Prove $Tx=x$, for $xin H$, if and only if $(Tx,x)=|x|^2$ and $ker(I-T)=ker(I-T^*)$
$begingroup$
Let $H$ be a complex Hilbert space and $T:Hrightarrow H$ an operation such that $|T|leq 1$. Show that
$Tx=x$ if and only if $(Tx,x)=|x|^2$
$ker(I-T)=ker(I-T^*)$.
My attempt
1. $(Tx,x)=(x,x)=|x|^2$ if $Tx=x$. Conversely, WTS $|Tx-x|=0$ $forall xin H$.
We get $|Tx-x|=(Tx-x,Tx-x)=|Tx|^2-(x,Tx)$. But $|T|leq 1$ has not been used.
Please how do I proceed?
linear-algebra functional-analysis operator-theory hilbert-spaces adjoint-operators
asked Jan 6 at 15:17
Muhammad MubarakMuhammad Mubarak
9810
$endgroup$
add a comment |
$begingroup$
Let $H$ be a complex Hilbert space and $T:Hrightarrow H$ an operation such that $|T|leq 1$. Show that
$Tx=x$ if and only if $(Tx,x)=|x|^2$
$ker(I-T)=ker(I-T^*)$.
My attempt
1. $(Tx,x)=(x,x)=|x|^2$ if $Tx=x$. Conversely, WTS $|Tx-x|=0$ $forall xin H$.
We get $|Tx-x|=(Tx-x,Tx-x)=|Tx|^2-(x,Tx)$. But $|T|leq 1$ has not been used.
Please how do I proceed?
linear-algebra functional-analysis operator-theory hilbert-spaces adjoint-operators
asked Jan 6 at 15:17
Muhammad MubarakMuhammad Mubarak
9810
$endgroup$
add a comment |
$begingroup$
Let $H$ be a complex Hilbert space and $T:Hrightarrow H$ an operation such that $|T|leq 1$. Show that
$Tx=x$ if and only if $(Tx,x)=|x|^2$
$ker(I-T)=ker(I-T^*)$.
My attempt
1. $(Tx,x)=(x,x)=|x|^2$ if $Tx=x$. Conversely, WTS $|Tx-x|=0$ $forall xin H$.
We get $|Tx-x|=(Tx-x,Tx-x)=|Tx|^2-(x,Tx)$. But $|T|leq 1$ has not been used.
Please how do I proceed?
linear-algebra functional-analysis operator-theory hilbert-spaces adjoint-operators
asked Jan 6 at 15:17
Muhammad MubarakMuhammad Mubarak
9810
$endgroup$
Let $H$ be a complex Hilbert space and $T:Hrightarrow H$ an operation such that $|T|leq 1$. Show that
$Tx=x$ if and only if $(Tx,x)=|x|^2$
$ker(I-T)=ker(I-T^*)$.
My attempt
1. $(Tx,x)=(x,x)=|x|^2$ if $Tx=x$. Conversely, WTS $|Tx-x|=0$ $forall xin H$.
We get $|Tx-x|=(Tx-x,Tx-x)=|Tx|^2-(x,Tx)$. But $|T|leq 1$ has not been used.
Please how do I proceed?
linear-algebra functional-analysis operator-theory hilbert-spaces adjoint-operators
linear-algebra functional-analysis operator-theory hilbert-spaces adjoint-operators
asked Jan 6 at 15:17
Muhammad MubarakMuhammad Mubarak
9810
asked Jan 6 at 15:17
Muhammad MubarakMuhammad Mubarak
9810
edited Jan 6 at 16:36
Muhammad Mubarak
asked Jan 6 at 15:17
Muhammad MubarakMuhammad Mubarak
9810
asked Jan 6 at 15:17
Muhammad MubarakMuhammad Mubarak
9810
asked Jan 6 at 15:17
Muhammad MubarakMuhammad Mubarak
9810
9810
add a comment |
add a comment |
1 Answer
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$begingroup$
For the first point, first off since $left langle Tx,xrightrangle=|x|^2$ you also have $left langle x,Txrightrangle = overline{|x|^2}=|x|^2$. Thus
$$0 leqleftlangle Tx-x,Tx-xrightrangle =|Tx|^2+|x|^2-leftlangle Tx, xrightrangle-leftlangle x,Txrightrangle= |Tx|^2-|x|^2 $$
Since $|T|leq 1$,
$$0 leq |Tx|^2-|x|^2 leq |x|^2-|x|^2=0 $$
and hence $Tx=x$.
For the second point, using the first point
begin{align*}xin ker(I-T)iff Tx=x iff left langle Tx,xrightrangle =|x|^2end{align*}
Now observe that if $|T|leq 1$ then also $|T^*|leq 1$. Indeed,
begin{align*}|T^*|&=sup_{xneq 0}frac{|T^*x|}{|x|}=sup_{xneq 0}frac{leftlangle T^*x,T^*xrightrangle^{1/2}}{|x|}=sup_{xneq 0}frac{leftlangle TT^*x,xrightrangle^{1/2}}{|x|}leq sup_{xneq 0}frac{|TT^*x|^{1/2}}{|x|^{1/2}}leq\
&leqsup_{xneq 0}frac{|T^*x|^{1/2}}{|x|^{1/2}}=|T^*|^{1/2}end{align*}
i.e. $|T^*|leq |T^*|^{1/2}$ and thus $|T^*|leq 1$ (remark: this argument can be easily extended to prove that in general $|T|=|T^*|$ for any bounded linear operator $T$).
Therefore the first point may be applied to $T^*$, too.
begin{align*}
xin ker (I-T^*)iff T^*x=xiff left langle T^*x,xrightrangle = |x|^2iff left langle x,Txrightrangle =|x|^2
end{align*}
But $left langle Tx,xrightrangle =|x|^2 $ implies $left langle x, Txrightrangle = left langle Tx,xrightrangle =|x|^2$, as shown above, and viceversa, thus the equivalence is proved.
answered Jan 6 at 15:49
Lorenzo QuarisaLorenzo Quarisa
3,775623
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For the first point, first off since $left langle Tx,xrightrangle=|x|^2$ you also have $left langle x,Txrightrangle = overline{|x|^2}=|x|^2$. Thus
$$0 leqleftlangle Tx-x,Tx-xrightrangle =|Tx|^2+|x|^2-leftlangle Tx, xrightrangle-leftlangle x,Txrightrangle= |Tx|^2-|x|^2 $$
Since $|T|leq 1$,
$$0 leq |Tx|^2-|x|^2 leq |x|^2-|x|^2=0 $$
and hence $Tx=x$.
For the second point, using the first point
begin{align*}xin ker(I-T)iff Tx=x iff left langle Tx,xrightrangle =|x|^2end{align*}
Now observe that if $|T|leq 1$ then also $|T^*|leq 1$. Indeed,
begin{align*}|T^*|&=sup_{xneq 0}frac{|T^*x|}{|x|}=sup_{xneq 0}frac{leftlangle T^*x,T^*xrightrangle^{1/2}}{|x|}=sup_{xneq 0}frac{leftlangle TT^*x,xrightrangle^{1/2}}{|x|}leq sup_{xneq 0}frac{|TT^*x|^{1/2}}{|x|^{1/2}}leq\
&leqsup_{xneq 0}frac{|T^*x|^{1/2}}{|x|^{1/2}}=|T^*|^{1/2}end{align*}
i.e. $|T^*|leq |T^*|^{1/2}$ and thus $|T^*|leq 1$ (remark: this argument can be easily extended to prove that in general $|T|=|T^*|$ for any bounded linear operator $T$).
Therefore the first point may be applied to $T^*$, too.
begin{align*}
xin ker (I-T^*)iff T^*x=xiff left langle T^*x,xrightrangle = |x|^2iff left langle x,Txrightrangle =|x|^2
end{align*}
But $left langle Tx,xrightrangle =|x|^2 $ implies $left langle x, Txrightrangle = left langle Tx,xrightrangle =|x|^2$, as shown above, and viceversa, thus the equivalence is proved.
answered Jan 6 at 15:49
Lorenzo QuarisaLorenzo Quarisa
3,775623
$endgroup$
add a comment |
$begingroup$
For the first point, first off since $left langle Tx,xrightrangle=|x|^2$ you also have $left langle x,Txrightrangle = overline{|x|^2}=|x|^2$. Thus
$$0 leqleftlangle Tx-x,Tx-xrightrangle =|Tx|^2+|x|^2-leftlangle Tx, xrightrangle-leftlangle x,Txrightrangle= |Tx|^2-|x|^2 $$
Since $|T|leq 1$,
$$0 leq |Tx|^2-|x|^2 leq |x|^2-|x|^2=0 $$
and hence $Tx=x$.
For the second point, using the first point
begin{align*}xin ker(I-T)iff Tx=x iff left langle Tx,xrightrangle =|x|^2end{align*}
Now observe that if $|T|leq 1$ then also $|T^*|leq 1$. Indeed,
begin{align*}|T^*|&=sup_{xneq 0}frac{|T^*x|}{|x|}=sup_{xneq 0}frac{leftlangle T^*x,T^*xrightrangle^{1/2}}{|x|}=sup_{xneq 0}frac{leftlangle TT^*x,xrightrangle^{1/2}}{|x|}leq sup_{xneq 0}frac{|TT^*x|^{1/2}}{|x|^{1/2}}leq\
&leqsup_{xneq 0}frac{|T^*x|^{1/2}}{|x|^{1/2}}=|T^*|^{1/2}end{align*}
i.e. $|T^*|leq |T^*|^{1/2}$ and thus $|T^*|leq 1$ (remark: this argument can be easily extended to prove that in general $|T|=|T^*|$ for any bounded linear operator $T$).
Therefore the first point may be applied to $T^*$, too.
begin{align*}
xin ker (I-T^*)iff T^*x=xiff left langle T^*x,xrightrangle = |x|^2iff left langle x,Txrightrangle =|x|^2
end{align*}
But $left langle Tx,xrightrangle =|x|^2 $ implies $left langle x, Txrightrangle = left langle Tx,xrightrangle =|x|^2$, as shown above, and viceversa, thus the equivalence is proved.
answered Jan 6 at 15:49
Lorenzo QuarisaLorenzo Quarisa
3,775623
$endgroup$
add a comment |
$begingroup$
For the first point, first off since $left langle Tx,xrightrangle=|x|^2$ you also have $left langle x,Txrightrangle = overline{|x|^2}=|x|^2$. Thus
$$0 leqleftlangle Tx-x,Tx-xrightrangle =|Tx|^2+|x|^2-leftlangle Tx, xrightrangle-leftlangle x,Txrightrangle= |Tx|^2-|x|^2 $$
Since $|T|leq 1$,
$$0 leq |Tx|^2-|x|^2 leq |x|^2-|x|^2=0 $$
and hence $Tx=x$.
For the second point, using the first point
begin{align*}xin ker(I-T)iff Tx=x iff left langle Tx,xrightrangle =|x|^2end{align*}
Now observe that if $|T|leq 1$ then also $|T^*|leq 1$. Indeed,
begin{align*}|T^*|&=sup_{xneq 0}frac{|T^*x|}{|x|}=sup_{xneq 0}frac{leftlangle T^*x,T^*xrightrangle^{1/2}}{|x|}=sup_{xneq 0}frac{leftlangle TT^*x,xrightrangle^{1/2}}{|x|}leq sup_{xneq 0}frac{|TT^*x|^{1/2}}{|x|^{1/2}}leq\
&leqsup_{xneq 0}frac{|T^*x|^{1/2}}{|x|^{1/2}}=|T^*|^{1/2}end{align*}
i.e. $|T^*|leq |T^*|^{1/2}$ and thus $|T^*|leq 1$ (remark: this argument can be easily extended to prove that in general $|T|=|T^*|$ for any bounded linear operator $T$).
Therefore the first point may be applied to $T^*$, too.
begin{align*}
xin ker (I-T^*)iff T^*x=xiff left langle T^*x,xrightrangle = |x|^2iff left langle x,Txrightrangle =|x|^2
end{align*}
But $left langle Tx,xrightrangle =|x|^2 $ implies $left langle x, Txrightrangle = left langle Tx,xrightrangle =|x|^2$, as shown above, and viceversa, thus the equivalence is proved.
answered Jan 6 at 15:49
Lorenzo QuarisaLorenzo Quarisa
3,775623
$endgroup$
For the first point, first off since $left langle Tx,xrightrangle=|x|^2$ you also have $left langle x,Txrightrangle = overline{|x|^2}=|x|^2$. Thus
$$0 leqleftlangle Tx-x,Tx-xrightrangle =|Tx|^2+|x|^2-leftlangle Tx, xrightrangle-leftlangle x,Txrightrangle= |Tx|^2-|x|^2 $$
Since $|T|leq 1$,
$$0 leq |Tx|^2-|x|^2 leq |x|^2-|x|^2=0 $$
and hence $Tx=x$.
For the second point, using the first point
begin{align*}xin ker(I-T)iff Tx=x iff left langle Tx,xrightrangle =|x|^2end{align*}
Now observe that if $|T|leq 1$ then also $|T^*|leq 1$. Indeed,
begin{align*}|T^*|&=sup_{xneq 0}frac{|T^*x|}{|x|}=sup_{xneq 0}frac{leftlangle T^*x,T^*xrightrangle^{1/2}}{|x|}=sup_{xneq 0}frac{leftlangle TT^*x,xrightrangle^{1/2}}{|x|}leq sup_{xneq 0}frac{|TT^*x|^{1/2}}{|x|^{1/2}}leq\
&leqsup_{xneq 0}frac{|T^*x|^{1/2}}{|x|^{1/2}}=|T^*|^{1/2}end{align*}
i.e. $|T^*|leq |T^*|^{1/2}$ and thus $|T^*|leq 1$ (remark: this argument can be easily extended to prove that in general $|T|=|T^*|$ for any bounded linear operator $T$).
Therefore the first point may be applied to $T^*$, too.
begin{align*}
xin ker (I-T^*)iff T^*x=xiff left langle T^*x,xrightrangle = |x|^2iff left langle x,Txrightrangle =|x|^2
end{align*}
But $left langle Tx,xrightrangle =|x|^2 $ implies $left langle x, Txrightrangle = left langle Tx,xrightrangle =|x|^2$, as shown above, and viceversa, thus the equivalence is proved.
answered Jan 6 at 15:49
Lorenzo QuarisaLorenzo Quarisa
3,775623
edited Jan 6 at 16:18
answered Jan 6 at 15:49
Lorenzo QuarisaLorenzo Quarisa
3,775623
answered Jan 6 at 15:49
Lorenzo QuarisaLorenzo Quarisa
3,775623
answered Jan 6 at 15:49
Lorenzo QuarisaLorenzo Quarisa
3,775623
3,775623
add a comment |
add a comment |
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