Is it oblique projection matrix $P=x.y*$
$begingroup$
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.
eigenvalues-eigenvectors projection-matrices
$endgroup$
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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.
eigenvalues-eigenvectors projection-matrices
$endgroup$
add a comment |
$begingroup$
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.
eigenvalues-eigenvectors projection-matrices
$endgroup$
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
eigenvalues-eigenvectors projection-matrices
asked Jan 19 at 14:32
Şenay ErdinçŞenay Erdinç
62
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