectors numerical-linear-algebra eigenfunctions
asked Jan 28 '14 at 13:34
Saurabh Uday ShringarpureSaurabh Uday Shringarpure
173137
$endgroup$
How do you show that eigenvectors of a Hermitian matrix form a complete set of basis?
linear-algebra matrices eigenvalues-eigenvectors numerical-linear-algebra eigenfunctions
linear-algebra matrices eigenvalues-eigenvectors numerical-linear-algebra eigenfunctions
asked Jan 28 '14 at 13:34
Saurabh Uday ShringarpureSaurabh Uday Shringarpure
173137
asked Jan 28 '14 at 13:34
Saurabh Uday ShringarpureSaurabh Uday Shringarpure
173137
asked Jan 28 '14 at 13:34
Saurabh Uday ShringarpureSaurabh Uday Shringarpure
173137
asked Jan 28 '14 at 13:34
Saurabh Uday ShringarpureSaurabh Uday Shringarpure
173137
asked Jan 28 '14 at 13:34
Saurabh Uday ShringarpureSaurabh Uday Shringarpure
173137
173137
bumped to the homepage by Community♦ 17 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 17 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
add a comment |
1 Answer
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$begingroup$
In the same way how you do that for normal matrices.
Use or prove the Schur decomposition: for any square matrix $AinmathbbC^ntimes n$, there is a unitary $Q$ and triangular $T$ such that $A=QTQ^*$. A simple but very strong result which can be shown quite simply by induction.
Show that $A$ is Hermitian iff $T$ is Hermitian. A Hermitian triangular matrix is necessarily diagonal.
The eigenvectors can be then found stuffed in the columns of $Q$.
answered Jan 28 '14 at 13:38
Algebraic PavelAlgebraic Pavel
16.5k31840
$endgroup$
add a comment |
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$begingroup$
In the same way how you do that for normal matrices.
Use or prove the Schur decomposition: for any square matrix $AinmathbbC^ntimes n$, there is a unitary $Q$ and triangular $T$ such that $A=QTQ^*$. A simple but very strong result which can be shown quite simply by induction.
Show that $A$ is Hermitian iff $T$ is Hermitian. A Hermitian triangular matrix is necessarily diagonal.
The eigenvectors can be then found stuffed in the columns of $Q$.
answered Jan 28 '14 at 13:38
Algebraic PavelAlgebraic Pavel
16.5k31840
$endgroup$
add a comment |
$begingroup$
In the same way how you do that for normal matrices.
Use or prove the Schur decomposition: for any square matrix $AinmathbbC^ntimes n$, there is a unitary $Q$ and triangular $T$ such that $A=QTQ^*$. A simple but very strong result which can be shown quite simply by induction.
Show that $A$ is Hermitian iff $T$ is Hermitian. A Hermitian triangular matrix is necessarily diagonal.
The eigenvectors can be then found stuffed in the columns of $Q$.
answered Jan 28 '14 at 13:38
Algebraic PavelAlgebraic Pavel
16.5k31840
$endgroup$
add a comment |
$begingroup$
In the same way how you do that for normal matrices.
Use or prove the Schur decomposition: for any square matrix $AinmathbbC^ntimes n$, there is a unitary $Q$ and triangular $T$ such that $A=QTQ^*$. A simple but very strong result which can be shown quite simply by induction.
Show that $A$ is Hermitian iff $T$ is Hermitian. A Hermitian triangular matrix is necessarily diagonal.
The eigenvectors can be then found stuffed in the columns of $Q$.
answered Jan 28 '14 at 13:38
Algebraic PavelAlgebraic Pavel
16.5k31840
$endgroup$
In the same way how you do that for normal matrices.
Use or prove the Schur decomposition: for any square matrix $AinmathbbC^ntimes n$, there is a unitary $Q$ and triangular $T$ such that $A=QTQ^*$. A simple but very strong result which can be shown quite simply by induction.
Show that $A$ is Hermitian iff $T$ is Hermitian. A Hermitian triangular matrix is necessarily diagonal.
The eigenvectors can be then found stuffed in the columns of $Q$.
answered Jan 28 '14 at 13:38
Algebraic PavelAlgebraic Pavel
16.5k31840
answered Jan 28 '14 at 13:38
Algebraic PavelAlgebraic Pavel
16.5k31840
answered Jan 28 '14 at 13:38
Algebraic PavelAlgebraic Pavel
16.5k31840
answered Jan 28 '14 at 13:38
Algebraic PavelAlgebraic Pavel
16.5k31840
16.5k31840
add a comment |
add a comment |
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