Determinant is linear as a function of each of the rows of the matrix.












2












$begingroup$


Today I heard in a lecture (some video on YouTube) that the determinant is linear as a function of each of the rows of the matrix.



I am not able to understand the above statement. I know that determinant is a special function which assign to each $x$ in $mathbb K^{n times n}$ a scalar. This is the intuitive idea. And this map is not linear as well. One way to see this is to consider the fact that determinant of $cA$ is $c^ndet(A)$



Can someone please explain what did the person mean by saying that the determinant is linear as a function of each of the rows of matrix?






linear-algebra matrices determinant







share|cite|improve this question











$endgroup$












  • $begingroup$
    I got it. It means that elementary row operations have a linear effect on determinant. Say $A=(r1,r2,...,r_n)$ is a matrix then det of $(r_1,..,cr_j +r_i,..,r_n)$ is nothing but determinant of $(r_1,..,cr_j,..,r_n)$ plus determinant of $(r_1,..,r_i,..,r_n)$.Am I right?
    $endgroup$
    – StammeringMathematician
    1 hour ago










  • $begingroup$
    Yes. The fact that is is linear in each row separately gives rise to the combinatorial formula for the determinant.
    $endgroup$
    – copper.hat
    1 hour ago


















2












$begingroup$


Today I heard in a lecture (some video on YouTube) that the determinant is linear as a function of each of the rows of the matrix.



I am not able to understand the above statement. I know that determinant is a special function which assign to each $x$ in $mathbb K^{n times n}$ a scalar. This is the intuitive idea. And this map is not linear as well. One way to see this is to consider the fact that determinant of $cA$ is $c^ndet(A)$



Can someone please explain what did the person mean by saying that the determinant is linear as a function of each of the rows of matrix?






linear-algebra matrices determinant







share|cite|improve this question











$endgroup$












  • $begingroup$
    I got it. It means that elementary row operations have a linear effect on determinant. Say $A=(r1,r2,...,r_n)$ is a matrix then det of $(r_1,..,cr_j +r_i,..,r_n)$ is nothing but determinant of $(r_1,..,cr_j,..,r_n)$ plus determinant of $(r_1,..,r_i,..,r_n)$.Am I right?
    $endgroup$
    – StammeringMathematician
    1 hour ago










  • $begingroup$
    Yes. The fact that is is linear in each row separately gives rise to the combinatorial formula for the determinant.
    $endgroup$
    – copper.hat
    1 hour ago
















2












2








2





$begingroup$


Today I heard in a lecture (some video on YouTube) that the determinant is linear as a function of each of the rows of the matrix.



I am not able to understand the above statement. I know that determinant is a special function which assign to each $x$ in $mathbb K^{n times n}$ a scalar. This is the intuitive idea. And this map is not linear as well. One way to see this is to consider the fact that determinant of $cA$ is $c^ndet(A)$



Can someone please explain what did the person mean by saying that the determinant is linear as a function of each of the rows of matrix?






linear-algebra matrices determinant







share|cite|improve this question











$endgroup$




Today I heard in a lecture (some video on YouTube) that the determinant is linear as a function of each of the rows of the matrix.



I am not able to understand the above statement. I know that determinant is a special function which assign to each $x$ in $mathbb K^{n times n}$ a scalar. This is the intuitive idea. And this map is not linear as well. One way to see this is to consider the fact that determinant of $cA$ is $c^ndet(A)$



Can someone please explain what did the person mean by saying that the determinant is linear as a function of each of the rows of matrix?






linear-algebra matrices determinant




linear-algebra matrices determinant






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









Rodrigo de Azevedo

13.2k41961




13.2k41961










asked 1 hour ago









StammeringMathematicianStammeringMathematician

2,8121324




2,8121324












  • $begingroup$
    I got it. It means that elementary row operations have a linear effect on determinant. Say $A=(r1,r2,...,r_n)$ is a matrix then det of $(r_1,..,cr_j +r_i,..,r_n)$ is nothing but determinant of $(r_1,..,cr_j,..,r_n)$ plus determinant of $(r_1,..,r_i,..,r_n)$.Am I right?
    $endgroup$
    – StammeringMathematician
    1 hour ago










  • $begingroup$
    Yes. The fact that is is linear in each row separately gives rise to the combinatorial formula for the determinant.
    $endgroup$
    – copper.hat
    1 hour ago




















  • $begingroup$
    I got it. It means that elementary row operations have a linear effect on determinant. Say $A=(r1,r2,...,r_n)$ is a matrix then det of $(r_1,..,cr_j +r_i,..,r_n)$ is nothing but determinant of $(r_1,..,cr_j,..,r_n)$ plus determinant of $(r_1,..,r_i,..,r_n)$.Am I right?
    $endgroup$
    – StammeringMathematician
    1 hour ago










  • $begingroup$
    Yes. The fact that is is linear in each row separately gives rise to the combinatorial formula for the determinant.
    $endgroup$
    – copper.hat
    1 hour ago


















$begingroup$
I got it. It means that elementary row operations have a linear effect on determinant. Say $A=(r1,r2,...,r_n)$ is a matrix then det of $(r_1,..,cr_j +r_i,..,r_n)$ is nothing but determinant of $(r_1,..,cr_j,..,r_n)$ plus determinant of $(r_1,..,r_i,..,r_n)$.Am I right?
$endgroup$
– StammeringMathematician
1 hour ago




$begingroup$
I got it. It means that elementary row operations have a linear effect on determinant. Say $A=(r1,r2,...,r_n)$ is a matrix then det of $(r_1,..,cr_j +r_i,..,r_n)$ is nothing but determinant of $(r_1,..,cr_j,..,r_n)$ plus determinant of $(r_1,..,r_i,..,r_n)$.Am I right?
$endgroup$
– StammeringMathematician
1 hour ago












$begingroup$
Yes. The fact that is is linear in each row separately gives rise to the combinatorial formula for the determinant.
$endgroup$
– copper.hat
1 hour ago






$begingroup$
Yes. The fact that is is linear in each row separately gives rise to the combinatorial formula for the determinant.
$endgroup$
– copper.hat
1 hour ago












2 Answers
2






active

oldest

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3












$begingroup$

If $r_1, ldots r_n$ are the rows of the matrix and $r_i = sa+tb$, where $s,t$ are scalars and $a,b$ are row vectors, then you have



$$detbegin{pmatrix}r_1 \ vdots \r_i \ vdots \ r_nend{pmatrix} = detbegin{pmatrix}r_1 \ vdots \ sa+tb \ vdots \ r_nend{pmatrix} = sdetbegin{pmatrix}r_1 \ vdots \ a \ vdots \ r_nend{pmatrix} + tdetbegin{pmatrix}r_1 \ vdots \ b \ vdots \ r_nend{pmatrix}$$



This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Let $M$ be an $ntimes n$ matrix with rows $mathbf{r}_1,dots,mathbf{r}_n$. Then we may think of the determinant as a function of the rows
    $$
    det(M)=det(mathbf{r}_1,dots,mathbf{r}_n).
    $$
    To say that $det$ is a linear function of the rows means that if we scale a single row by $c$, the result is scaled by $c$; that is,
    $$
    det(mathbf{r}_1,dots,mathbf{r}_{i-1},cmathbf{r}_i,mathbf{r}_{i+1}dotsmathbf{r}_n)=cdet(mathbf{r}_1,dots,mathbf{r}_n).
    $$
    Similarly if we fix all but one row (say the first), we obtain
    $$
    det(mathbf{x}+mathbf{r}_1,mathbf{r}_2,dots,mathbf{r}_n)=det(mathbf{x},dots,mathbf{r}_n)+det(mathbf{r}_1,dots,mathbf{r}_n).
    $$
    Your mistake was that you scale all the rows at once; to be linear, you can only do things "one at a time"






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

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      3












      $begingroup$

      If $r_1, ldots r_n$ are the rows of the matrix and $r_i = sa+tb$, where $s,t$ are scalars and $a,b$ are row vectors, then you have



      $$detbegin{pmatrix}r_1 \ vdots \r_i \ vdots \ r_nend{pmatrix} = detbegin{pmatrix}r_1 \ vdots \ sa+tb \ vdots \ r_nend{pmatrix} = sdetbegin{pmatrix}r_1 \ vdots \ a \ vdots \ r_nend{pmatrix} + tdetbegin{pmatrix}r_1 \ vdots \ b \ vdots \ r_nend{pmatrix}$$



      This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        If $r_1, ldots r_n$ are the rows of the matrix and $r_i = sa+tb$, where $s,t$ are scalars and $a,b$ are row vectors, then you have



        $$detbegin{pmatrix}r_1 \ vdots \r_i \ vdots \ r_nend{pmatrix} = detbegin{pmatrix}r_1 \ vdots \ sa+tb \ vdots \ r_nend{pmatrix} = sdetbegin{pmatrix}r_1 \ vdots \ a \ vdots \ r_nend{pmatrix} + tdetbegin{pmatrix}r_1 \ vdots \ b \ vdots \ r_nend{pmatrix}$$



        This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          If $r_1, ldots r_n$ are the rows of the matrix and $r_i = sa+tb$, where $s,t$ are scalars and $a,b$ are row vectors, then you have



          $$detbegin{pmatrix}r_1 \ vdots \r_i \ vdots \ r_nend{pmatrix} = detbegin{pmatrix}r_1 \ vdots \ sa+tb \ vdots \ r_nend{pmatrix} = sdetbegin{pmatrix}r_1 \ vdots \ a \ vdots \ r_nend{pmatrix} + tdetbegin{pmatrix}r_1 \ vdots \ b \ vdots \ r_nend{pmatrix}$$



          This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.






          share|cite|improve this answer









          $endgroup$



          If $r_1, ldots r_n$ are the rows of the matrix and $r_i = sa+tb$, where $s,t$ are scalars and $a,b$ are row vectors, then you have



          $$detbegin{pmatrix}r_1 \ vdots \r_i \ vdots \ r_nend{pmatrix} = detbegin{pmatrix}r_1 \ vdots \ sa+tb \ vdots \ r_nend{pmatrix} = sdetbegin{pmatrix}r_1 \ vdots \ a \ vdots \ r_nend{pmatrix} + tdetbegin{pmatrix}r_1 \ vdots \ b \ vdots \ r_nend{pmatrix}$$



          This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          trancelocationtrancelocation

          14.1k1829




          14.1k1829























              2












              $begingroup$

              Let $M$ be an $ntimes n$ matrix with rows $mathbf{r}_1,dots,mathbf{r}_n$. Then we may think of the determinant as a function of the rows
              $$
              det(M)=det(mathbf{r}_1,dots,mathbf{r}_n).
              $$
              To say that $det$ is a linear function of the rows means that if we scale a single row by $c$, the result is scaled by $c$; that is,
              $$
              det(mathbf{r}_1,dots,mathbf{r}_{i-1},cmathbf{r}_i,mathbf{r}_{i+1}dotsmathbf{r}_n)=cdet(mathbf{r}_1,dots,mathbf{r}_n).
              $$
              Similarly if we fix all but one row (say the first), we obtain
              $$
              det(mathbf{x}+mathbf{r}_1,mathbf{r}_2,dots,mathbf{r}_n)=det(mathbf{x},dots,mathbf{r}_n)+det(mathbf{r}_1,dots,mathbf{r}_n).
              $$
              Your mistake was that you scale all the rows at once; to be linear, you can only do things "one at a time"






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Let $M$ be an $ntimes n$ matrix with rows $mathbf{r}_1,dots,mathbf{r}_n$. Then we may think of the determinant as a function of the rows
                $$
                det(M)=det(mathbf{r}_1,dots,mathbf{r}_n).
                $$
                To say that $det$ is a linear function of the rows means that if we scale a single row by $c$, the result is scaled by $c$; that is,
                $$
                det(mathbf{r}_1,dots,mathbf{r}_{i-1},cmathbf{r}_i,mathbf{r}_{i+1}dotsmathbf{r}_n)=cdet(mathbf{r}_1,dots,mathbf{r}_n).
                $$
                Similarly if we fix all but one row (say the first), we obtain
                $$
                det(mathbf{x}+mathbf{r}_1,mathbf{r}_2,dots,mathbf{r}_n)=det(mathbf{x},dots,mathbf{r}_n)+det(mathbf{r}_1,dots,mathbf{r}_n).
                $$
                Your mistake was that you scale all the rows at once; to be linear, you can only do things "one at a time"






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Let $M$ be an $ntimes n$ matrix with rows $mathbf{r}_1,dots,mathbf{r}_n$. Then we may think of the determinant as a function of the rows
                  $$
                  det(M)=det(mathbf{r}_1,dots,mathbf{r}_n).
                  $$
                  To say that $det$ is a linear function of the rows means that if we scale a single row by $c$, the result is scaled by $c$; that is,
                  $$
                  det(mathbf{r}_1,dots,mathbf{r}_{i-1},cmathbf{r}_i,mathbf{r}_{i+1}dotsmathbf{r}_n)=cdet(mathbf{r}_1,dots,mathbf{r}_n).
                  $$
                  Similarly if we fix all but one row (say the first), we obtain
                  $$
                  det(mathbf{x}+mathbf{r}_1,mathbf{r}_2,dots,mathbf{r}_n)=det(mathbf{x},dots,mathbf{r}_n)+det(mathbf{r}_1,dots,mathbf{r}_n).
                  $$
                  Your mistake was that you scale all the rows at once; to be linear, you can only do things "one at a time"






                  share|cite|improve this answer









                  $endgroup$



                  Let $M$ be an $ntimes n$ matrix with rows $mathbf{r}_1,dots,mathbf{r}_n$. Then we may think of the determinant as a function of the rows
                  $$
                  det(M)=det(mathbf{r}_1,dots,mathbf{r}_n).
                  $$
                  To say that $det$ is a linear function of the rows means that if we scale a single row by $c$, the result is scaled by $c$; that is,
                  $$
                  det(mathbf{r}_1,dots,mathbf{r}_{i-1},cmathbf{r}_i,mathbf{r}_{i+1}dotsmathbf{r}_n)=cdet(mathbf{r}_1,dots,mathbf{r}_n).
                  $$
                  Similarly if we fix all but one row (say the first), we obtain
                  $$
                  det(mathbf{x}+mathbf{r}_1,mathbf{r}_2,dots,mathbf{r}_n)=det(mathbf{x},dots,mathbf{r}_n)+det(mathbf{r}_1,dots,mathbf{r}_n).
                  $$
                  Your mistake was that you scale all the rows at once; to be linear, you can only do things "one at a time"







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  TomGrubbTomGrubb

                  11.2k11639




                  11.2k11639






























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