Projection Matrixes, $A,B$ such that $text{Id} - (A+B)$ is invertible












3












$begingroup$


This question was on an old qual exam and I have been stuck on it:



Let $A,B$ be two real $5x5$ matrices such that $A^2=A , B^2 = B$ and $text{Id} - (A+B)$ is invertible. Show Rank($A$)=Rank($B$).



My biggest question is why is $A,B$ being $5x5$ needed? I don't see why this wouldn't generalize to any square matrix with the above properties.



I have tried to use that $mathbb{R}^5 = text{Im}(A)bigoplustext{Ker}(A)$ and similarly for $B$ and tried to show nullity(A) = nullity(B) to no avail. I believe the strongest assumption given is that $text{Id} - (A+B)$ is invertible, but I am unsure how to use this to prove Rank($A$) = Rank($B$).






linear-algebra matrix-rank projection-matrices







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  • 1




    $begingroup$
    see the possible Jordan's form for $A$ and $B$
    $endgroup$
    – Martín Vacas Vignolo
    Dec 23 '18 at 5:30
















3












$begingroup$


This question was on an old qual exam and I have been stuck on it:



Let $A,B$ be two real $5x5$ matrices such that $A^2=A , B^2 = B$ and $text{Id} - (A+B)$ is invertible. Show Rank($A$)=Rank($B$).



My biggest question is why is $A,B$ being $5x5$ needed? I don't see why this wouldn't generalize to any square matrix with the above properties.



I have tried to use that $mathbb{R}^5 = text{Im}(A)bigoplustext{Ker}(A)$ and similarly for $B$ and tried to show nullity(A) = nullity(B) to no avail. I believe the strongest assumption given is that $text{Id} - (A+B)$ is invertible, but I am unsure how to use this to prove Rank($A$) = Rank($B$).






linear-algebra matrix-rank projection-matrices







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    see the possible Jordan's form for $A$ and $B$
    $endgroup$
    – Martín Vacas Vignolo
    Dec 23 '18 at 5:30














3












3








3





$begingroup$


This question was on an old qual exam and I have been stuck on it:



Let $A,B$ be two real $5x5$ matrices such that $A^2=A , B^2 = B$ and $text{Id} - (A+B)$ is invertible. Show Rank($A$)=Rank($B$).



My biggest question is why is $A,B$ being $5x5$ needed? I don't see why this wouldn't generalize to any square matrix with the above properties.



I have tried to use that $mathbb{R}^5 = text{Im}(A)bigoplustext{Ker}(A)$ and similarly for $B$ and tried to show nullity(A) = nullity(B) to no avail. I believe the strongest assumption given is that $text{Id} - (A+B)$ is invertible, but I am unsure how to use this to prove Rank($A$) = Rank($B$).






linear-algebra matrix-rank projection-matrices







share|cite|improve this question









$endgroup$




This question was on an old qual exam and I have been stuck on it:



Let $A,B$ be two real $5x5$ matrices such that $A^2=A , B^2 = B$ and $text{Id} - (A+B)$ is invertible. Show Rank($A$)=Rank($B$).



My biggest question is why is $A,B$ being $5x5$ needed? I don't see why this wouldn't generalize to any square matrix with the above properties.



I have tried to use that $mathbb{R}^5 = text{Im}(A)bigoplustext{Ker}(A)$ and similarly for $B$ and tried to show nullity(A) = nullity(B) to no avail. I believe the strongest assumption given is that $text{Id} - (A+B)$ is invertible, but I am unsure how to use this to prove Rank($A$) = Rank($B$).






linear-algebra matrix-rank projection-matrices




linear-algebra matrix-rank projection-matrices






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asked Dec 23 '18 at 5:27









Story123Story123

18519




18519








  • 1




    $begingroup$
    see the possible Jordan's form for $A$ and $B$
    $endgroup$
    – Martín Vacas Vignolo
    Dec 23 '18 at 5:30














  • 1




    $begingroup$
    see the possible Jordan's form for $A$ and $B$
    $endgroup$
    – Martín Vacas Vignolo
    Dec 23 '18 at 5:30








1




1




$begingroup$
see the possible Jordan's form for $A$ and $B$
$endgroup$
– Martín Vacas Vignolo
Dec 23 '18 at 5:30




$begingroup$
see the possible Jordan's form for $A$ and $B$
$endgroup$
– Martín Vacas Vignolo
Dec 23 '18 at 5:30










2 Answers
2






active

oldest

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4












$begingroup$

The $5times 5$ assumption is indeed irrelevant. As a hint, try writing $Id-(A+B)$ as $(Id-A)-B$.



A full proof is hidden below.




Let's say our matrices are $ntimes n$, and $A$ has rank $a$ and $B$ has rank $b$. Note then that $Id-A$ is a projection of rank $n-a$, and so $(Id-A)-B$ has rank at most $n-a+b$ (its image is contained in the sum of the image of $Id-A$ and the image of $B$). Since $(Id-A)-B$ is invertible, we have $nleq n-a+b$ and so $aleq b$. But swapping the roles of $A$ and $B$, we also get $bleq a$, and so $a=b$.







share|cite|improve this answer









$endgroup$





















    5












    $begingroup$

    You don't need $5times 5$, $ntimes n$ works.



    You also don't need $Bbb R$, any field will work.



    As you say, consider $Bbb R^n=text{im},A+ker A$.
    On $text{im}, A$, $I-A$ vanishes, so for $I-A-B$ to be invertible,
    $B$ must be injective on $text{im}, A$. Therefore $text{rank},Bgedim
    text{im},A=text{rank},A$    .






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      The $5times 5$ assumption is indeed irrelevant. As a hint, try writing $Id-(A+B)$ as $(Id-A)-B$.



      A full proof is hidden below.




      Let's say our matrices are $ntimes n$, and $A$ has rank $a$ and $B$ has rank $b$. Note then that $Id-A$ is a projection of rank $n-a$, and so $(Id-A)-B$ has rank at most $n-a+b$ (its image is contained in the sum of the image of $Id-A$ and the image of $B$). Since $(Id-A)-B$ is invertible, we have $nleq n-a+b$ and so $aleq b$. But swapping the roles of $A$ and $B$, we also get $bleq a$, and so $a=b$.







      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        The $5times 5$ assumption is indeed irrelevant. As a hint, try writing $Id-(A+B)$ as $(Id-A)-B$.



        A full proof is hidden below.




        Let's say our matrices are $ntimes n$, and $A$ has rank $a$ and $B$ has rank $b$. Note then that $Id-A$ is a projection of rank $n-a$, and so $(Id-A)-B$ has rank at most $n-a+b$ (its image is contained in the sum of the image of $Id-A$ and the image of $B$). Since $(Id-A)-B$ is invertible, we have $nleq n-a+b$ and so $aleq b$. But swapping the roles of $A$ and $B$, we also get $bleq a$, and so $a=b$.







        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          The $5times 5$ assumption is indeed irrelevant. As a hint, try writing $Id-(A+B)$ as $(Id-A)-B$.



          A full proof is hidden below.




          Let's say our matrices are $ntimes n$, and $A$ has rank $a$ and $B$ has rank $b$. Note then that $Id-A$ is a projection of rank $n-a$, and so $(Id-A)-B$ has rank at most $n-a+b$ (its image is contained in the sum of the image of $Id-A$ and the image of $B$). Since $(Id-A)-B$ is invertible, we have $nleq n-a+b$ and so $aleq b$. But swapping the roles of $A$ and $B$, we also get $bleq a$, and so $a=b$.







          share|cite|improve this answer









          $endgroup$



          The $5times 5$ assumption is indeed irrelevant. As a hint, try writing $Id-(A+B)$ as $(Id-A)-B$.



          A full proof is hidden below.




          Let's say our matrices are $ntimes n$, and $A$ has rank $a$ and $B$ has rank $b$. Note then that $Id-A$ is a projection of rank $n-a$, and so $(Id-A)-B$ has rank at most $n-a+b$ (its image is contained in the sum of the image of $Id-A$ and the image of $B$). Since $(Id-A)-B$ is invertible, we have $nleq n-a+b$ and so $aleq b$. But swapping the roles of $A$ and $B$, we also get $bleq a$, and so $a=b$.








          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 23 '18 at 5:42









          Eric WofseyEric Wofsey

          193k14221352




          193k14221352























              5












              $begingroup$

              You don't need $5times 5$, $ntimes n$ works.



              You also don't need $Bbb R$, any field will work.



              As you say, consider $Bbb R^n=text{im},A+ker A$.
              On $text{im}, A$, $I-A$ vanishes, so for $I-A-B$ to be invertible,
              $B$ must be injective on $text{im}, A$. Therefore $text{rank},Bgedim
              text{im},A=text{rank},A$              .






              share|cite|improve this answer









              $endgroup$


















                5












                $begingroup$

                You don't need $5times 5$, $ntimes n$ works.



                You also don't need $Bbb R$, any field will work.



                As you say, consider $Bbb R^n=text{im},A+ker A$.
                On $text{im}, A$, $I-A$ vanishes, so for $I-A-B$ to be invertible,
                $B$ must be injective on $text{im}, A$. Therefore $text{rank},Bgedim
                text{im},A=text{rank},A$                .






                share|cite|improve this answer









                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  You don't need $5times 5$, $ntimes n$ works.



                  You also don't need $Bbb R$, any field will work.



                  As you say, consider $Bbb R^n=text{im},A+ker A$.
                  On $text{im}, A$, $I-A$ vanishes, so for $I-A-B$ to be invertible,
                  $B$ must be injective on $text{im}, A$. Therefore $text{rank},Bgedim
                  text{im},A=text{rank},A$                  .






                  share|cite|improve this answer









                  $endgroup$



                  You don't need $5times 5$, $ntimes n$ works.



                  You also don't need $Bbb R$, any field will work.



                  As you say, consider $Bbb R^n=text{im},A+ker A$.
                  On $text{im}, A$, $I-A$ vanishes, so for $I-A-B$ to be invertible,
                  $B$ must be injective on $text{im}, A$. Therefore $text{rank},Bgedim
                  text{im},A=text{rank},A$                  .







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 23 '18 at 5:42









                  Lord Shark the UnknownLord Shark the Unknown

                  109k1163136




                  109k1163136






























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