Special property of matrix $Ainmathbb{R}^{n,n}$ and it's determinant












0












$begingroup$


How to prove for matrix $Ainmathbb{R}^{n,n}$ that if in places where $x$ columns crosses with $y$ rows are placed $0$, then we can be sure, that
$$
det A = 0qquad qquad text{if }quad x+y > n
$$
.






matrices determinant







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  • $begingroup$
    what are $x$ and $y$?
    $endgroup$
    – Lau
    Dec 7 '18 at 8:27










  • $begingroup$
    @Lau $x$ is a number of columns and $y$ is number of rows of which "crosses places" are placed zeros.
    $endgroup$
    – avan1235
    Dec 7 '18 at 8:29
















0












$begingroup$


How to prove for matrix $Ainmathbb{R}^{n,n}$ that if in places where $x$ columns crosses with $y$ rows are placed $0$, then we can be sure, that
$$
det A = 0qquad qquad text{if }quad x+y > n
$$
.






matrices determinant







share|cite|improve this question









$endgroup$












  • $begingroup$
    what are $x$ and $y$?
    $endgroup$
    – Lau
    Dec 7 '18 at 8:27










  • $begingroup$
    @Lau $x$ is a number of columns and $y$ is number of rows of which "crosses places" are placed zeros.
    $endgroup$
    – avan1235
    Dec 7 '18 at 8:29














0












0








0





$begingroup$


How to prove for matrix $Ainmathbb{R}^{n,n}$ that if in places where $x$ columns crosses with $y$ rows are placed $0$, then we can be sure, that
$$
det A = 0qquad qquad text{if }quad x+y > n
$$
.






matrices determinant







share|cite|improve this question









$endgroup$




How to prove for matrix $Ainmathbb{R}^{n,n}$ that if in places where $x$ columns crosses with $y$ rows are placed $0$, then we can be sure, that
$$
det A = 0qquad qquad text{if }quad x+y > n
$$
.






matrices determinant




matrices determinant






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asked Dec 7 '18 at 8:19









avan1235avan1235

3197




3197












  • $begingroup$
    what are $x$ and $y$?
    $endgroup$
    – Lau
    Dec 7 '18 at 8:27










  • $begingroup$
    @Lau $x$ is a number of columns and $y$ is number of rows of which "crosses places" are placed zeros.
    $endgroup$
    – avan1235
    Dec 7 '18 at 8:29


















  • $begingroup$
    what are $x$ and $y$?
    $endgroup$
    – Lau
    Dec 7 '18 at 8:27










  • $begingroup$
    @Lau $x$ is a number of columns and $y$ is number of rows of which "crosses places" are placed zeros.
    $endgroup$
    – avan1235
    Dec 7 '18 at 8:29
















$begingroup$
what are $x$ and $y$?
$endgroup$
– Lau
Dec 7 '18 at 8:27




$begingroup$
what are $x$ and $y$?
$endgroup$
– Lau
Dec 7 '18 at 8:27












$begingroup$
@Lau $x$ is a number of columns and $y$ is number of rows of which "crosses places" are placed zeros.
$endgroup$
– avan1235
Dec 7 '18 at 8:29




$begingroup$
@Lau $x$ is a number of columns and $y$ is number of rows of which "crosses places" are placed zeros.
$endgroup$
– avan1235
Dec 7 '18 at 8:29










2 Answers
2






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$begingroup$

Let $A=[a_{ij}]_n=begin{bmatrix}a_{11}&a_{12}&a_{13}&...&a_{1n}\a_{21}&a_{22}&a_{23}&...&a_{2n}\vdots&vdots&vdots&&vdots\a_{n1}&a_{n2}&a_{n3}&...&a_{nn}end{bmatrix}_n$



Now, $a_{ij}=0$ where the $x$ columns cross the $y$ rows. In totality, we have $xy$ number of $0$s. We know that the determinant of a matrix gets multiplied by $-1$ on interchange of any two rows/columns, but its magnitude is unchanged. We can use elementary row and column interchanges to gather/cluster all the above zeroes together. The transformed matrix $A'$ looks like this:



$A'=begin{bmatrix}0_{ytimes x}&B_{ytimes(n-x)}\C_{(n-y)times x}&D_{(n-y)times(n-x)}end{bmatrix}_n$



where $0_{ytimes x}$ is the zero matrix of indicated order.



Also note that $det A'=(-1)^kcdotdet A$ or $|det A'|=|det A|$



Observe that the rows of $B$ are linearly dependent, since $y$ rows/vectors in $n-x(<y)$ dimesional vector space cannot be linearly independent. This means the rows/vectors of the submatrix $begin{bmatrix}0&Bend{bmatrix}_{ytimes n}$ are linearly dependent. This means $det A'=0impliesdet A=0$.






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    If I rightly understand your question, I have a solution:



    Consider
    $$det A = det
    begin{pmatrix}
    O& B\
    C& D
    end{pmatrix}
    $$
    where $O$ is a $xtimes y$ zero matrix, $B$ is a $(n-x)times y$ matrix, $C$ is a $xtimes(n-y)$ matrix and $D$ is a $(n-x)times(n-y)$ matrix. Notice that $y>n-x$, so the rank of $B$ is no more than $n-x$ and hence the first $y$ lines of $A$ must be linearly dependent, which implies that $det{A}=0$.



    In general, by alternating rows and columns, every determine of $A$ satisfies the given condition can be transfer to the form above (or the opposite number of it).






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

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      active

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      active

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      2












      $begingroup$

      Let $A=[a_{ij}]_n=begin{bmatrix}a_{11}&a_{12}&a_{13}&...&a_{1n}\a_{21}&a_{22}&a_{23}&...&a_{2n}\vdots&vdots&vdots&&vdots\a_{n1}&a_{n2}&a_{n3}&...&a_{nn}end{bmatrix}_n$



      Now, $a_{ij}=0$ where the $x$ columns cross the $y$ rows. In totality, we have $xy$ number of $0$s. We know that the determinant of a matrix gets multiplied by $-1$ on interchange of any two rows/columns, but its magnitude is unchanged. We can use elementary row and column interchanges to gather/cluster all the above zeroes together. The transformed matrix $A'$ looks like this:



      $A'=begin{bmatrix}0_{ytimes x}&B_{ytimes(n-x)}\C_{(n-y)times x}&D_{(n-y)times(n-x)}end{bmatrix}_n$



      where $0_{ytimes x}$ is the zero matrix of indicated order.



      Also note that $det A'=(-1)^kcdotdet A$ or $|det A'|=|det A|$



      Observe that the rows of $B$ are linearly dependent, since $y$ rows/vectors in $n-x(<y)$ dimesional vector space cannot be linearly independent. This means the rows/vectors of the submatrix $begin{bmatrix}0&Bend{bmatrix}_{ytimes n}$ are linearly dependent. This means $det A'=0impliesdet A=0$.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        Let $A=[a_{ij}]_n=begin{bmatrix}a_{11}&a_{12}&a_{13}&...&a_{1n}\a_{21}&a_{22}&a_{23}&...&a_{2n}\vdots&vdots&vdots&&vdots\a_{n1}&a_{n2}&a_{n3}&...&a_{nn}end{bmatrix}_n$



        Now, $a_{ij}=0$ where the $x$ columns cross the $y$ rows. In totality, we have $xy$ number of $0$s. We know that the determinant of a matrix gets multiplied by $-1$ on interchange of any two rows/columns, but its magnitude is unchanged. We can use elementary row and column interchanges to gather/cluster all the above zeroes together. The transformed matrix $A'$ looks like this:



        $A'=begin{bmatrix}0_{ytimes x}&B_{ytimes(n-x)}\C_{(n-y)times x}&D_{(n-y)times(n-x)}end{bmatrix}_n$



        where $0_{ytimes x}$ is the zero matrix of indicated order.



        Also note that $det A'=(-1)^kcdotdet A$ or $|det A'|=|det A|$



        Observe that the rows of $B$ are linearly dependent, since $y$ rows/vectors in $n-x(<y)$ dimesional vector space cannot be linearly independent. This means the rows/vectors of the submatrix $begin{bmatrix}0&Bend{bmatrix}_{ytimes n}$ are linearly dependent. This means $det A'=0impliesdet A=0$.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Let $A=[a_{ij}]_n=begin{bmatrix}a_{11}&a_{12}&a_{13}&...&a_{1n}\a_{21}&a_{22}&a_{23}&...&a_{2n}\vdots&vdots&vdots&&vdots\a_{n1}&a_{n2}&a_{n3}&...&a_{nn}end{bmatrix}_n$



          Now, $a_{ij}=0$ where the $x$ columns cross the $y$ rows. In totality, we have $xy$ number of $0$s. We know that the determinant of a matrix gets multiplied by $-1$ on interchange of any two rows/columns, but its magnitude is unchanged. We can use elementary row and column interchanges to gather/cluster all the above zeroes together. The transformed matrix $A'$ looks like this:



          $A'=begin{bmatrix}0_{ytimes x}&B_{ytimes(n-x)}\C_{(n-y)times x}&D_{(n-y)times(n-x)}end{bmatrix}_n$



          where $0_{ytimes x}$ is the zero matrix of indicated order.



          Also note that $det A'=(-1)^kcdotdet A$ or $|det A'|=|det A|$



          Observe that the rows of $B$ are linearly dependent, since $y$ rows/vectors in $n-x(<y)$ dimesional vector space cannot be linearly independent. This means the rows/vectors of the submatrix $begin{bmatrix}0&Bend{bmatrix}_{ytimes n}$ are linearly dependent. This means $det A'=0impliesdet A=0$.






          share|cite|improve this answer











          $endgroup$



          Let $A=[a_{ij}]_n=begin{bmatrix}a_{11}&a_{12}&a_{13}&...&a_{1n}\a_{21}&a_{22}&a_{23}&...&a_{2n}\vdots&vdots&vdots&&vdots\a_{n1}&a_{n2}&a_{n3}&...&a_{nn}end{bmatrix}_n$



          Now, $a_{ij}=0$ where the $x$ columns cross the $y$ rows. In totality, we have $xy$ number of $0$s. We know that the determinant of a matrix gets multiplied by $-1$ on interchange of any two rows/columns, but its magnitude is unchanged. We can use elementary row and column interchanges to gather/cluster all the above zeroes together. The transformed matrix $A'$ looks like this:



          $A'=begin{bmatrix}0_{ytimes x}&B_{ytimes(n-x)}\C_{(n-y)times x}&D_{(n-y)times(n-x)}end{bmatrix}_n$



          where $0_{ytimes x}$ is the zero matrix of indicated order.



          Also note that $det A'=(-1)^kcdotdet A$ or $|det A'|=|det A|$



          Observe that the rows of $B$ are linearly dependent, since $y$ rows/vectors in $n-x(<y)$ dimesional vector space cannot be linearly independent. This means the rows/vectors of the submatrix $begin{bmatrix}0&Bend{bmatrix}_{ytimes n}$ are linearly dependent. This means $det A'=0impliesdet A=0$.







          share|cite|improve this answer














          share|cite|improve this answer



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          edited Dec 7 '18 at 9:54

























          answered Dec 7 '18 at 8:57









          Shubham JohriShubham Johri

          5,102717




          5,102717























              3












              $begingroup$

              If I rightly understand your question, I have a solution:



              Consider
              $$det A = det
              begin{pmatrix}
              O& B\
              C& D
              end{pmatrix}
              $$
              where $O$ is a $xtimes y$ zero matrix, $B$ is a $(n-x)times y$ matrix, $C$ is a $xtimes(n-y)$ matrix and $D$ is a $(n-x)times(n-y)$ matrix. Notice that $y>n-x$, so the rank of $B$ is no more than $n-x$ and hence the first $y$ lines of $A$ must be linearly dependent, which implies that $det{A}=0$.



              In general, by alternating rows and columns, every determine of $A$ satisfies the given condition can be transfer to the form above (or the opposite number of it).






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                If I rightly understand your question, I have a solution:



                Consider
                $$det A = det
                begin{pmatrix}
                O& B\
                C& D
                end{pmatrix}
                $$
                where $O$ is a $xtimes y$ zero matrix, $B$ is a $(n-x)times y$ matrix, $C$ is a $xtimes(n-y)$ matrix and $D$ is a $(n-x)times(n-y)$ matrix. Notice that $y>n-x$, so the rank of $B$ is no more than $n-x$ and hence the first $y$ lines of $A$ must be linearly dependent, which implies that $det{A}=0$.



                In general, by alternating rows and columns, every determine of $A$ satisfies the given condition can be transfer to the form above (or the opposite number of it).






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  If I rightly understand your question, I have a solution:



                  Consider
                  $$det A = det
                  begin{pmatrix}
                  O& B\
                  C& D
                  end{pmatrix}
                  $$
                  where $O$ is a $xtimes y$ zero matrix, $B$ is a $(n-x)times y$ matrix, $C$ is a $xtimes(n-y)$ matrix and $D$ is a $(n-x)times(n-y)$ matrix. Notice that $y>n-x$, so the rank of $B$ is no more than $n-x$ and hence the first $y$ lines of $A$ must be linearly dependent, which implies that $det{A}=0$.



                  In general, by alternating rows and columns, every determine of $A$ satisfies the given condition can be transfer to the form above (or the opposite number of it).






                  share|cite|improve this answer









                  $endgroup$



                  If I rightly understand your question, I have a solution:



                  Consider
                  $$det A = det
                  begin{pmatrix}
                  O& B\
                  C& D
                  end{pmatrix}
                  $$
                  where $O$ is a $xtimes y$ zero matrix, $B$ is a $(n-x)times y$ matrix, $C$ is a $xtimes(n-y)$ matrix and $D$ is a $(n-x)times(n-y)$ matrix. Notice that $y>n-x$, so the rank of $B$ is no more than $n-x$ and hence the first $y$ lines of $A$ must be linearly dependent, which implies that $det{A}=0$.



                  In general, by alternating rows and columns, every determine of $A$ satisfies the given condition can be transfer to the form above (or the opposite number of it).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 7 '18 at 8:52









                  LauLau

                  527315




                  527315






























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