Complementary minor of a kernel of given full-rank matrix $A$ is equal to minor of $A$.

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Let $d leq n$, $d,n inmathbbN$. Suppose $beginpmatrixA & B endpmatrix,beginpmatrixC \ D endpmatrix$ be a block matrix with $A in Mat_mathbbZ^dtimes d,B in Mat_mathbbZ^dtimes (n-d), C in Mat_mathbbZ^dtimes (n-d), D in Mat_mathbbZ^(n-d)times (n-)d$ such that 1) $beginpmatrixA & B endpmatrix$ has rank $d$ and 2) $$beginpmatrixA & B endpmatrixbeginpmatrixC \ D endpmatrix = 0 $$
which is equivalent to say
$$AC + BD=0.$$
Also suppose that each columns of $beginpmatrixC \ D endpmatrix$ consists of a $mathbbZ$-basis of $ker beginpmatrixA & B endpmatrix$.
Then can we say $det(A) = det(D)$? It is mentioned in some book as a basic fact of linear algebra, but I don't know how to prove it. Any hint will be appreciated.




linear-algebra change-of-basis




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    Let $d leq n$, $d,n inmathbbN$. Suppose $beginpmatrixA & B endpmatrix,beginpmatrixC \ D endpmatrix$ be a block matrix with $A in Mat_mathbbZ^dtimes d,B in Mat_mathbbZ^dtimes (n-d), C in Mat_mathbbZ^dtimes (n-d), D in Mat_mathbbZ^(n-d)times (n-)d$ such that 1) $beginpmatrixA & B endpmatrix$ has rank $d$ and 2) $$beginpmatrixA & B endpmatrixbeginpmatrixC \ D endpmatrix = 0 $$
    which is equivalent to say
    $$AC + BD=0.$$
    Also suppose that each columns of $beginpmatrixC \ D endpmatrix$ consists of a $mathbbZ$-basis of $ker beginpmatrixA & B endpmatrix$.
    Then can we say $det(A) = det(D)$? It is mentioned in some book as a basic fact of linear algebra, but I don't know how to prove it. Any hint will be appreciated.




    linear-algebra change-of-basis




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      Let $d leq n$, $d,n inmathbbN$. Suppose $beginpmatrixA & B endpmatrix,beginpmatrixC \ D endpmatrix$ be a block matrix with $A in Mat_mathbbZ^dtimes d,B in Mat_mathbbZ^dtimes (n-d), C in Mat_mathbbZ^dtimes (n-d), D in Mat_mathbbZ^(n-d)times (n-)d$ such that 1) $beginpmatrixA & B endpmatrix$ has rank $d$ and 2) $$beginpmatrixA & B endpmatrixbeginpmatrixC \ D endpmatrix = 0 $$
      which is equivalent to say
      $$AC + BD=0.$$
      Also suppose that each columns of $beginpmatrixC \ D endpmatrix$ consists of a $mathbbZ$-basis of $ker beginpmatrixA & B endpmatrix$.
      Then can we say $det(A) = det(D)$? It is mentioned in some book as a basic fact of linear algebra, but I don't know how to prove it. Any hint will be appreciated.




      linear-algebra change-of-basis




      share|cite|improve this question













      Let $d leq n$, $d,n inmathbbN$. Suppose $beginpmatrixA & B endpmatrix,beginpmatrixC \ D endpmatrix$ be a block matrix with $A in Mat_mathbbZ^dtimes d,B in Mat_mathbbZ^dtimes (n-d), C in Mat_mathbbZ^dtimes (n-d), D in Mat_mathbbZ^(n-d)times (n-)d$ such that 1) $beginpmatrixA & B endpmatrix$ has rank $d$ and 2) $$beginpmatrixA & B endpmatrixbeginpmatrixC \ D endpmatrix = 0 $$
      which is equivalent to say
      $$AC + BD=0.$$
      Also suppose that each columns of $beginpmatrixC \ D endpmatrix$ consists of a $mathbbZ$-basis of $ker beginpmatrixA & B endpmatrix$.
      Then can we say $det(A) = det(D)$? It is mentioned in some book as a basic fact of linear algebra, but I don't know how to prove it. Any hint will be appreciated.





      linear-algebra change-of-basis





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      edited Aug 6 at 3:43
























      asked Aug 6 at 2:57









      user124697

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          Take $d=1$ and $n=2$; let $A=3$, $B=0$, $C=0$, $D=1$. These satisfy the hypotheses, yet $det Anot= det D$.






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          • Thank you for very simple counterexample :)
            – user124697
            Aug 6 at 18:54










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          up vote
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          accepted










          Take $d=1$ and $n=2$; let $A=3$, $B=0$, $C=0$, $D=1$. These satisfy the hypotheses, yet $det Anot= det D$.






          share|cite|improve this answer























          • Thank you for very simple counterexample :)
            – user124697
            Aug 6 at 18:54














          up vote
          1
          down vote



          accepted










          Take $d=1$ and $n=2$; let $A=3$, $B=0$, $C=0$, $D=1$. These satisfy the hypotheses, yet $det Anot= det D$.






          share|cite|improve this answer























          • Thank you for very simple counterexample :)
            – user124697
            Aug 6 at 18:54












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Take $d=1$ and $n=2$; let $A=3$, $B=0$, $C=0$, $D=1$. These satisfy the hypotheses, yet $det Anot= det D$.






          share|cite|improve this answer















          Take $d=1$ and $n=2$; let $A=3$, $B=0$, $C=0$, $D=1$. These satisfy the hypotheses, yet $det Anot= det D$.







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 6 at 8:14


























          answered Aug 6 at 6:25









          ancientmathematician

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          • Thank you for very simple counterexample :)
            – user124697
            Aug 6 at 18:54
















          • Thank you for very simple counterexample :)
            – user124697
            Aug 6 at 18:54















          Thank you for very simple counterexample :)
          – user124697
          Aug 6 at 18:54




          Thank you for very simple counterexample :)
          – user124697
          Aug 6 at 18:54












           

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