Projector equal to identity matrix

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I have a simple question. If I take a projector f with matrix $M$, then I have :



$$M^2=M$$, so by multiplying by $M^-1$ on the right, I get :



$MMM^-1= I_d$, so $M=I_d$, but we can find matrices $M$ which are not identity and represent a projector.



Where is my error ?



Regards




matrices projection




share|cite|improve this question

























    up vote
    0
    down vote

    favorite












    I have a simple question. If I take a projector f with matrix $M$, then I have :



    $$M^2=M$$, so by multiplying by $M^-1$ on the right, I get :



    $MMM^-1= I_d$, so $M=I_d$, but we can find matrices $M$ which are not identity and represent a projector.



    Where is my error ?



    Regards




    matrices projection




    share|cite|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I have a simple question. If I take a projector f with matrix $M$, then I have :



      $$M^2=M$$, so by multiplying by $M^-1$ on the right, I get :



      $MMM^-1= I_d$, so $M=I_d$, but we can find matrices $M$ which are not identity and represent a projector.



      Where is my error ?



      Regards




      matrices projection




      share|cite|improve this question













      I have a simple question. If I take a projector f with matrix $M$, then I have :



      $$M^2=M$$, so by multiplying by $M^-1$ on the right, I get :



      $MMM^-1= I_d$, so $M=I_d$, but we can find matrices $M$ which are not identity and represent a projector.



      Where is my error ?



      Regards





      matrices projection





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Jul 27 at 10:07









      José Carlos Santos

      113k1696173




      113k1696173









      asked Jul 27 at 10:04









      youpilat13

      6911




      6911




















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          Your error lies in assuming that $M$ is invertible. Actually, you proved that the only invertible projection is the identity.






          share|cite|improve this answer





















          • -@José Carlos Santos. thanks, so when we get in generally a projector matrix (different from Identity), it is not invertible, is it ?
            – youpilat13
            Jul 27 at 10:12










          • Think about what a projection does geometrically. It maps vectors to a subspace, like the shadow of a figure on a wall. You cannot reproduce the figure from the shadow.
            – Paul
            Jul 27 at 10:16










          • Yes. If we are projecting over a subspace which is not the whole space, then the projection map is not surjective and therefore it is not invertible.
            – José Carlos Santos
            Jul 27 at 10:17

















          up vote
          0
          down vote













          If $M^2=M$, then we have:



          $M$ is invertible $ iff M=Id$.



          Hence (if $M^2=M$), the



          $M$ is not invertible $ iff M ne Id$.



          Try also to prove: if $M^2=M$, then the only possible eigenvalues of $M$ are $0$ and $1$.






          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








            up vote
            1
            down vote



            accepted










            Your error lies in assuming that $M$ is invertible. Actually, you proved that the only invertible projection is the identity.






            share|cite|improve this answer





















            • -@José Carlos Santos. thanks, so when we get in generally a projector matrix (different from Identity), it is not invertible, is it ?
              – youpilat13
              Jul 27 at 10:12










            • Think about what a projection does geometrically. It maps vectors to a subspace, like the shadow of a figure on a wall. You cannot reproduce the figure from the shadow.
              – Paul
              Jul 27 at 10:16










            • Yes. If we are projecting over a subspace which is not the whole space, then the projection map is not surjective and therefore it is not invertible.
              – José Carlos Santos
              Jul 27 at 10:17














            up vote
            1
            down vote



            accepted










            Your error lies in assuming that $M$ is invertible. Actually, you proved that the only invertible projection is the identity.






            share|cite|improve this answer





















            • -@José Carlos Santos. thanks, so when we get in generally a projector matrix (different from Identity), it is not invertible, is it ?
              – youpilat13
              Jul 27 at 10:12










            • Think about what a projection does geometrically. It maps vectors to a subspace, like the shadow of a figure on a wall. You cannot reproduce the figure from the shadow.
              – Paul
              Jul 27 at 10:16










            • Yes. If we are projecting over a subspace which is not the whole space, then the projection map is not surjective and therefore it is not invertible.
              – José Carlos Santos
              Jul 27 at 10:17












            up vote
            1
            down vote



            accepted







            up vote
            1
            down vote



            accepted






            Your error lies in assuming that $M$ is invertible. Actually, you proved that the only invertible projection is the identity.






            share|cite|improve this answer













            Your error lies in assuming that $M$ is invertible. Actually, you proved that the only invertible projection is the identity.







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            answered Jul 27 at 10:06









            José Carlos Santos

            113k1696173




            113k1696173











            • -@José Carlos Santos. thanks, so when we get in generally a projector matrix (different from Identity), it is not invertible, is it ?
              – youpilat13
              Jul 27 at 10:12










            • Think about what a projection does geometrically. It maps vectors to a subspace, like the shadow of a figure on a wall. You cannot reproduce the figure from the shadow.
              – Paul
              Jul 27 at 10:16










            • Yes. If we are projecting over a subspace which is not the whole space, then the projection map is not surjective and therefore it is not invertible.
              – José Carlos Santos
              Jul 27 at 10:17
















            • -@José Carlos Santos. thanks, so when we get in generally a projector matrix (different from Identity), it is not invertible, is it ?
              – youpilat13
              Jul 27 at 10:12










            • Think about what a projection does geometrically. It maps vectors to a subspace, like the shadow of a figure on a wall. You cannot reproduce the figure from the shadow.
              – Paul
              Jul 27 at 10:16










            • Yes. If we are projecting over a subspace which is not the whole space, then the projection map is not surjective and therefore it is not invertible.
              – José Carlos Santos
              Jul 27 at 10:17















            -@José Carlos Santos. thanks, so when we get in generally a projector matrix (different from Identity), it is not invertible, is it ?
            – youpilat13
            Jul 27 at 10:12




            -@José Carlos Santos. thanks, so when we get in generally a projector matrix (different from Identity), it is not invertible, is it ?
            – youpilat13
            Jul 27 at 10:12












            Think about what a projection does geometrically. It maps vectors to a subspace, like the shadow of a figure on a wall. You cannot reproduce the figure from the shadow.
            – Paul
            Jul 27 at 10:16




            Think about what a projection does geometrically. It maps vectors to a subspace, like the shadow of a figure on a wall. You cannot reproduce the figure from the shadow.
            – Paul
            Jul 27 at 10:16












            Yes. If we are projecting over a subspace which is not the whole space, then the projection map is not surjective and therefore it is not invertible.
            – José Carlos Santos
            Jul 27 at 10:17




            Yes. If we are projecting over a subspace which is not the whole space, then the projection map is not surjective and therefore it is not invertible.
            – José Carlos Santos
            Jul 27 at 10:17










            up vote
            0
            down vote













            If $M^2=M$, then we have:



            $M$ is invertible $ iff M=Id$.



            Hence (if $M^2=M$), the



            $M$ is not invertible $ iff M ne Id$.



            Try also to prove: if $M^2=M$, then the only possible eigenvalues of $M$ are $0$ and $1$.






            share|cite|improve this answer

























              up vote
              0
              down vote













              If $M^2=M$, then we have:



              $M$ is invertible $ iff M=Id$.



              Hence (if $M^2=M$), the



              $M$ is not invertible $ iff M ne Id$.



              Try also to prove: if $M^2=M$, then the only possible eigenvalues of $M$ are $0$ and $1$.






              share|cite|improve this answer























                up vote
                0
                down vote










                up vote
                0
                down vote









                If $M^2=M$, then we have:



                $M$ is invertible $ iff M=Id$.



                Hence (if $M^2=M$), the



                $M$ is not invertible $ iff M ne Id$.



                Try also to prove: if $M^2=M$, then the only possible eigenvalues of $M$ are $0$ and $1$.






                share|cite|improve this answer













                If $M^2=M$, then we have:



                $M$ is invertible $ iff M=Id$.



                Hence (if $M^2=M$), the



                $M$ is not invertible $ iff M ne Id$.



                Try also to prove: if $M^2=M$, then the only possible eigenvalues of $M$ are $0$ and $1$.







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 27 at 10:10









                Fred

                37.2k1237




                37.2k1237






















                     

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