a question on partial isometry
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This is a statement from wikipedia.I don't understand why can we deduce that $C$ is a partial isometry if $A^*A=B^*B$.
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Nov 26 at 1:32
mathrookie
765512
add a comment |
up vote
1
down vote
favorite
This is a statement from wikipedia.I don't understand why can we deduce that $C$ is a partial isometry if $A^*A=B^*B$.
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Nov 26 at 1:32
mathrookie
765512
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This is a statement from wikipedia.I don't understand why can we deduce that $C$ is a partial isometry if $A^*A=B^*B$.
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Nov 26 at 1:32
mathrookie
765512
This is a statement from wikipedia.I don't understand why can we deduce that $C$ is a partial isometry if $A^*A=B^*B$.
operator-theory operator-algebras c-star-algebras von-neumann-algebras
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Nov 26 at 1:32
mathrookie
765512
asked Nov 26 at 1:32
mathrookie
765512
asked Nov 26 at 1:32
mathrookie
765512
asked Nov 26 at 1:32
mathrookie
765512
asked Nov 26 at 1:32
mathrookie
765512
765512
add a comment |
add a comment |
1 Answer
1
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oldest
votes
up vote
1
down vote
accepted
You have
$$
|CBx|^2=langle B^*C^*CBx,xrangle=langle A^*Ax,xrangle=langle B^*Bx,xrangle=|Bx|^2.
$$
So $C$ is isometric on $operatorname{ran}B$, and so on $overline{operatorname{ran}B}$. On the orthogonal complement, it is $0$ by definition. So $C$ is a partial isometry.
answered Nov 26 at 2:34
Martin Argerami
122k1175173
You mean that $kerC=bar{ranB}^perp$ ,but if $xin bar{ran(B)}$,why can we have $Cxneq 0$?
– mathrookie
Dec 5 at 0:52
Because $C$ is isometric on $operatorname{ran} B$.
– Martin Argerami
Dec 5 at 1:56
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You have
$$
|CBx|^2=langle B^*C^*CBx,xrangle=langle A^*Ax,xrangle=langle B^*Bx,xrangle=|Bx|^2.
$$
So $C$ is isometric on $operatorname{ran}B$, and so on $overline{operatorname{ran}B}$. On the orthogonal complement, it is $0$ by definition. So $C$ is a partial isometry.
answered Nov 26 at 2:34
Martin Argerami
122k1175173
You mean that $kerC=bar{ranB}^perp$ ,but if $xin bar{ran(B)}$,why can we have $Cxneq 0$?
– mathrookie
Dec 5 at 0:52
Because $C$ is isometric on $operatorname{ran} B$.
– Martin Argerami
Dec 5 at 1:56
add a comment |
up vote
1
down vote
accepted
You have
$$
|CBx|^2=langle B^*C^*CBx,xrangle=langle A^*Ax,xrangle=langle B^*Bx,xrangle=|Bx|^2.
$$
So $C$ is isometric on $operatorname{ran}B$, and so on $overline{operatorname{ran}B}$. On the orthogonal complement, it is $0$ by definition. So $C$ is a partial isometry.
answered Nov 26 at 2:34
Martin Argerami
122k1175173
You mean that $kerC=bar{ranB}^perp$ ,but if $xin bar{ran(B)}$,why can we have $Cxneq 0$?
– mathrookie
Dec 5 at 0:52
Because $C$ is isometric on $operatorname{ran} B$.
– Martin Argerami
Dec 5 at 1:56
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You have
$$
|CBx|^2=langle B^*C^*CBx,xrangle=langle A^*Ax,xrangle=langle B^*Bx,xrangle=|Bx|^2.
$$
So $C$ is isometric on $operatorname{ran}B$, and so on $overline{operatorname{ran}B}$. On the orthogonal complement, it is $0$ by definition. So $C$ is a partial isometry.
answered Nov 26 at 2:34
Martin Argerami
122k1175173
You have
$$
|CBx|^2=langle B^*C^*CBx,xrangle=langle A^*Ax,xrangle=langle B^*Bx,xrangle=|Bx|^2.
$$
So $C$ is isometric on $operatorname{ran}B$, and so on $overline{operatorname{ran}B}$. On the orthogonal complement, it is $0$ by definition. So $C$ is a partial isometry.
answered Nov 26 at 2:34
Martin Argerami
122k1175173
answered Nov 26 at 2:34
Martin Argerami
122k1175173
answered Nov 26 at 2:34
Martin Argerami
122k1175173
answered Nov 26 at 2:34
Martin Argerami
122k1175173
122k1175173
You mean that $kerC=bar{ranB}^perp$ ,but if $xin bar{ran(B)}$,why can we have $Cxneq 0$?
– mathrookie
Dec 5 at 0:52
Because $C$ is isometric on $operatorname{ran} B$.
– Martin Argerami
Dec 5 at 1:56
add a comment |
You mean that $kerC=bar{ranB}^perp$ ,but if $xin bar{ran(B)}$,why can we have $Cxneq 0$?
– mathrookie
Dec 5 at 0:52
Because $C$ is isometric on $operatorname{ran} B$.
– Martin Argerami
Dec 5 at 1:56
You mean that $kerC=bar{ranB}^perp$ ,but if $xin bar{ran(B)}$,why can we have $Cxneq 0$?
– mathrookie
Dec 5 at 0:52
You mean that $kerC=bar{ranB}^perp$ ,but if $xin bar{ran(B)}$,why can we have $Cxneq 0$?
– mathrookie
Dec 5 at 0:52
Because $C$ is isometric on $operatorname{ran} B$.
– Martin Argerami
Dec 5 at 1:56
Because $C$ is isometric on $operatorname{ran} B$.
– Martin Argerami
Dec 5 at 1:56
add a comment |
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