Mixing
Matrices properties discussion
$begingroup$
I have been thinking about the most interesting property of the matrices that I have learnt and I finally concluded that the one that is the most interesting one is the Cauchy Theorem of determinant which states that for $A, Binmathbb{C}^{n, n} $
$$
det (AB) =det(A) cdot det(B)
$$
My question is what is Your favourite and the most beautiful theorem (or maybe some trick) for matrices?
linear-algebra matrices
asked Feb 3 at 8:52
avan1235avan1235
3578
$endgroup$
add a comment |
$begingroup$
I have been thinking about the most interesting property of the matrices that I have learnt and I finally concluded that the one that is the most interesting one is the Cauchy Theorem of determinant which states that for $A, Binmathbb{C}^{n, n} $
$$
det (AB) =det(A) cdot det(B)
$$
My question is what is Your favourite and the most beautiful theorem (or maybe some trick) for matrices?
linear-algebra matrices
asked Feb 3 at 8:52
avan1235avan1235
3578
$endgroup$
$begingroup$
I got two favorites: SVD and Schur decomposition!
$endgroup$
– Mostafa Ayaz
Feb 3 at 9:10
$begingroup$
@Mostafa Ayaz Can you post the description of them?
$endgroup$
– avan1235
Feb 3 at 9:26
$begingroup$
Sure! Just gimme a second :)
$endgroup$
– Mostafa Ayaz
Feb 3 at 9:26
$begingroup$
"Most interesting ones" : on which creteria ? Is it Usefulness ? There are hundreds of them... Is it Aesthetics ? It depends of one's sense of (mathematical) beauty...
$endgroup$
– Jean Marie
Feb 3 at 11:28
$begingroup$
Any which you would like to choose @Jean Marie
$endgroup$
– avan1235
Feb 3 at 12:24
add a comment |
$begingroup$
I have been thinking about the most interesting property of the matrices that I have learnt and I finally concluded that the one that is the most interesting one is the Cauchy Theorem of determinant which states that for $A, Binmathbb{C}^{n, n} $
$$
det (AB) =det(A) cdot det(B)
$$
My question is what is Your favourite and the most beautiful theorem (or maybe some trick) for matrices?
linear-algebra matrices
asked Feb 3 at 8:52
avan1235avan1235
3578
$endgroup$
I have been thinking about the most interesting property of the matrices that I have learnt and I finally concluded that the one that is the most interesting one is the Cauchy Theorem of determinant which states that for $A, Binmathbb{C}^{n, n} $
$$
det (AB) =det(A) cdot det(B)
$$
My question is what is Your favourite and the most beautiful theorem (or maybe some trick) for matrices?
linear-algebra matrices
linear-algebra matrices
asked Feb 3 at 8:52
avan1235avan1235
3578
asked Feb 3 at 8:52
avan1235avan1235
3578
asked Feb 3 at 8:52
avan1235avan1235
3578
asked Feb 3 at 8:52
avan1235avan1235
3578
asked Feb 3 at 8:52
avan1235avan1235
3578
3578
$begingroup$
I got two favorites: SVD and Schur decomposition!
$endgroup$
– Mostafa Ayaz
Feb 3 at 9:10
$begingroup$
@Mostafa Ayaz Can you post the description of them?
$endgroup$
– avan1235
Feb 3 at 9:26
$begingroup$
Sure! Just gimme a second :)
$endgroup$
– Mostafa Ayaz
Feb 3 at 9:26
$begingroup$
"Most interesting ones" : on which creteria ? Is it Usefulness ? There are hundreds of them... Is it Aesthetics ? It depends of one's sense of (mathematical) beauty...
$endgroup$
– Jean Marie
Feb 3 at 11:28
$begingroup$
Any which you would like to choose @Jean Marie
$endgroup$
– avan1235
Feb 3 at 12:24
add a comment |
$begingroup$
I got two favorites: SVD and Schur decomposition!
$endgroup$
– Mostafa Ayaz
Feb 3 at 9:10
$begingroup$
@Mostafa Ayaz Can you post the description of them?
$endgroup$
– avan1235
Feb 3 at 9:26
$begingroup$
Sure! Just gimme a second :)
$endgroup$
– Mostafa Ayaz
Feb 3 at 9:26
$begingroup$
"Most interesting ones" : on which creteria ? Is it Usefulness ? There are hundreds of them... Is it Aesthetics ? It depends of one's sense of (mathematical) beauty...
$endgroup$
– Jean Marie
Feb 3 at 11:28
$begingroup$
Any which you would like to choose @Jean Marie
$endgroup$
– avan1235
Feb 3 at 12:24
$begingroup$
I got two favorites: SVD and Schur decomposition!
$endgroup$
– Mostafa Ayaz
Feb 3 at 9:10
$begingroup$
I got two favorites: SVD and Schur decomposition!
$endgroup$
– Mostafa Ayaz
Feb 3 at 9:10
$begingroup$
@Mostafa Ayaz Can you post the description of them?
$endgroup$
– avan1235
Feb 3 at 9:26
$begingroup$
@Mostafa Ayaz Can you post the description of them?
$endgroup$
– avan1235
Feb 3 at 9:26
$begingroup$
Sure! Just gimme a second :)
$endgroup$
– Mostafa Ayaz
Feb 3 at 9:26
$begingroup$
Sure! Just gimme a second :)
$endgroup$
– Mostafa Ayaz
Feb 3 at 9:26
$begingroup$
"Most interesting ones" : on which creteria ? Is it Usefulness ? There are hundreds of them... Is it Aesthetics ? It depends of one's sense of (mathematical) beauty...
$endgroup$
– Jean Marie
Feb 3 at 11:28
$begingroup$
"Most interesting ones" : on which creteria ? Is it Usefulness ? There are hundreds of them... Is it Aesthetics ? It depends of one's sense of (mathematical) beauty...
$endgroup$
– Jean Marie
Feb 3 at 11:28
$begingroup$
Any which you would like to choose @Jean Marie
$endgroup$
– avan1235
Feb 3 at 12:24
$begingroup$
Any which you would like to choose @Jean Marie
$endgroup$
– avan1235
Feb 3 at 12:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The Schur decomposition is defined for any square matrix as follows:
For any $ntimes n$ matrix $A$, there exist (non necessarily unique) unitary $Q_{ntimes n}$ and upper-triangular $U_{ntimes n}$ such that:$$A=QUQ^H$$where $^H$ denotes the Hermitian operator (conjugate-transpose operator) and a unitary matrix is any square matrix $Q$ with the property $QQ^H=I$.
Also the SVD is a very important and beautiful matrix decomposition (probably the best ever!) defined for any arbitrary matrix as follows:
The Singular Value Decomposition of an $mtimes n$ matrix $A$ is any triple $(U_{mtimes m},D_{mtimes n},V_{ntimes n})$ such that $$A=UDV^H$$where $U,V$ are unitary and $D$ is diagonal as the non-diagonal entries (i.e. entries $d_{ij}$ with $ine j$) are all zero.
Such a decomposition exists for any matrix.
The existence of SVD for a matrix $A$ is followed a very interesting background. As any matrix $A$ can be interpreted as a linear operator from $Bbb R^n$ onto $Bbb R^m$, one can split the $A$-operation of a vector as below:
Operation 1 : a vector is first rotated in $Bbb R^n$ by multiplying in $V$ ($V$-operation).
Operation 2 : the outcome vector is then multiplied in $D$ which can be interpreted as a scale and transformation from $Bbb R^n$ to $Bbb R^m$ ($D$-operation).
Operation 3 : the final vector is once more rotated in $Bbb R^m$ to yield the result.
SVD enjoys a widespread use in mathematical frameworks. For more information about these two and more useful decompositions (such as Cholesky decomposition for a symmetric non-negative definite matrix), I suggest you the following links:
https://en.wikipedia.org/wiki/Singular_value_decomposition
https://en.wikipedia.org/wiki/Schur_decomposition
https://en.wikipedia.org/wiki/Matrix_decomposition
answered Feb 3 at 9:56
Mostafa AyazMostafa Ayaz
18.1k31040
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add a comment |
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$begingroup$
The Schur decomposition is defined for any square matrix as follows:
For any $ntimes n$ matrix $A$, there exist (non necessarily unique) unitary $Q_{ntimes n}$ and upper-triangular $U_{ntimes n}$ such that:$$A=QUQ^H$$where $^H$ denotes the Hermitian operator (conjugate-transpose operator) and a unitary matrix is any square matrix $Q$ with the property $QQ^H=I$.
Also the SVD is a very important and beautiful matrix decomposition (probably the best ever!) defined for any arbitrary matrix as follows:
The Singular Value Decomposition of an $mtimes n$ matrix $A$ is any triple $(U_{mtimes m},D_{mtimes n},V_{ntimes n})$ such that $$A=UDV^H$$where $U,V$ are unitary and $D$ is diagonal as the non-diagonal entries (i.e. entries $d_{ij}$ with $ine j$) are all zero.
Such a decomposition exists for any matrix.
The existence of SVD for a matrix $A$ is followed a very interesting background. As any matrix $A$ can be interpreted as a linear operator from $Bbb R^n$ onto $Bbb R^m$, one can split the $A$-operation of a vector as below:
Operation 1 : a vector is first rotated in $Bbb R^n$ by multiplying in $V$ ($V$-operation).
Operation 2 : the outcome vector is then multiplied in $D$ which can be interpreted as a scale and transformation from $Bbb R^n$ to $Bbb R^m$ ($D$-operation).
Operation 3 : the final vector is once more rotated in $Bbb R^m$ to yield the result.
SVD enjoys a widespread use in mathematical frameworks. For more information about these two and more useful decompositions (such as Cholesky decomposition for a symmetric non-negative definite matrix), I suggest you the following links:
https://en.wikipedia.org/wiki/Singular_value_decomposition
https://en.wikipedia.org/wiki/Schur_decomposition
https://en.wikipedia.org/wiki/Matrix_decomposition
answered Feb 3 at 9:56
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
The Schur decomposition is defined for any square matrix as follows:
For any $ntimes n$ matrix $A$, there exist (non necessarily unique) unitary $Q_{ntimes n}$ and upper-triangular $U_{ntimes n}$ such that:$$A=QUQ^H$$where $^H$ denotes the Hermitian operator (conjugate-transpose operator) and a unitary matrix is any square matrix $Q$ with the property $QQ^H=I$.
Also the SVD is a very important and beautiful matrix decomposition (probably the best ever!) defined for any arbitrary matrix as follows:
The Singular Value Decomposition of an $mtimes n$ matrix $A$ is any triple $(U_{mtimes m},D_{mtimes n},V_{ntimes n})$ such that $$A=UDV^H$$where $U,V$ are unitary and $D$ is diagonal as the non-diagonal entries (i.e. entries $d_{ij}$ with $ine j$) are all zero.
Such a decomposition exists for any matrix.
The existence of SVD for a matrix $A$ is followed a very interesting background. As any matrix $A$ can be interpreted as a linear operator from $Bbb R^n$ onto $Bbb R^m$, one can split the $A$-operation of a vector as below:
Operation 1 : a vector is first rotated in $Bbb R^n$ by multiplying in $V$ ($V$-operation).
Operation 2 : the outcome vector is then multiplied in $D$ which can be interpreted as a scale and transformation from $Bbb R^n$ to $Bbb R^m$ ($D$-operation).
Operation 3 : the final vector is once more rotated in $Bbb R^m$ to yield the result.
SVD enjoys a widespread use in mathematical frameworks. For more information about these two and more useful decompositions (such as Cholesky decomposition for a symmetric non-negative definite matrix), I suggest you the following links:
https://en.wikipedia.org/wiki/Singular_value_decomposition
https://en.wikipedia.org/wiki/Schur_decomposition
https://en.wikipedia.org/wiki/Matrix_decomposition
answered Feb 3 at 9:56
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
The Schur decomposition is defined for any square matrix as follows:
For any $ntimes n$ matrix $A$, there exist (non necessarily unique) unitary $Q_{ntimes n}$ and upper-triangular $U_{ntimes n}$ such that:$$A=QUQ^H$$where $^H$ denotes the Hermitian operator (conjugate-transpose operator) and a unitary matrix is any square matrix $Q$ with the property $QQ^H=I$.
Also the SVD is a very important and beautiful matrix decomposition (probably the best ever!) defined for any arbitrary matrix as follows:
The Singular Value Decomposition of an $mtimes n$ matrix $A$ is any triple $(U_{mtimes m},D_{mtimes n},V_{ntimes n})$ such that $$A=UDV^H$$where $U,V$ are unitary and $D$ is diagonal as the non-diagonal entries (i.e. entries $d_{ij}$ with $ine j$) are all zero.
Such a decomposition exists for any matrix.
The existence of SVD for a matrix $A$ is followed a very interesting background. As any matrix $A$ can be interpreted as a linear operator from $Bbb R^n$ onto $Bbb R^m$, one can split the $A$-operation of a vector as below:
Operation 1 : a vector is first rotated in $Bbb R^n$ by multiplying in $V$ ($V$-operation).
Operation 2 : the outcome vector is then multiplied in $D$ which can be interpreted as a scale and transformation from $Bbb R^n$ to $Bbb R^m$ ($D$-operation).
Operation 3 : the final vector is once more rotated in $Bbb R^m$ to yield the result.
SVD enjoys a widespread use in mathematical frameworks. For more information about these two and more useful decompositions (such as Cholesky decomposition for a symmetric non-negative definite matrix), I suggest you the following links:
https://en.wikipedia.org/wiki/Singular_value_decomposition
https://en.wikipedia.org/wiki/Schur_decomposition
https://en.wikipedia.org/wiki/Matrix_decomposition
answered Feb 3 at 9:56
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
The Schur decomposition is defined for any square matrix as follows:
For any $ntimes n$ matrix $A$, there exist (non necessarily unique) unitary $Q_{ntimes n}$ and upper-triangular $U_{ntimes n}$ such that:$$A=QUQ^H$$where $^H$ denotes the Hermitian operator (conjugate-transpose operator) and a unitary matrix is any square matrix $Q$ with the property $QQ^H=I$.
Also the SVD is a very important and beautiful matrix decomposition (probably the best ever!) defined for any arbitrary matrix as follows:
The Singular Value Decomposition of an $mtimes n$ matrix $A$ is any triple $(U_{mtimes m},D_{mtimes n},V_{ntimes n})$ such that $$A=UDV^H$$where $U,V$ are unitary and $D$ is diagonal as the non-diagonal entries (i.e. entries $d_{ij}$ with $ine j$) are all zero.
Such a decomposition exists for any matrix.
The existence of SVD for a matrix $A$ is followed a very interesting background. As any matrix $A$ can be interpreted as a linear operator from $Bbb R^n$ onto $Bbb R^m$, one can split the $A$-operation of a vector as below:
Operation 1 : a vector is first rotated in $Bbb R^n$ by multiplying in $V$ ($V$-operation).
Operation 2 : the outcome vector is then multiplied in $D$ which can be interpreted as a scale and transformation from $Bbb R^n$ to $Bbb R^m$ ($D$-operation).
Operation 3 : the final vector is once more rotated in $Bbb R^m$ to yield the result.
SVD enjoys a widespread use in mathematical frameworks. For more information about these two and more useful decompositions (such as Cholesky decomposition for a symmetric non-negative definite matrix), I suggest you the following links:
https://en.wikipedia.org/wiki/Singular_value_decomposition
https://en.wikipedia.org/wiki/Schur_decomposition
https://en.wikipedia.org/wiki/Matrix_decomposition
answered Feb 3 at 9:56
Mostafa AyazMostafa Ayaz
18.1k31040
answered Feb 3 at 9:56
Mostafa AyazMostafa Ayaz
18.1k31040
answered Feb 3 at 9:56
Mostafa AyazMostafa Ayaz
18.1k31040
answered Feb 3 at 9:56
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
add a comment |
add a comment |
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$begingroup$
I got two favorites: SVD and Schur decomposition!
$endgroup$
– Mostafa Ayaz
Feb 3 at 9:10
$begingroup$
@Mostafa Ayaz Can you post the description of them?
$endgroup$
– avan1235
Feb 3 at 9:26
$begingroup$
Sure! Just gimme a second :)
$endgroup$
– Mostafa Ayaz
Feb 3 at 9:26
$begingroup$
"Most interesting ones" : on which creteria ? Is it Usefulness ? There are hundreds of them... Is it Aesthetics ? It depends of one's sense of (mathematical) beauty...
$endgroup$
– Jean Marie
Feb 3 at 11:28
$begingroup$
Any which you would like to choose @Jean Marie
$endgroup$
– avan1235
Feb 3 at 12:24