Find the rank of matrix $B$












1












$begingroup$


If $A_{n times n}$ is a non-singular matrix, such that $A times B$ is the zero matrix ($ntimes n$), then the rank of $B$ is?



(A) $n-1$



(B) $n-2$



(C) $1$



(D) $0$



The answer to this question was given to be (D).



But I am confused why only (D).



$|AB|=0$



$|A|.|B|=0, ;|A| neq 0 rightarrow |B| = 0$.



Surely rank of $B$ is less than $n.$ But why exactly $0$?






linear-algebra







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  • 1




    $begingroup$
    Your argument only works if $B$ is square, otherwise it doesn't make sense to talk about its determinant.
    $endgroup$
    – pwerth
    Dec 25 '18 at 3:03










  • $begingroup$
    Yes, please guide now.
    $endgroup$
    – user3767495
    Dec 25 '18 at 3:04
















1












$begingroup$


If $A_{n times n}$ is a non-singular matrix, such that $A times B$ is the zero matrix ($ntimes n$), then the rank of $B$ is?



(A) $n-1$



(B) $n-2$



(C) $1$



(D) $0$



The answer to this question was given to be (D).



But I am confused why only (D).



$|AB|=0$



$|A|.|B|=0, ;|A| neq 0 rightarrow |B| = 0$.



Surely rank of $B$ is less than $n.$ But why exactly $0$?






linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Your argument only works if $B$ is square, otherwise it doesn't make sense to talk about its determinant.
    $endgroup$
    – pwerth
    Dec 25 '18 at 3:03










  • $begingroup$
    Yes, please guide now.
    $endgroup$
    – user3767495
    Dec 25 '18 at 3:04














1












1








1





$begingroup$


If $A_{n times n}$ is a non-singular matrix, such that $A times B$ is the zero matrix ($ntimes n$), then the rank of $B$ is?



(A) $n-1$



(B) $n-2$



(C) $1$



(D) $0$



The answer to this question was given to be (D).



But I am confused why only (D).



$|AB|=0$



$|A|.|B|=0, ;|A| neq 0 rightarrow |B| = 0$.



Surely rank of $B$ is less than $n.$ But why exactly $0$?






linear-algebra







share|cite|improve this question











$endgroup$




If $A_{n times n}$ is a non-singular matrix, such that $A times B$ is the zero matrix ($ntimes n$), then the rank of $B$ is?



(A) $n-1$



(B) $n-2$



(C) $1$



(D) $0$



The answer to this question was given to be (D).



But I am confused why only (D).



$|AB|=0$



$|A|.|B|=0, ;|A| neq 0 rightarrow |B| = 0$.



Surely rank of $B$ is less than $n.$ But why exactly $0$?






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 3:07









Namaste

1




1










asked Dec 25 '18 at 2:59









user3767495user3767495

40218




40218








  • 1




    $begingroup$
    Your argument only works if $B$ is square, otherwise it doesn't make sense to talk about its determinant.
    $endgroup$
    – pwerth
    Dec 25 '18 at 3:03










  • $begingroup$
    Yes, please guide now.
    $endgroup$
    – user3767495
    Dec 25 '18 at 3:04














  • 1




    $begingroup$
    Your argument only works if $B$ is square, otherwise it doesn't make sense to talk about its determinant.
    $endgroup$
    – pwerth
    Dec 25 '18 at 3:03










  • $begingroup$
    Yes, please guide now.
    $endgroup$
    – user3767495
    Dec 25 '18 at 3:04








1




1




$begingroup$
Your argument only works if $B$ is square, otherwise it doesn't make sense to talk about its determinant.
$endgroup$
– pwerth
Dec 25 '18 at 3:03




$begingroup$
Your argument only works if $B$ is square, otherwise it doesn't make sense to talk about its determinant.
$endgroup$
– pwerth
Dec 25 '18 at 3:03












$begingroup$
Yes, please guide now.
$endgroup$
– user3767495
Dec 25 '18 at 3:04




$begingroup$
Yes, please guide now.
$endgroup$
– user3767495
Dec 25 '18 at 3:04










2 Answers
2






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3












$begingroup$

If $A$ is non-singular, $A^{-1}$ exists. Then, $B = A^{-1}AB = A^{-1}0 = 0$.



Since $B$ is the zero-matrix (of whatever size), $text{rank}(B) = 0$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Multiplication $B$ with a non-singular matrix doesn't change its rank.



    This can be seen from a nonsingular matrix can be decomposed into a product of elementary matrices, and we know that performing elementary operation doesn't change the rank of a matrix.



    Hence if $A$ is nonsingular and $AB=C$, we have $rank(B)=rank(C)$.






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      If $A$ is non-singular, $A^{-1}$ exists. Then, $B = A^{-1}AB = A^{-1}0 = 0$.



      Since $B$ is the zero-matrix (of whatever size), $text{rank}(B) = 0$.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        If $A$ is non-singular, $A^{-1}$ exists. Then, $B = A^{-1}AB = A^{-1}0 = 0$.



        Since $B$ is the zero-matrix (of whatever size), $text{rank}(B) = 0$.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          If $A$ is non-singular, $A^{-1}$ exists. Then, $B = A^{-1}AB = A^{-1}0 = 0$.



          Since $B$ is the zero-matrix (of whatever size), $text{rank}(B) = 0$.






          share|cite|improve this answer









          $endgroup$



          If $A$ is non-singular, $A^{-1}$ exists. Then, $B = A^{-1}AB = A^{-1}0 = 0$.



          Since $B$ is the zero-matrix (of whatever size), $text{rank}(B) = 0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 25 '18 at 3:06









          JimmyK4542JimmyK4542

          41.8k248110




          41.8k248110























              1












              $begingroup$

              Multiplication $B$ with a non-singular matrix doesn't change its rank.



              This can be seen from a nonsingular matrix can be decomposed into a product of elementary matrices, and we know that performing elementary operation doesn't change the rank of a matrix.



              Hence if $A$ is nonsingular and $AB=C$, we have $rank(B)=rank(C)$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Multiplication $B$ with a non-singular matrix doesn't change its rank.



                This can be seen from a nonsingular matrix can be decomposed into a product of elementary matrices, and we know that performing elementary operation doesn't change the rank of a matrix.



                Hence if $A$ is nonsingular and $AB=C$, we have $rank(B)=rank(C)$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Multiplication $B$ with a non-singular matrix doesn't change its rank.



                  This can be seen from a nonsingular matrix can be decomposed into a product of elementary matrices, and we know that performing elementary operation doesn't change the rank of a matrix.



                  Hence if $A$ is nonsingular and $AB=C$, we have $rank(B)=rank(C)$.






                  share|cite|improve this answer









                  $endgroup$



                  Multiplication $B$ with a non-singular matrix doesn't change its rank.



                  This can be seen from a nonsingular matrix can be decomposed into a product of elementary matrices, and we know that performing elementary operation doesn't change the rank of a matrix.



                  Hence if $A$ is nonsingular and $AB=C$, we have $rank(B)=rank(C)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 25 '18 at 3:05









                  Siong Thye GohSiong Thye Goh

                  104k1469121




                  104k1469121






























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