How to properly represent a matrix function.












1












$begingroup$


Given the function $f_{h}(x,y,z)=(x-z,y+hz,x+y+3z)$, what is the correct way to represent the matrix function in respect to the standard basis?



With the representation theorem, I would write the matrix in columns as:
$$F_{h|S_3}=(f_h(e_1)|{S_3} quad f_h(e_2)|{S_3} quad f_h(e_3)|{S_3})=begin{bmatrix}1 & 0 & 1\ 0 & 1 & 1\ -1 & h & 3end{bmatrix}$$
But in my textbook it is written in rows as:
$$F_{h|S_3}=(f_h(e_1)|{S_3} quad f_h(e_2)|{S_3} quad f_h(e_3)|{S_3})=begin{bmatrix}1 & 0 & -1\ 0 & 1 & h\ 1 & 1 & 3end{bmatrix}$$



What is the difference between them?






linear-algebra matrices linear-transformations







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Given the function $f_{h}(x,y,z)=(x-z,y+hz,x+y+3z)$, what is the correct way to represent the matrix function in respect to the standard basis?



    With the representation theorem, I would write the matrix in columns as:
    $$F_{h|S_3}=(f_h(e_1)|{S_3} quad f_h(e_2)|{S_3} quad f_h(e_3)|{S_3})=begin{bmatrix}1 & 0 & 1\ 0 & 1 & 1\ -1 & h & 3end{bmatrix}$$
    But in my textbook it is written in rows as:
    $$F_{h|S_3}=(f_h(e_1)|{S_3} quad f_h(e_2)|{S_3} quad f_h(e_3)|{S_3})=begin{bmatrix}1 & 0 & -1\ 0 & 1 & h\ 1 & 1 & 3end{bmatrix}$$



    What is the difference between them?






    linear-algebra matrices linear-transformations







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Given the function $f_{h}(x,y,z)=(x-z,y+hz,x+y+3z)$, what is the correct way to represent the matrix function in respect to the standard basis?



      With the representation theorem, I would write the matrix in columns as:
      $$F_{h|S_3}=(f_h(e_1)|{S_3} quad f_h(e_2)|{S_3} quad f_h(e_3)|{S_3})=begin{bmatrix}1 & 0 & 1\ 0 & 1 & 1\ -1 & h & 3end{bmatrix}$$
      But in my textbook it is written in rows as:
      $$F_{h|S_3}=(f_h(e_1)|{S_3} quad f_h(e_2)|{S_3} quad f_h(e_3)|{S_3})=begin{bmatrix}1 & 0 & -1\ 0 & 1 & h\ 1 & 1 & 3end{bmatrix}$$



      What is the difference between them?






      linear-algebra matrices linear-transformations







      share|cite|improve this question











      $endgroup$




      Given the function $f_{h}(x,y,z)=(x-z,y+hz,x+y+3z)$, what is the correct way to represent the matrix function in respect to the standard basis?



      With the representation theorem, I would write the matrix in columns as:
      $$F_{h|S_3}=(f_h(e_1)|{S_3} quad f_h(e_2)|{S_3} quad f_h(e_3)|{S_3})=begin{bmatrix}1 & 0 & 1\ 0 & 1 & 1\ -1 & h & 3end{bmatrix}$$
      But in my textbook it is written in rows as:
      $$F_{h|S_3}=(f_h(e_1)|{S_3} quad f_h(e_2)|{S_3} quad f_h(e_3)|{S_3})=begin{bmatrix}1 & 0 & -1\ 0 & 1 & h\ 1 & 1 & 3end{bmatrix}$$



      What is the difference between them?






      linear-algebra matrices linear-transformations




      linear-algebra matrices linear-transformations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 13 at 17:06









      Kevin

      13311




      13311










      asked Jan 13 at 16:15









      Kevin MooreKevin Moore

      16918




      16918






















          1 Answer
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          $begingroup$

          The correct way to represent the function $f_h$ in matrix form depends on the convention that you want to use to represent it. Let $(x,y,z) in mathbb{R}^3$. The matrix which you computed is useful for expressing $f_h$ as $$f_h(x,y,z) = begin{bmatrix} x& y & z end{bmatrix} begin{bmatrix}
          1& 0 & 1 \
          0 & 1& 1\
          -1 & h & 3
          end{bmatrix}.$$           We can verify this by expanding the matrix product:
          begin{align}
          begin{bmatrix} x& y & z end{bmatrix} begin{bmatrix}
          1& 0 & 2 \
          0 & 1& 1\
          -1 & h & 3
          end{bmatrix} &= begin{bmatrix}
          x cdot 1 + y cdot 0 + z cdot (-1) \
          x cdot 0 + y cdot 1 + z cdot h \
          x cdot 1 + y cdot 1 + z cdot 3
          end{bmatrix} \ &= begin{bmatrix}
          x - z & y + hz & x + y + 3z end{bmatrix}.
          end{align}           The result is a $1 times 3$ matrix, which we can interpret as a row vector in $mathbb{R}^3$.
          The matrix written in your textbook is useful for the following representation:
          $$f_h(x,y,z) = begin{bmatrix}
          1& 0 & -1 \
          0 & 1& h\
          1 & 1 & 3
          end{bmatrix} begin{bmatrix}
          x \ y \ z end{bmatrix} = begin{bmatrix} x-z \ y + hz \ x + y + 3z end{bmatrix}.$$           The result is a $3 times 1 $ matrix, which we can regard as a column vector in $mathbb{R}^3$.



          The two representations look different in the sense that the former yields a row vector and the latter a column vector, however they are the same in the sense that they can both be regarded as lists of three real numbers, i.e. as $(x-z,y+hz,x+y+3z) in mathbb{R}^3.$ The accepted answer here sums this idea up pretty well.






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            The correct way to represent the function $f_h$ in matrix form depends on the convention that you want to use to represent it. Let $(x,y,z) in mathbb{R}^3$. The matrix which you computed is useful for expressing $f_h$ as $$f_h(x,y,z) = begin{bmatrix} x& y & z end{bmatrix} begin{bmatrix}
            1& 0 & 1 \
            0 & 1& 1\
            -1 & h & 3
            end{bmatrix}.$$             We can verify this by expanding the matrix product:
            begin{align}
            begin{bmatrix} x& y & z end{bmatrix} begin{bmatrix}
            1& 0 & 2 \
            0 & 1& 1\
            -1 & h & 3
            end{bmatrix} &= begin{bmatrix}
            x cdot 1 + y cdot 0 + z cdot (-1) \
            x cdot 0 + y cdot 1 + z cdot h \
            x cdot 1 + y cdot 1 + z cdot 3
            end{bmatrix} \ &= begin{bmatrix}
            x - z & y + hz & x + y + 3z end{bmatrix}.
            end{align}             The result is a $1 times 3$ matrix, which we can interpret as a row vector in $mathbb{R}^3$.
            The matrix written in your textbook is useful for the following representation:
            $$f_h(x,y,z) = begin{bmatrix}
            1& 0 & -1 \
            0 & 1& h\
            1 & 1 & 3
            end{bmatrix} begin{bmatrix}
            x \ y \ z end{bmatrix} = begin{bmatrix} x-z \ y + hz \ x + y + 3z end{bmatrix}.$$             The result is a $3 times 1 $ matrix, which we can regard as a column vector in $mathbb{R}^3$.



            The two representations look different in the sense that the former yields a row vector and the latter a column vector, however they are the same in the sense that they can both be regarded as lists of three real numbers, i.e. as $(x-z,y+hz,x+y+3z) in mathbb{R}^3.$ The accepted answer here sums this idea up pretty well.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              The correct way to represent the function $f_h$ in matrix form depends on the convention that you want to use to represent it. Let $(x,y,z) in mathbb{R}^3$. The matrix which you computed is useful for expressing $f_h$ as $$f_h(x,y,z) = begin{bmatrix} x& y & z end{bmatrix} begin{bmatrix}
              1& 0 & 1 \
              0 & 1& 1\
              -1 & h & 3
              end{bmatrix}.$$               We can verify this by expanding the matrix product:
              begin{align}
              begin{bmatrix} x& y & z end{bmatrix} begin{bmatrix}
              1& 0 & 2 \
              0 & 1& 1\
              -1 & h & 3
              end{bmatrix} &= begin{bmatrix}
              x cdot 1 + y cdot 0 + z cdot (-1) \
              x cdot 0 + y cdot 1 + z cdot h \
              x cdot 1 + y cdot 1 + z cdot 3
              end{bmatrix} \ &= begin{bmatrix}
              x - z & y + hz & x + y + 3z end{bmatrix}.
              end{align}               The result is a $1 times 3$ matrix, which we can interpret as a row vector in $mathbb{R}^3$.
              The matrix written in your textbook is useful for the following representation:
              $$f_h(x,y,z) = begin{bmatrix}
              1& 0 & -1 \
              0 & 1& h\
              1 & 1 & 3
              end{bmatrix} begin{bmatrix}
              x \ y \ z end{bmatrix} = begin{bmatrix} x-z \ y + hz \ x + y + 3z end{bmatrix}.$$               The result is a $3 times 1 $ matrix, which we can regard as a column vector in $mathbb{R}^3$.



              The two representations look different in the sense that the former yields a row vector and the latter a column vector, however they are the same in the sense that they can both be regarded as lists of three real numbers, i.e. as $(x-z,y+hz,x+y+3z) in mathbb{R}^3.$ The accepted answer here sums this idea up pretty well.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                The correct way to represent the function $f_h$ in matrix form depends on the convention that you want to use to represent it. Let $(x,y,z) in mathbb{R}^3$. The matrix which you computed is useful for expressing $f_h$ as $$f_h(x,y,z) = begin{bmatrix} x& y & z end{bmatrix} begin{bmatrix}
                1& 0 & 1 \
                0 & 1& 1\
                -1 & h & 3
                end{bmatrix}.$$                 We can verify this by expanding the matrix product:
                begin{align}
                begin{bmatrix} x& y & z end{bmatrix} begin{bmatrix}
                1& 0 & 2 \
                0 & 1& 1\
                -1 & h & 3
                end{bmatrix} &= begin{bmatrix}
                x cdot 1 + y cdot 0 + z cdot (-1) \
                x cdot 0 + y cdot 1 + z cdot h \
                x cdot 1 + y cdot 1 + z cdot 3
                end{bmatrix} \ &= begin{bmatrix}
                x - z & y + hz & x + y + 3z end{bmatrix}.
                end{align}                 The result is a $1 times 3$ matrix, which we can interpret as a row vector in $mathbb{R}^3$.
                The matrix written in your textbook is useful for the following representation:
                $$f_h(x,y,z) = begin{bmatrix}
                1& 0 & -1 \
                0 & 1& h\
                1 & 1 & 3
                end{bmatrix} begin{bmatrix}
                x \ y \ z end{bmatrix} = begin{bmatrix} x-z \ y + hz \ x + y + 3z end{bmatrix}.$$                 The result is a $3 times 1 $ matrix, which we can regard as a column vector in $mathbb{R}^3$.



                The two representations look different in the sense that the former yields a row vector and the latter a column vector, however they are the same in the sense that they can both be regarded as lists of three real numbers, i.e. as $(x-z,y+hz,x+y+3z) in mathbb{R}^3.$ The accepted answer here sums this idea up pretty well.






                share|cite|improve this answer











                $endgroup$



                The correct way to represent the function $f_h$ in matrix form depends on the convention that you want to use to represent it. Let $(x,y,z) in mathbb{R}^3$. The matrix which you computed is useful for expressing $f_h$ as $$f_h(x,y,z) = begin{bmatrix} x& y & z end{bmatrix} begin{bmatrix}
                1& 0 & 1 \
                0 & 1& 1\
                -1 & h & 3
                end{bmatrix}.$$                 We can verify this by expanding the matrix product:
                begin{align}
                begin{bmatrix} x& y & z end{bmatrix} begin{bmatrix}
                1& 0 & 2 \
                0 & 1& 1\
                -1 & h & 3
                end{bmatrix} &= begin{bmatrix}
                x cdot 1 + y cdot 0 + z cdot (-1) \
                x cdot 0 + y cdot 1 + z cdot h \
                x cdot 1 + y cdot 1 + z cdot 3
                end{bmatrix} \ &= begin{bmatrix}
                x - z & y + hz & x + y + 3z end{bmatrix}.
                end{align}                 The result is a $1 times 3$ matrix, which we can interpret as a row vector in $mathbb{R}^3$.
                The matrix written in your textbook is useful for the following representation:
                $$f_h(x,y,z) = begin{bmatrix}
                1& 0 & -1 \
                0 & 1& h\
                1 & 1 & 3
                end{bmatrix} begin{bmatrix}
                x \ y \ z end{bmatrix} = begin{bmatrix} x-z \ y + hz \ x + y + 3z end{bmatrix}.$$                 The result is a $3 times 1 $ matrix, which we can regard as a column vector in $mathbb{R}^3$.



                The two representations look different in the sense that the former yields a row vector and the latter a column vector, however they are the same in the sense that they can both be regarded as lists of three real numbers, i.e. as $(x-z,y+hz,x+y+3z) in mathbb{R}^3.$ The accepted answer here sums this idea up pretty well.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 13 at 20:03

























                answered Jan 13 at 19:46









                E-muE-mu

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