Is there a name for this 'multiplication table' operation of vectors?
Is there a name for an operation or a transformation that generates a matrix from two vectors, creating a multiplication table from its elements? This might be like multiplying a vector $mathbf{s}$ to another column vector $mathbf{v}$ as if $mathbf{s}$ were a scalar.
$$Mleft( begin{bmatrix} v_1 \ v_2 \ v_3 end{bmatrix}, begin{bmatrix} s_1 \ s_2 \ s_3 end{bmatrix} right) = begin{bmatrix} mathbf{s}v_1 \ mathbf{s}v_2 \ mathbf{s}v_3 end{bmatrix} = begin{bmatrix} s_1v_1 & s_2v_1& s_3v_1 \ s_1v_2 & s_2v_2 & s_3v_2 \ s_1v_3 & s_2v_3 & s_3v_3end {bmatrix}$$
"Multiplication table matrix" and "vector as a scalar" gave me nothing useful in google. If there is no such name I would also appreciate suggestions on a better way to notate this besides writing out the entire matrix.
linear-algebra matrices vectors products
add a comment |
Is there a name for an operation or a transformation that generates a matrix from two vectors, creating a multiplication table from its elements? This might be like multiplying a vector $mathbf{s}$ to another column vector $mathbf{v}$ as if $mathbf{s}$ were a scalar.
$$Mleft( begin{bmatrix} v_1 \ v_2 \ v_3 end{bmatrix}, begin{bmatrix} s_1 \ s_2 \ s_3 end{bmatrix} right) = begin{bmatrix} mathbf{s}v_1 \ mathbf{s}v_2 \ mathbf{s}v_3 end{bmatrix} = begin{bmatrix} s_1v_1 & s_2v_1& s_3v_1 \ s_1v_2 & s_2v_2 & s_3v_2 \ s_1v_3 & s_2v_3 & s_3v_3end {bmatrix}$$
"Multiplication table matrix" and "vector as a scalar" gave me nothing useful in google. If there is no such name I would also appreciate suggestions on a better way to notate this besides writing out the entire matrix.
linear-algebra matrices vectors products
2
Kronecker product
– Michael Burr
Nov 28 '18 at 2:48
Thank you sir, this is what I was looking for.
– Matthew11
Nov 28 '18 at 2:52
1
Your matrix $M(mathbf{v}, mathbf{s})$ also has the form of $mathbf{v} mathbf{s}^{rm T}$.
– Rócherz
Nov 28 '18 at 2:52
1
Actually, for a pair of vectors, this is called the outer product. The Kronecker product is a generalization of it to matrices.
– amd
Nov 28 '18 at 2:58
Yes it seems you are correct Mr. Amd, thank you for the clarification.
– Matthew11
Nov 28 '18 at 3:09
add a comment |
Is there a name for an operation or a transformation that generates a matrix from two vectors, creating a multiplication table from its elements? This might be like multiplying a vector $mathbf{s}$ to another column vector $mathbf{v}$ as if $mathbf{s}$ were a scalar.
$$Mleft( begin{bmatrix} v_1 \ v_2 \ v_3 end{bmatrix}, begin{bmatrix} s_1 \ s_2 \ s_3 end{bmatrix} right) = begin{bmatrix} mathbf{s}v_1 \ mathbf{s}v_2 \ mathbf{s}v_3 end{bmatrix} = begin{bmatrix} s_1v_1 & s_2v_1& s_3v_1 \ s_1v_2 & s_2v_2 & s_3v_2 \ s_1v_3 & s_2v_3 & s_3v_3end {bmatrix}$$
"Multiplication table matrix" and "vector as a scalar" gave me nothing useful in google. If there is no such name I would also appreciate suggestions on a better way to notate this besides writing out the entire matrix.
linear-algebra matrices vectors products
Is there a name for an operation or a transformation that generates a matrix from two vectors, creating a multiplication table from its elements? This might be like multiplying a vector $mathbf{s}$ to another column vector $mathbf{v}$ as if $mathbf{s}$ were a scalar.
$$Mleft( begin{bmatrix} v_1 \ v_2 \ v_3 end{bmatrix}, begin{bmatrix} s_1 \ s_2 \ s_3 end{bmatrix} right) = begin{bmatrix} mathbf{s}v_1 \ mathbf{s}v_2 \ mathbf{s}v_3 end{bmatrix} = begin{bmatrix} s_1v_1 & s_2v_1& s_3v_1 \ s_1v_2 & s_2v_2 & s_3v_2 \ s_1v_3 & s_2v_3 & s_3v_3end {bmatrix}$$
"Multiplication table matrix" and "vector as a scalar" gave me nothing useful in google. If there is no such name I would also appreciate suggestions on a better way to notate this besides writing out the entire matrix.
linear-algebra matrices vectors products
linear-algebra matrices vectors products
asked Nov 28 '18 at 2:43
Matthew11
312110
312110
2
Kronecker product
– Michael Burr
Nov 28 '18 at 2:48
Thank you sir, this is what I was looking for.
– Matthew11
Nov 28 '18 at 2:52
1
Your matrix $M(mathbf{v}, mathbf{s})$ also has the form of $mathbf{v} mathbf{s}^{rm T}$.
– Rócherz
Nov 28 '18 at 2:52
1
Actually, for a pair of vectors, this is called the outer product. The Kronecker product is a generalization of it to matrices.
– amd
Nov 28 '18 at 2:58
Yes it seems you are correct Mr. Amd, thank you for the clarification.
– Matthew11
Nov 28 '18 at 3:09
add a comment |
2
Kronecker product
– Michael Burr
Nov 28 '18 at 2:48
Thank you sir, this is what I was looking for.
– Matthew11
Nov 28 '18 at 2:52
1
Your matrix $M(mathbf{v}, mathbf{s})$ also has the form of $mathbf{v} mathbf{s}^{rm T}$.
– Rócherz
Nov 28 '18 at 2:52
1
Actually, for a pair of vectors, this is called the outer product. The Kronecker product is a generalization of it to matrices.
– amd
Nov 28 '18 at 2:58
Yes it seems you are correct Mr. Amd, thank you for the clarification.
– Matthew11
Nov 28 '18 at 3:09
2
2
Kronecker product
– Michael Burr
Nov 28 '18 at 2:48
Kronecker product
– Michael Burr
Nov 28 '18 at 2:48
Thank you sir, this is what I was looking for.
– Matthew11
Nov 28 '18 at 2:52
Thank you sir, this is what I was looking for.
– Matthew11
Nov 28 '18 at 2:52
1
1
Your matrix $M(mathbf{v}, mathbf{s})$ also has the form of $mathbf{v} mathbf{s}^{rm T}$.
– Rócherz
Nov 28 '18 at 2:52
Your matrix $M(mathbf{v}, mathbf{s})$ also has the form of $mathbf{v} mathbf{s}^{rm T}$.
– Rócherz
Nov 28 '18 at 2:52
1
1
Actually, for a pair of vectors, this is called the outer product. The Kronecker product is a generalization of it to matrices.
– amd
Nov 28 '18 at 2:58
Actually, for a pair of vectors, this is called the outer product. The Kronecker product is a generalization of it to matrices.
– amd
Nov 28 '18 at 2:58
Yes it seems you are correct Mr. Amd, thank you for the clarification.
– Matthew11
Nov 28 '18 at 3:09
Yes it seems you are correct Mr. Amd, thank you for the clarification.
– Matthew11
Nov 28 '18 at 3:09
add a comment |
1 Answer
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Let $mathbf{v} =begin{bmatrix} v_1 \ v_2 \ v_3 end{bmatrix}$ and $mathbf{s} =begin{bmatrix} s_1 \ s_2 \ s_3 end{bmatrix}$. Define
$$M(mathbf{v}, mathbf{s}) =begin{bmatrix} v_1mathbf{s} \ v_2mathbf{s} \ v_3mathbf{s} end{bmatrix} = begin{bmatrix} v_1s_1 &v_1s_2& v_1s_3 \ v_2s_1 & v_2s_2 & v_2s_3 \ v_3s_1 & v_3s_2 & v_3s_3end{bmatrix}.$$
Your matrix $M(mathbf{v}, mathbf{s})$ happens to be algebraically equal to these expressions:
- the outer product $mathbf{v} mathbf{s}^{rm T}$,
- the unnamed product $mathbf{v} operatorname{diag}(mathbf{s})$ where $operatorname{diag}(mathbf{s})$ is a diagonal matrix with the entries of $mathbf{s}$ along the diagonal, and
- a Kronecker matrix product of the form $mathbf{v} otimes mathbf{s}^{rm T}$.
add a comment |
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1 Answer
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1 Answer
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Let $mathbf{v} =begin{bmatrix} v_1 \ v_2 \ v_3 end{bmatrix}$ and $mathbf{s} =begin{bmatrix} s_1 \ s_2 \ s_3 end{bmatrix}$. Define
$$M(mathbf{v}, mathbf{s}) =begin{bmatrix} v_1mathbf{s} \ v_2mathbf{s} \ v_3mathbf{s} end{bmatrix} = begin{bmatrix} v_1s_1 &v_1s_2& v_1s_3 \ v_2s_1 & v_2s_2 & v_2s_3 \ v_3s_1 & v_3s_2 & v_3s_3end{bmatrix}.$$
Your matrix $M(mathbf{v}, mathbf{s})$ happens to be algebraically equal to these expressions:
- the outer product $mathbf{v} mathbf{s}^{rm T}$,
- the unnamed product $mathbf{v} operatorname{diag}(mathbf{s})$ where $operatorname{diag}(mathbf{s})$ is a diagonal matrix with the entries of $mathbf{s}$ along the diagonal, and
- a Kronecker matrix product of the form $mathbf{v} otimes mathbf{s}^{rm T}$.
add a comment |
Let $mathbf{v} =begin{bmatrix} v_1 \ v_2 \ v_3 end{bmatrix}$ and $mathbf{s} =begin{bmatrix} s_1 \ s_2 \ s_3 end{bmatrix}$. Define
$$M(mathbf{v}, mathbf{s}) =begin{bmatrix} v_1mathbf{s} \ v_2mathbf{s} \ v_3mathbf{s} end{bmatrix} = begin{bmatrix} v_1s_1 &v_1s_2& v_1s_3 \ v_2s_1 & v_2s_2 & v_2s_3 \ v_3s_1 & v_3s_2 & v_3s_3end{bmatrix}.$$
Your matrix $M(mathbf{v}, mathbf{s})$ happens to be algebraically equal to these expressions:
- the outer product $mathbf{v} mathbf{s}^{rm T}$,
- the unnamed product $mathbf{v} operatorname{diag}(mathbf{s})$ where $operatorname{diag}(mathbf{s})$ is a diagonal matrix with the entries of $mathbf{s}$ along the diagonal, and
- a Kronecker matrix product of the form $mathbf{v} otimes mathbf{s}^{rm T}$.
add a comment |
Let $mathbf{v} =begin{bmatrix} v_1 \ v_2 \ v_3 end{bmatrix}$ and $mathbf{s} =begin{bmatrix} s_1 \ s_2 \ s_3 end{bmatrix}$. Define
$$M(mathbf{v}, mathbf{s}) =begin{bmatrix} v_1mathbf{s} \ v_2mathbf{s} \ v_3mathbf{s} end{bmatrix} = begin{bmatrix} v_1s_1 &v_1s_2& v_1s_3 \ v_2s_1 & v_2s_2 & v_2s_3 \ v_3s_1 & v_3s_2 & v_3s_3end{bmatrix}.$$
Your matrix $M(mathbf{v}, mathbf{s})$ happens to be algebraically equal to these expressions:
- the outer product $mathbf{v} mathbf{s}^{rm T}$,
- the unnamed product $mathbf{v} operatorname{diag}(mathbf{s})$ where $operatorname{diag}(mathbf{s})$ is a diagonal matrix with the entries of $mathbf{s}$ along the diagonal, and
- a Kronecker matrix product of the form $mathbf{v} otimes mathbf{s}^{rm T}$.
Let $mathbf{v} =begin{bmatrix} v_1 \ v_2 \ v_3 end{bmatrix}$ and $mathbf{s} =begin{bmatrix} s_1 \ s_2 \ s_3 end{bmatrix}$. Define
$$M(mathbf{v}, mathbf{s}) =begin{bmatrix} v_1mathbf{s} \ v_2mathbf{s} \ v_3mathbf{s} end{bmatrix} = begin{bmatrix} v_1s_1 &v_1s_2& v_1s_3 \ v_2s_1 & v_2s_2 & v_2s_3 \ v_3s_1 & v_3s_2 & v_3s_3end{bmatrix}.$$
Your matrix $M(mathbf{v}, mathbf{s})$ happens to be algebraically equal to these expressions:
- the outer product $mathbf{v} mathbf{s}^{rm T}$,
- the unnamed product $mathbf{v} operatorname{diag}(mathbf{s})$ where $operatorname{diag}(mathbf{s})$ is a diagonal matrix with the entries of $mathbf{s}$ along the diagonal, and
- a Kronecker matrix product of the form $mathbf{v} otimes mathbf{s}^{rm T}$.
answered Nov 28 '18 at 3:15
Rócherz
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2,7762721
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2
Kronecker product
– Michael Burr
Nov 28 '18 at 2:48
Thank you sir, this is what I was looking for.
– Matthew11
Nov 28 '18 at 2:52
1
Your matrix $M(mathbf{v}, mathbf{s})$ also has the form of $mathbf{v} mathbf{s}^{rm T}$.
– Rócherz
Nov 28 '18 at 2:52
1
Actually, for a pair of vectors, this is called the outer product. The Kronecker product is a generalization of it to matrices.
– amd
Nov 28 '18 at 2:58
Yes it seems you are correct Mr. Amd, thank you for the clarification.
– Matthew11
Nov 28 '18 at 3:09