partial isometry











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If $Ain M_{rtimes m}(mathbb{C}),Bin M_{mtimes r}(mathbb{C})$,$AB=Id_r$,then $A,B$ is a partial isometry.



My question:how to show $A^*A,B^*B$ is a projection?






linear-algebra operator-theory matrix-calculus c-star-algebras







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    up vote
    1
    down vote

    favorite












    If $Ain M_{rtimes m}(mathbb{C}),Bin M_{mtimes r}(mathbb{C})$,$AB=Id_r$,then $A,B$ is a partial isometry.



    My question:how to show $A^*A,B^*B$ is a projection?






    linear-algebra operator-theory matrix-calculus c-star-algebras







    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      If $Ain M_{rtimes m}(mathbb{C}),Bin M_{mtimes r}(mathbb{C})$,$AB=Id_r$,then $A,B$ is a partial isometry.



      My question:how to show $A^*A,B^*B$ is a projection?






      linear-algebra operator-theory matrix-calculus c-star-algebras







      share|cite|improve this question













      If $Ain M_{rtimes m}(mathbb{C}),Bin M_{mtimes r}(mathbb{C})$,$AB=Id_r$,then $A,B$ is a partial isometry.



      My question:how to show $A^*A,B^*B$ is a projection?






      linear-algebra operator-theory matrix-calculus c-star-algebras




      linear-algebra operator-theory matrix-calculus c-star-algebras






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked yesterday









      mathrookie

      696512




      696512






















          1 Answer
          1






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          up vote
          2
          down vote



          accepted










          This is not true. If $m=r=1$, you can take $A=2$, $B=1/2$, and none of them are partial isometries. Even if you require $|A|leq1$, $|B|leq1$, then statements fails: take
          $$
          A=begin{bmatrix} tfrac12&tfrac12end{bmatrix} B=begin{bmatrix} 1\ 1end{bmatrix}.
          $$
          What you can tell from $AB=I$ is that $BA$ is an idempotent:
          $$
          (BA)^2=B(AB)A=BA.
          $$
          It is not necessary that $BA$ is selfadjoint, though. For instance take
          $$
          A=begin{bmatrix} tfrac12&1end{bmatrix} B=begin{bmatrix} 1\ tfrac12end{bmatrix}.
          $$
          then $AB=1$, and $BA=begin{bmatrix} 1/2 & 1\ 1/4& 1/2end{bmatrix}$.



          You do have that $B^*A^*AB=I_r$, so $AB$ is a partial isometry, and $ABB^*A^*$ is a projection, since
          $$
          (ABB^*A^*)^2=ABB^*A^*ABB^*A^*=ABI_rB^*A^*=ABB^*A^*.
          $$






          share|cite|improve this answer





















          • I saw this conclusion from a book. It states that we can conclude $B$ is injective(view it as a map from $mathbb{C}^r$ to $mathbb{C}^m$) and $B$ is a partial isometry and hence $B$ is an isometry.I feel quite confused about it.
            – mathrookie
            2 hours ago












          • That $B$ is injective, yes. Also, $A$ is surjective. The rest of the conclusions do not hold, as you can see from the examples.
            – Martin Argerami
            1 hour ago











          Your Answer





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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          This is not true. If $m=r=1$, you can take $A=2$, $B=1/2$, and none of them are partial isometries. Even if you require $|A|leq1$, $|B|leq1$, then statements fails: take
          $$
          A=begin{bmatrix} tfrac12&tfrac12end{bmatrix} B=begin{bmatrix} 1\ 1end{bmatrix}.
          $$
          What you can tell from $AB=I$ is that $BA$ is an idempotent:
          $$
          (BA)^2=B(AB)A=BA.
          $$
          It is not necessary that $BA$ is selfadjoint, though. For instance take
          $$
          A=begin{bmatrix} tfrac12&1end{bmatrix} B=begin{bmatrix} 1\ tfrac12end{bmatrix}.
          $$
          then $AB=1$, and $BA=begin{bmatrix} 1/2 & 1\ 1/4& 1/2end{bmatrix}$.



          You do have that $B^*A^*AB=I_r$, so $AB$ is a partial isometry, and $ABB^*A^*$ is a projection, since
          $$
          (ABB^*A^*)^2=ABB^*A^*ABB^*A^*=ABI_rB^*A^*=ABB^*A^*.
          $$






          share|cite|improve this answer





















          • I saw this conclusion from a book. It states that we can conclude $B$ is injective(view it as a map from $mathbb{C}^r$ to $mathbb{C}^m$) and $B$ is a partial isometry and hence $B$ is an isometry.I feel quite confused about it.
            – mathrookie
            2 hours ago












          • That $B$ is injective, yes. Also, $A$ is surjective. The rest of the conclusions do not hold, as you can see from the examples.
            – Martin Argerami
            1 hour ago















          up vote
          2
          down vote



          accepted










          This is not true. If $m=r=1$, you can take $A=2$, $B=1/2$, and none of them are partial isometries. Even if you require $|A|leq1$, $|B|leq1$, then statements fails: take
          $$
          A=begin{bmatrix} tfrac12&tfrac12end{bmatrix} B=begin{bmatrix} 1\ 1end{bmatrix}.
          $$
          What you can tell from $AB=I$ is that $BA$ is an idempotent:
          $$
          (BA)^2=B(AB)A=BA.
          $$
          It is not necessary that $BA$ is selfadjoint, though. For instance take
          $$
          A=begin{bmatrix} tfrac12&1end{bmatrix} B=begin{bmatrix} 1\ tfrac12end{bmatrix}.
          $$
          then $AB=1$, and $BA=begin{bmatrix} 1/2 & 1\ 1/4& 1/2end{bmatrix}$.



          You do have that $B^*A^*AB=I_r$, so $AB$ is a partial isometry, and $ABB^*A^*$ is a projection, since
          $$
          (ABB^*A^*)^2=ABB^*A^*ABB^*A^*=ABI_rB^*A^*=ABB^*A^*.
          $$






          share|cite|improve this answer





















          • I saw this conclusion from a book. It states that we can conclude $B$ is injective(view it as a map from $mathbb{C}^r$ to $mathbb{C}^m$) and $B$ is a partial isometry and hence $B$ is an isometry.I feel quite confused about it.
            – mathrookie
            2 hours ago












          • That $B$ is injective, yes. Also, $A$ is surjective. The rest of the conclusions do not hold, as you can see from the examples.
            – Martin Argerami
            1 hour ago













          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          This is not true. If $m=r=1$, you can take $A=2$, $B=1/2$, and none of them are partial isometries. Even if you require $|A|leq1$, $|B|leq1$, then statements fails: take
          $$
          A=begin{bmatrix} tfrac12&tfrac12end{bmatrix} B=begin{bmatrix} 1\ 1end{bmatrix}.
          $$
          What you can tell from $AB=I$ is that $BA$ is an idempotent:
          $$
          (BA)^2=B(AB)A=BA.
          $$
          It is not necessary that $BA$ is selfadjoint, though. For instance take
          $$
          A=begin{bmatrix} tfrac12&1end{bmatrix} B=begin{bmatrix} 1\ tfrac12end{bmatrix}.
          $$
          then $AB=1$, and $BA=begin{bmatrix} 1/2 & 1\ 1/4& 1/2end{bmatrix}$.



          You do have that $B^*A^*AB=I_r$, so $AB$ is a partial isometry, and $ABB^*A^*$ is a projection, since
          $$
          (ABB^*A^*)^2=ABB^*A^*ABB^*A^*=ABI_rB^*A^*=ABB^*A^*.
          $$






          share|cite|improve this answer












          This is not true. If $m=r=1$, you can take $A=2$, $B=1/2$, and none of them are partial isometries. Even if you require $|A|leq1$, $|B|leq1$, then statements fails: take
          $$
          A=begin{bmatrix} tfrac12&tfrac12end{bmatrix} B=begin{bmatrix} 1\ 1end{bmatrix}.
          $$
          What you can tell from $AB=I$ is that $BA$ is an idempotent:
          $$
          (BA)^2=B(AB)A=BA.
          $$
          It is not necessary that $BA$ is selfadjoint, though. For instance take
          $$
          A=begin{bmatrix} tfrac12&1end{bmatrix} B=begin{bmatrix} 1\ tfrac12end{bmatrix}.
          $$
          then $AB=1$, and $BA=begin{bmatrix} 1/2 & 1\ 1/4& 1/2end{bmatrix}$.



          You do have that $B^*A^*AB=I_r$, so $AB$ is a partial isometry, and $ABB^*A^*$ is a projection, since
          $$
          (ABB^*A^*)^2=ABB^*A^*ABB^*A^*=ABI_rB^*A^*=ABB^*A^*.
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          Martin Argerami

          121k1073172




          121k1073172












          • I saw this conclusion from a book. It states that we can conclude $B$ is injective(view it as a map from $mathbb{C}^r$ to $mathbb{C}^m$) and $B$ is a partial isometry and hence $B$ is an isometry.I feel quite confused about it.
            – mathrookie
            2 hours ago












          • That $B$ is injective, yes. Also, $A$ is surjective. The rest of the conclusions do not hold, as you can see from the examples.
            – Martin Argerami
            1 hour ago


















          • I saw this conclusion from a book. It states that we can conclude $B$ is injective(view it as a map from $mathbb{C}^r$ to $mathbb{C}^m$) and $B$ is a partial isometry and hence $B$ is an isometry.I feel quite confused about it.
            – mathrookie
            2 hours ago












          • That $B$ is injective, yes. Also, $A$ is surjective. The rest of the conclusions do not hold, as you can see from the examples.
            – Martin Argerami
            1 hour ago
















          I saw this conclusion from a book. It states that we can conclude $B$ is injective(view it as a map from $mathbb{C}^r$ to $mathbb{C}^m$) and $B$ is a partial isometry and hence $B$ is an isometry.I feel quite confused about it.
          – mathrookie
          2 hours ago






          I saw this conclusion from a book. It states that we can conclude $B$ is injective(view it as a map from $mathbb{C}^r$ to $mathbb{C}^m$) and $B$ is a partial isometry and hence $B$ is an isometry.I feel quite confused about it.
          – mathrookie
          2 hours ago














          That $B$ is injective, yes. Also, $A$ is surjective. The rest of the conclusions do not hold, as you can see from the examples.
          – Martin Argerami
          1 hour ago




          That $B$ is injective, yes. Also, $A$ is surjective. The rest of the conclusions do not hold, as you can see from the examples.
          – Martin Argerami
          1 hour ago


















           

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