Is it oblique projection matrix $P=x.y*$












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$begingroup$


Let $A in C^{n times n}$ be nondefective matrix and let $x$ and $y$ be right and left eigenvectors of $A$, (with corresponding simple eigenvalue $l$). Then we have that $y^*.x=1$, where $y^*$ is the conjugate transpose of $y$.
Are all the following statements true?:



1) The matrix $P=x.y^*$ is oblique projection matrix, and for any $vin C^n$ we have $P.v$ is in the direction of $x$;



2) The matrix $I-P$ is oblique projection matrix on the complement of $x$;



3) Any vector $vin C^n (vnot=0)$ can be presented uniquely in the form
$$
v=x'+d,
$$

where $d$ is orthogonal to $y$, (i.e. $y^*.d=0$) and $x'$ is in the direction of $x$.



4) Moreover, in the upper expression we can say that:
$$
x'=Pv
$$

and
$$
d=(I-P)v.
$$



6) Because of $y^*.(I-P)=y^*.(I-x.y^*)=y^*-y^*.x.y^*=y^*-y^*=0$, it follows that the vector $(I-P).w$ is orthogonal to $y$ for any $w in C^n$.



Please correct me if I have inaccuracies in my conclusions.










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$endgroup$

















    0












    $begingroup$


    Let $A in C^{n times n}$ be nondefective matrix and let $x$ and $y$ be right and left eigenvectors of $A$, (with corresponding simple eigenvalue $l$). Then we have that $y^*.x=1$, where $y^*$ is the conjugate transpose of $y$.
    Are all the following statements true?:



    1) The matrix $P=x.y^*$ is oblique projection matrix, and for any $vin C^n$ we have $P.v$ is in the direction of $x$;



    2) The matrix $I-P$ is oblique projection matrix on the complement of $x$;



    3) Any vector $vin C^n (vnot=0)$ can be presented uniquely in the form
    $$
    v=x'+d,
    $$

    where $d$ is orthogonal to $y$, (i.e. $y^*.d=0$) and $x'$ is in the direction of $x$.



    4) Moreover, in the upper expression we can say that:
    $$
    x'=Pv
    $$

    and
    $$
    d=(I-P)v.
    $$



    6) Because of $y^*.(I-P)=y^*.(I-x.y^*)=y^*-y^*.x.y^*=y^*-y^*=0$, it follows that the vector $(I-P).w$ is orthogonal to $y$ for any $w in C^n$.



    Please correct me if I have inaccuracies in my conclusions.










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $A in C^{n times n}$ be nondefective matrix and let $x$ and $y$ be right and left eigenvectors of $A$, (with corresponding simple eigenvalue $l$). Then we have that $y^*.x=1$, where $y^*$ is the conjugate transpose of $y$.
      Are all the following statements true?:



      1) The matrix $P=x.y^*$ is oblique projection matrix, and for any $vin C^n$ we have $P.v$ is in the direction of $x$;



      2) The matrix $I-P$ is oblique projection matrix on the complement of $x$;



      3) Any vector $vin C^n (vnot=0)$ can be presented uniquely in the form
      $$
      v=x'+d,
      $$

      where $d$ is orthogonal to $y$, (i.e. $y^*.d=0$) and $x'$ is in the direction of $x$.



      4) Moreover, in the upper expression we can say that:
      $$
      x'=Pv
      $$

      and
      $$
      d=(I-P)v.
      $$



      6) Because of $y^*.(I-P)=y^*.(I-x.y^*)=y^*-y^*.x.y^*=y^*-y^*=0$, it follows that the vector $(I-P).w$ is orthogonal to $y$ for any $w in C^n$.



      Please correct me if I have inaccuracies in my conclusions.










      share|cite|improve this question









      $endgroup$




      Let $A in C^{n times n}$ be nondefective matrix and let $x$ and $y$ be right and left eigenvectors of $A$, (with corresponding simple eigenvalue $l$). Then we have that $y^*.x=1$, where $y^*$ is the conjugate transpose of $y$.
      Are all the following statements true?:



      1) The matrix $P=x.y^*$ is oblique projection matrix, and for any $vin C^n$ we have $P.v$ is in the direction of $x$;



      2) The matrix $I-P$ is oblique projection matrix on the complement of $x$;



      3) Any vector $vin C^n (vnot=0)$ can be presented uniquely in the form
      $$
      v=x'+d,
      $$

      where $d$ is orthogonal to $y$, (i.e. $y^*.d=0$) and $x'$ is in the direction of $x$.



      4) Moreover, in the upper expression we can say that:
      $$
      x'=Pv
      $$

      and
      $$
      d=(I-P)v.
      $$



      6) Because of $y^*.(I-P)=y^*.(I-x.y^*)=y^*-y^*.x.y^*=y^*-y^*=0$, it follows that the vector $(I-P).w$ is orthogonal to $y$ for any $w in C^n$.



      Please correct me if I have inaccuracies in my conclusions.







      eigenvalues-eigenvectors projection-matrices






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      asked Jan 19 at 14:32









      Şenay ErdinçŞenay Erdinç

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