Mixing
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
asked yesterday
mathrookie
696512
add a comment |
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
asked yesterday
mathrookie
696512
add a comment |
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
asked yesterday
mathrookie
696512
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
asked yesterday
mathrookie
696512
asked yesterday
mathrookie
696512
asked yesterday
mathrookie
696512
asked yesterday
mathrookie
696512
asked yesterday
mathrookie
696512
696512
add a comment |
add a comment |
1 Answer
1
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oldest
votes
up vote
2
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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
$$
(ABB^*A^*)^2=ABB^*A^*ABB^*A^*=ABI_rB^*A^*=ABB^*A^*.
$$
answered yesterday
Martin Argerami
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
add a comment |
1 Answer
1
active
oldest
votes
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
$$
(ABB^*A^*)^2=ABB^*A^*ABB^*A^*=ABI_rB^*A^*=ABB^*A^*.
$$
answered yesterday
Martin Argerami
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
add a comment |
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
$$
(ABB^*A^*)^2=ABB^*A^*ABB^*A^*=ABI_rB^*A^*=ABB^*A^*.
$$
answered yesterday
Martin Argerami
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
add a comment |
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
$$
(ABB^*A^*)^2=ABB^*A^*ABB^*A^*=ABI_rB^*A^*=ABB^*A^*.
$$
answered yesterday
Martin Argerami
121k1073172
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
$$
(ABB^*A^*)^2=ABB^*A^*ABB^*A^*=ABI_rB^*A^*=ABB^*A^*.
$$
answered yesterday
Martin Argerami
121k1073172
answered yesterday
Martin Argerami
121k1073172
answered yesterday
Martin Argerami
121k1073172
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
add a comment |
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
add a comment |
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