show that $ker(Id-T) = ker(Id-T)^{star}$
$begingroup$
$H$ is a hilbert space and $T$ is a bounded linear operator on $H$, also $|T| leq 1$
by calculating $|Tx-x|^2$ I have shown the following string of equivalences $$Tx = x iff langle,Tx, xrangle = |x|^2 iff langle,x, Txrangle = |x|^2 $$
it's supposed to help me prove $$ker(Id-T) = ker(Id-T)^{star}$$
but I fail to see how one must proceed.
functional-analysis operator-theory hilbert-spaces
asked Jan 7 at 21:24
rapidracimrapidracim
1,7291419
$endgroup$
add a comment |
$begingroup$
$H$ is a hilbert space and $T$ is a bounded linear operator on $H$, also $|T| leq 1$
by calculating $|Tx-x|^2$ I have shown the following string of equivalences $$Tx = x iff langle,Tx, xrangle = |x|^2 iff langle,x, Txrangle = |x|^2 $$
it's supposed to help me prove $$ker(Id-T) = ker(Id-T)^{star}$$
but I fail to see how one must proceed.
functional-analysis operator-theory hilbert-spaces
asked Jan 7 at 21:24
rapidracimrapidracim
1,7291419
$endgroup$
$begingroup$
Note that if $|T| < 1$ then $I-T$ and $I-T^*$ are both invertible so in particular they both have trivial kernels.
$endgroup$
– mechanodroid
Jan 7 at 21:59
add a comment |
$begingroup$
$H$ is a hilbert space and $T$ is a bounded linear operator on $H$, also $|T| leq 1$
by calculating $|Tx-x|^2$ I have shown the following string of equivalences $$Tx = x iff langle,Tx, xrangle = |x|^2 iff langle,x, Txrangle = |x|^2 $$
it's supposed to help me prove $$ker(Id-T) = ker(Id-T)^{star}$$
but I fail to see how one must proceed.
functional-analysis operator-theory hilbert-spaces
asked Jan 7 at 21:24
rapidracimrapidracim
1,7291419
$endgroup$
$H$ is a hilbert space and $T$ is a bounded linear operator on $H$, also $|T| leq 1$
by calculating $|Tx-x|^2$ I have shown the following string of equivalences $$Tx = x iff langle,Tx, xrangle = |x|^2 iff langle,x, Txrangle = |x|^2 $$
it's supposed to help me prove $$ker(Id-T) = ker(Id-T)^{star}$$
but I fail to see how one must proceed.
functional-analysis operator-theory hilbert-spaces
functional-analysis operator-theory hilbert-spaces
asked Jan 7 at 21:24
rapidracimrapidracim
1,7291419
asked Jan 7 at 21:24
rapidracimrapidracim
1,7291419
asked Jan 7 at 21:24
rapidracimrapidracim
1,7291419
asked Jan 7 at 21:24
rapidracimrapidracim
1,7291419
asked Jan 7 at 21:24
rapidracimrapidracim
1,7291419
1,7291419
$begingroup$
Note that if $|T| < 1$ then $I-T$ and $I-T^*$ are both invertible so in particular they both have trivial kernels.
$endgroup$
– mechanodroid
Jan 7 at 21:59
add a comment |
$begingroup$
Note that if $|T| < 1$ then $I-T$ and $I-T^*$ are both invertible so in particular they both have trivial kernels.
$endgroup$
– mechanodroid
Jan 7 at 21:59
$begingroup$
Note that if $|T| < 1$ then $I-T$ and $I-T^*$ are both invertible so in particular they both have trivial kernels.
$endgroup$
– mechanodroid
Jan 7 at 21:59
$begingroup$
Note that if $|T| < 1$ then $I-T$ and $I-T^*$ are both invertible so in particular they both have trivial kernels.
$endgroup$
– mechanodroid
Jan 7 at 21:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You made an error in your equivalence chain.
In fact, you have:
$$Tx = x iff langle,Tx, xrangle = |x|^2 iff langle,x, T^* xrangle = |x|^2.$$
You also have $Vert T^* Vert le Vert T Vert le 1$. In conjonction with $langle,x, T^* xrangle = |x|^2$
, this leads to
$$langle,x, T^* xrangle = |x|^2 iff T^*x=x$$
and concludes the proof.
answered Jan 7 at 21:39
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
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$begingroup$
You made an error in your equivalence chain.
In fact, you have:
$$Tx = x iff langle,Tx, xrangle = |x|^2 iff langle,x, T^* xrangle = |x|^2.$$
You also have $Vert T^* Vert le Vert T Vert le 1$. In conjonction with $langle,x, T^* xrangle = |x|^2$
, this leads to
$$langle,x, T^* xrangle = |x|^2 iff T^*x=x$$
and concludes the proof.
answered Jan 7 at 21:39
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
$begingroup$
You made an error in your equivalence chain.
In fact, you have:
$$Tx = x iff langle,Tx, xrangle = |x|^2 iff langle,x, T^* xrangle = |x|^2.$$
You also have $Vert T^* Vert le Vert T Vert le 1$. In conjonction with $langle,x, T^* xrangle = |x|^2$
, this leads to
$$langle,x, T^* xrangle = |x|^2 iff T^*x=x$$
and concludes the proof.
answered Jan 7 at 21:39
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
$begingroup$
You made an error in your equivalence chain.
In fact, you have:
$$Tx = x iff langle,Tx, xrangle = |x|^2 iff langle,x, T^* xrangle = |x|^2.$$
You also have $Vert T^* Vert le Vert T Vert le 1$. In conjonction with $langle,x, T^* xrangle = |x|^2$
, this leads to
$$langle,x, T^* xrangle = |x|^2 iff T^*x=x$$
and concludes the proof.
answered Jan 7 at 21:39
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
You made an error in your equivalence chain.
In fact, you have:
$$Tx = x iff langle,Tx, xrangle = |x|^2 iff langle,x, T^* xrangle = |x|^2.$$
You also have $Vert T^* Vert le Vert T Vert le 1$. In conjonction with $langle,x, T^* xrangle = |x|^2$
, this leads to
$$langle,x, T^* xrangle = |x|^2 iff T^*x=x$$
and concludes the proof.
answered Jan 7 at 21:39
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 7 at 21:39
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 7 at 21:39
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 7 at 21:39
mathcounterexamples.netmathcounterexamples.net
27k22158
27k22158
add a comment |
add a comment |
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$begingroup$
Note that if $|T| < 1$ then $I-T$ and $I-T^*$ are both invertible so in particular they both have trivial kernels.
$endgroup$
– mechanodroid
Jan 7 at 21:59