Let $A$ be $C ^ {*}$ algebra, $a ∈ A$,$ p$, $q ∈ A$ be orthogonal projections Show that if a is positive...
$begingroup$
Let $A$ be $C ^{ *}$ algebra, $a ∈ A$, $p$, $q ∈ A$ be orthogonal projections. Show that if a is positive and $pap = 0$, then $paq = 0$.
from orthogonal projection condition we have that $pq=0$ but how can this be used/related to solve the problem?
abstract-algebra
asked Dec 16 '18 at 15:33
katerinekaterine
326
$endgroup$
add a comment |
$begingroup$
Let $A$ be $C ^{ *}$ algebra, $a ∈ A$, $p$, $q ∈ A$ be orthogonal projections. Show that if a is positive and $pap = 0$, then $paq = 0$.
from orthogonal projection condition we have that $pq=0$ but how can this be used/related to solve the problem?
abstract-algebra
asked Dec 16 '18 at 15:33
katerinekaterine
326
$endgroup$
$begingroup$
Exactly how does $pq=0$ follow? I doubt it...
$endgroup$
– David C. Ullrich
Dec 16 '18 at 16:58
1
$begingroup$
if $a$ is positive and $p$ is self-adjoint (which a projection is), then $a=b^*b$ for some $b in A$ and $pap = (bp)^*(bp) = 0$....
$endgroup$
– Badam Baplan
Dec 16 '18 at 19:08
$begingroup$
Do you even need them to be orthogonal? Doesn't seem so to me. Give me a moment to sanity check this.
$endgroup$
– Munk
Dec 17 '18 at 16:45
add a comment |
$begingroup$
Let $A$ be $C ^{ *}$ algebra, $a ∈ A$, $p$, $q ∈ A$ be orthogonal projections. Show that if a is positive and $pap = 0$, then $paq = 0$.
from orthogonal projection condition we have that $pq=0$ but how can this be used/related to solve the problem?
abstract-algebra
asked Dec 16 '18 at 15:33
katerinekaterine
326
$endgroup$
Let $A$ be $C ^{ *}$ algebra, $a ∈ A$, $p$, $q ∈ A$ be orthogonal projections. Show that if a is positive and $pap = 0$, then $paq = 0$.
from orthogonal projection condition we have that $pq=0$ but how can this be used/related to solve the problem?
abstract-algebra
abstract-algebra
asked Dec 16 '18 at 15:33
katerinekaterine
326
asked Dec 16 '18 at 15:33
katerinekaterine
326
asked Dec 16 '18 at 15:33
katerinekaterine
326
asked Dec 16 '18 at 15:33
katerinekaterine
326
asked Dec 16 '18 at 15:33
katerinekaterine
326
326
$begingroup$
Exactly how does $pq=0$ follow? I doubt it...
$endgroup$
– David C. Ullrich
Dec 16 '18 at 16:58
1
$begingroup$
if $a$ is positive and $p$ is self-adjoint (which a projection is), then $a=b^*b$ for some $b in A$ and $pap = (bp)^*(bp) = 0$....
$endgroup$
– Badam Baplan
Dec 16 '18 at 19:08
$begingroup$
Do you even need them to be orthogonal? Doesn't seem so to me. Give me a moment to sanity check this.
$endgroup$
– Munk
Dec 17 '18 at 16:45
add a comment |
$begingroup$
Exactly how does $pq=0$ follow? I doubt it...
$endgroup$
– David C. Ullrich
Dec 16 '18 at 16:58
1
$begingroup$
if $a$ is positive and $p$ is self-adjoint (which a projection is), then $a=b^*b$ for some $b in A$ and $pap = (bp)^*(bp) = 0$....
$endgroup$
– Badam Baplan
Dec 16 '18 at 19:08
$begingroup$
Do you even need them to be orthogonal? Doesn't seem so to me. Give me a moment to sanity check this.
$endgroup$
– Munk
Dec 17 '18 at 16:45
$begingroup$
Exactly how does $pq=0$ follow? I doubt it...
$endgroup$
– David C. Ullrich
Dec 16 '18 at 16:58
$begingroup$
Exactly how does $pq=0$ follow? I doubt it...
$endgroup$
– David C. Ullrich
Dec 16 '18 at 16:58
1
1
$begingroup$
if $a$ is positive and $p$ is self-adjoint (which a projection is), then $a=b^*b$ for some $b in A$ and $pap = (bp)^*(bp) = 0$....
$endgroup$
– Badam Baplan
Dec 16 '18 at 19:08
$begingroup$
if $a$ is positive and $p$ is self-adjoint (which a projection is), then $a=b^*b$ for some $b in A$ and $pap = (bp)^*(bp) = 0$....
$endgroup$
– Badam Baplan
Dec 16 '18 at 19:08
$begingroup$
Do you even need them to be orthogonal? Doesn't seem so to me. Give me a moment to sanity check this.
$endgroup$
– Munk
Dec 17 '18 at 16:45
$begingroup$
Do you even need them to be orthogonal? Doesn't seem so to me. Give me a moment to sanity check this.
$endgroup$
– Munk
Dec 17 '18 at 16:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since $ageq 0$, then $a = b^*b$ for some element $b$ inside $A$. This allows us to write (by the $C^ast$-identity)
$$
0 = Vert pap Vert = Vert (bp)^*bp Vert= Vert bp Vert^2
$$
and thereby $bp=0$. Applying the adjoint yields $pb^*=0$. Hence $ paq = (pb^*)bq = 0$, and we're done.
edited Dec 17 '18 at 21:15
katerine
326
answered Dec 17 '18 at 16:50
MunkMunk
28518
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
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active
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$begingroup$
Since $ageq 0$, then $a = b^*b$ for some element $b$ inside $A$. This allows us to write (by the $C^ast$-identity)
$$
0 = Vert pap Vert = Vert (bp)^*bp Vert= Vert bp Vert^2
$$
and thereby $bp=0$. Applying the adjoint yields $pb^*=0$. Hence $ paq = (pb^*)bq = 0$, and we're done.
edited Dec 17 '18 at 21:15
katerine
326
answered Dec 17 '18 at 16:50
MunkMunk
28518
$endgroup$
add a comment |
$begingroup$
Since $ageq 0$, then $a = b^*b$ for some element $b$ inside $A$. This allows us to write (by the $C^ast$-identity)
$$
0 = Vert pap Vert = Vert (bp)^*bp Vert= Vert bp Vert^2
$$
and thereby $bp=0$. Applying the adjoint yields $pb^*=0$. Hence $ paq = (pb^*)bq = 0$, and we're done.
edited Dec 17 '18 at 21:15
katerine
326
answered Dec 17 '18 at 16:50
MunkMunk
28518
$endgroup$
add a comment |
$begingroup$
Since $ageq 0$, then $a = b^*b$ for some element $b$ inside $A$. This allows us to write (by the $C^ast$-identity)
$$
0 = Vert pap Vert = Vert (bp)^*bp Vert= Vert bp Vert^2
$$
and thereby $bp=0$. Applying the adjoint yields $pb^*=0$. Hence $ paq = (pb^*)bq = 0$, and we're done.
edited Dec 17 '18 at 21:15
katerine
326
answered Dec 17 '18 at 16:50
MunkMunk
28518
$endgroup$
Since $ageq 0$, then $a = b^*b$ for some element $b$ inside $A$. This allows us to write (by the $C^ast$-identity)
$$
0 = Vert pap Vert = Vert (bp)^*bp Vert= Vert bp Vert^2
$$
and thereby $bp=0$. Applying the adjoint yields $pb^*=0$. Hence $ paq = (pb^*)bq = 0$, and we're done.
edited Dec 17 '18 at 21:15
katerine
326
answered Dec 17 '18 at 16:50
MunkMunk
28518
edited Dec 17 '18 at 21:15
katerine
326
edited Dec 17 '18 at 21:15
katerine
326
edited Dec 17 '18 at 21:15
katerine
326
326
answered Dec 17 '18 at 16:50
MunkMunk
28518
answered Dec 17 '18 at 16:50
MunkMunk
28518
answered Dec 17 '18 at 16:50
MunkMunk
28518
28518
add a comment |
add a comment |
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$begingroup$
Exactly how does $pq=0$ follow? I doubt it...
$endgroup$
– David C. Ullrich
Dec 16 '18 at 16:58
1
$begingroup$
if $a$ is positive and $p$ is self-adjoint (which a projection is), then $a=b^*b$ for some $b in A$ and $pap = (bp)^*(bp) = 0$....
$endgroup$
– Badam Baplan
Dec 16 '18 at 19:08
$begingroup$
Do you even need them to be orthogonal? Doesn't seem so to me. Give me a moment to sanity check this.
$endgroup$
– Munk
Dec 17 '18 at 16:45