Is there a name for this 'multiplication table' operation of vectors?












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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







share|cite|improve this question


















  • 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
















0














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







share|cite|improve this question


















  • 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














0












0








0







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







share|cite|improve this question













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






share|cite|improve this question













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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














  • 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










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 $mathbf{v} otimes mathbf{s}^{rm T}$.






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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 $mathbf{v} otimes mathbf{s}^{rm T}$.






    share|cite|improve this answer


























      1














      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 $mathbf{v} otimes mathbf{s}^{rm T}$.






      share|cite|improve this 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 $mathbf{v} otimes mathbf{s}^{rm T}$.






        share|cite|improve this answer












        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 $mathbf{v} otimes mathbf{s}^{rm T}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 '18 at 3:15









        Rócherz

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        2,7762721






























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