Invertibility of the square root of an operator
$begingroup$
Let $mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$. Let $Min mathcal{B}(F)^+$ (i.e. $langle Mx;, ;xranglegeq 0$ for all $xin F$).
The square root of $M$ is defined to be the unique operator $N$ such that
$$N^2=M.$$In this case we write $N=M^{1/2}$
If $M$ is an invertible operators, is $M^{1/2}$ invertible?
functional-analysis operator-theory
edited Jan 5 at 14:05
Bernard
123k741116
asked Jan 5 at 14:01
SchülerSchüler
1,5491421
$endgroup$
add a comment |
$begingroup$
Let $mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$. Let $Min mathcal{B}(F)^+$ (i.e. $langle Mx;, ;xranglegeq 0$ for all $xin F$).
The square root of $M$ is defined to be the unique operator $N$ such that
$$N^2=M.$$In this case we write $N=M^{1/2}$
If $M$ is an invertible operators, is $M^{1/2}$ invertible?
functional-analysis operator-theory
edited Jan 5 at 14:05
Bernard
123k741116
asked Jan 5 at 14:01
SchülerSchüler
1,5491421
$endgroup$
add a comment |
$begingroup$
Let $mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$. Let $Min mathcal{B}(F)^+$ (i.e. $langle Mx;, ;xranglegeq 0$ for all $xin F$).
The square root of $M$ is defined to be the unique operator $N$ such that
$$N^2=M.$$In this case we write $N=M^{1/2}$
If $M$ is an invertible operators, is $M^{1/2}$ invertible?
functional-analysis operator-theory
edited Jan 5 at 14:05
Bernard
123k741116
asked Jan 5 at 14:01
SchülerSchüler
1,5491421
$endgroup$
Let $mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$. Let $Min mathcal{B}(F)^+$ (i.e. $langle Mx;, ;xranglegeq 0$ for all $xin F$).
The square root of $M$ is defined to be the unique operator $N$ such that
$$N^2=M.$$In this case we write $N=M^{1/2}$
If $M$ is an invertible operators, is $M^{1/2}$ invertible?
functional-analysis operator-theory
functional-analysis operator-theory
edited Jan 5 at 14:05
Bernard
123k741116
asked Jan 5 at 14:01
SchülerSchüler
1,5491421
edited Jan 5 at 14:05
Bernard
123k741116
asked Jan 5 at 14:01
SchülerSchüler
1,5491421
edited Jan 5 at 14:05
Bernard
123k741116
edited Jan 5 at 14:05
Bernard
123k741116
edited Jan 5 at 14:05
Bernard
123k741116
123k741116
asked Jan 5 at 14:01
SchülerSchüler
1,5491421
asked Jan 5 at 14:01
SchülerSchüler
1,5491421
asked Jan 5 at 14:01
SchülerSchüler
1,5491421
1,5491421
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that $M$ is invertible if and only if there is $epsilon>0$ such that
$$sigma(M)subset [epsilon, infty).$$ Thus $sigma(M^{1/2})subset [epsilon^{1/2},infty)$ and $M^{1/2}$ is invertible.
answered Jan 5 at 14:12
SongSong
18.3k21449
$endgroup$
$begingroup$
Thank you for the answer. Is your equivalence well known?
$endgroup$
– Schüler
Jan 5 at 14:33
$begingroup$
I think it is. It is in fact easier than it seems. Since $Mge 0$, we know $sigma(M)subset [0,infty)$. But if $M$ is invertible, we must have $epsilon =min sigma(M) >0$. The other direction is true since $0notin sigma(M)$ means $M$ is invertible.
$endgroup$
– Song
Jan 5 at 14:36
$begingroup$
According to our answer, the converse is also true. That is if $A^{1/2}$ is invertible, then $A$ is also invertible. Do you agree with me? Thanks.
$endgroup$
– Schüler
Jan 5 at 17:16
$begingroup$
Of course, I agree with you!
$endgroup$
– Song
Jan 5 at 17:20
$begingroup$
I think the equivalence is true only for positive operators. That is it is better to write:'' Let $M$ be a positive operator. Then $M$ is invertible if and only if there is $epsilon>0$ such that $$sigma(M)subset [epsilon, infty).$$ '' Do you agree with me? Thanks
$endgroup$
– Schüler
Feb 9 at 16:17
|
show 3 more comments
$begingroup$
$N(x)=N(y)$ implies $M(x)=N^2(x)=N^2(y)=M(y)$ since $M$ is injective, we deduce that $N$ is injective.
For every $y$, $y=M(x)=N(N(x))$ implies that $N$ is surjective and open (open mapping theorem)
This implies that the inverse of $N$ is continuous $N$ and $N$ is invertible.
answered Jan 5 at 14:06
Tsemo AristideTsemo Aristide
59.5k11446
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that $M$ is invertible if and only if there is $epsilon>0$ such that
$$sigma(M)subset [epsilon, infty).$$ Thus $sigma(M^{1/2})subset [epsilon^{1/2},infty)$ and $M^{1/2}$ is invertible.
answered Jan 5 at 14:12
SongSong
18.3k21449
$endgroup$
$begingroup$
Thank you for the answer. Is your equivalence well known?
$endgroup$
– Schüler
Jan 5 at 14:33
$begingroup$
I think it is. It is in fact easier than it seems. Since $Mge 0$, we know $sigma(M)subset [0,infty)$. But if $M$ is invertible, we must have $epsilon =min sigma(M) >0$. The other direction is true since $0notin sigma(M)$ means $M$ is invertible.
$endgroup$
– Song
Jan 5 at 14:36
$begingroup$
According to our answer, the converse is also true. That is if $A^{1/2}$ is invertible, then $A$ is also invertible. Do you agree with me? Thanks.
$endgroup$
– Schüler
Jan 5 at 17:16
$begingroup$
Of course, I agree with you!
$endgroup$
– Song
Jan 5 at 17:20
$begingroup$
I think the equivalence is true only for positive operators. That is it is better to write:'' Let $M$ be a positive operator. Then $M$ is invertible if and only if there is $epsilon>0$ such that $$sigma(M)subset [epsilon, infty).$$ '' Do you agree with me? Thanks
$endgroup$
– Schüler
Feb 9 at 16:17
|
show 3 more comments
$begingroup$
Note that $M$ is invertible if and only if there is $epsilon>0$ such that
$$sigma(M)subset [epsilon, infty).$$ Thus $sigma(M^{1/2})subset [epsilon^{1/2},infty)$ and $M^{1/2}$ is invertible.
answered Jan 5 at 14:12
SongSong
18.3k21449
$endgroup$
$begingroup$
Thank you for the answer. Is your equivalence well known?
$endgroup$
– Schüler
Jan 5 at 14:33
$begingroup$
I think it is. It is in fact easier than it seems. Since $Mge 0$, we know $sigma(M)subset [0,infty)$. But if $M$ is invertible, we must have $epsilon =min sigma(M) >0$. The other direction is true since $0notin sigma(M)$ means $M$ is invertible.
$endgroup$
– Song
Jan 5 at 14:36
$begingroup$
According to our answer, the converse is also true. That is if $A^{1/2}$ is invertible, then $A$ is also invertible. Do you agree with me? Thanks.
$endgroup$
– Schüler
Jan 5 at 17:16
$begingroup$
Of course, I agree with you!
$endgroup$
– Song
Jan 5 at 17:20
$begingroup$
I think the equivalence is true only for positive operators. That is it is better to write:'' Let $M$ be a positive operator. Then $M$ is invertible if and only if there is $epsilon>0$ such that $$sigma(M)subset [epsilon, infty).$$ '' Do you agree with me? Thanks
$endgroup$
– Schüler
Feb 9 at 16:17
|
show 3 more comments
$begingroup$
Note that $M$ is invertible if and only if there is $epsilon>0$ such that
$$sigma(M)subset [epsilon, infty).$$ Thus $sigma(M^{1/2})subset [epsilon^{1/2},infty)$ and $M^{1/2}$ is invertible.
answered Jan 5 at 14:12
SongSong
18.3k21449
$endgroup$
Note that $M$ is invertible if and only if there is $epsilon>0$ such that
$$sigma(M)subset [epsilon, infty).$$ Thus $sigma(M^{1/2})subset [epsilon^{1/2},infty)$ and $M^{1/2}$ is invertible.
answered Jan 5 at 14:12
SongSong
18.3k21449
answered Jan 5 at 14:12
SongSong
18.3k21449
answered Jan 5 at 14:12
SongSong
18.3k21449
answered Jan 5 at 14:12
SongSong
18.3k21449
18.3k21449
$begingroup$
Thank you for the answer. Is your equivalence well known?
$endgroup$
– Schüler
Jan 5 at 14:33
$begingroup$
I think it is. It is in fact easier than it seems. Since $Mge 0$, we know $sigma(M)subset [0,infty)$. But if $M$ is invertible, we must have $epsilon =min sigma(M) >0$. The other direction is true since $0notin sigma(M)$ means $M$ is invertible.
$endgroup$
– Song
Jan 5 at 14:36
$begingroup$
According to our answer, the converse is also true. That is if $A^{1/2}$ is invertible, then $A$ is also invertible. Do you agree with me? Thanks.
$endgroup$
– Schüler
Jan 5 at 17:16
$begingroup$
Of course, I agree with you!
$endgroup$
– Song
Jan 5 at 17:20
$begingroup$
I think the equivalence is true only for positive operators. That is it is better to write:'' Let $M$ be a positive operator. Then $M$ is invertible if and only if there is $epsilon>0$ such that $$sigma(M)subset [epsilon, infty).$$ '' Do you agree with me? Thanks
$endgroup$
– Schüler
Feb 9 at 16:17
|
show 3 more comments
$begingroup$
Thank you for the answer. Is your equivalence well known?
$endgroup$
– Schüler
Jan 5 at 14:33
$begingroup$
I think it is. It is in fact easier than it seems. Since $Mge 0$, we know $sigma(M)subset [0,infty)$. But if $M$ is invertible, we must have $epsilon =min sigma(M) >0$. The other direction is true since $0notin sigma(M)$ means $M$ is invertible.
$endgroup$
– Song
Jan 5 at 14:36
$begingroup$
According to our answer, the converse is also true. That is if $A^{1/2}$ is invertible, then $A$ is also invertible. Do you agree with me? Thanks.
$endgroup$
– Schüler
Jan 5 at 17:16
$begingroup$
Of course, I agree with you!
$endgroup$
– Song
Jan 5 at 17:20
$begingroup$
I think the equivalence is true only for positive operators. That is it is better to write:'' Let $M$ be a positive operator. Then $M$ is invertible if and only if there is $epsilon>0$ such that $$sigma(M)subset [epsilon, infty).$$ '' Do you agree with me? Thanks
$endgroup$
– Schüler
Feb 9 at 16:17
$begingroup$
Thank you for the answer. Is your equivalence well known?
$endgroup$
– Schüler
Jan 5 at 14:33
$begingroup$
Thank you for the answer. Is your equivalence well known?
$endgroup$
– Schüler
Jan 5 at 14:33
$begingroup$
I think it is. It is in fact easier than it seems. Since $Mge 0$, we know $sigma(M)subset [0,infty)$. But if $M$ is invertible, we must have $epsilon =min sigma(M) >0$. The other direction is true since $0notin sigma(M)$ means $M$ is invertible.
$endgroup$
– Song
Jan 5 at 14:36
$begingroup$
I think it is. It is in fact easier than it seems. Since $Mge 0$, we know $sigma(M)subset [0,infty)$. But if $M$ is invertible, we must have $epsilon =min sigma(M) >0$. The other direction is true since $0notin sigma(M)$ means $M$ is invertible.
$endgroup$
– Song
Jan 5 at 14:36
$begingroup$
According to our answer, the converse is also true. That is if $A^{1/2}$ is invertible, then $A$ is also invertible. Do you agree with me? Thanks.
$endgroup$
– Schüler
Jan 5 at 17:16
$begingroup$
According to our answer, the converse is also true. That is if $A^{1/2}$ is invertible, then $A$ is also invertible. Do you agree with me? Thanks.
$endgroup$
– Schüler
Jan 5 at 17:16
$begingroup$
Of course, I agree with you!
$endgroup$
– Song
Jan 5 at 17:20
$begingroup$
Of course, I agree with you!
$endgroup$
– Song
Jan 5 at 17:20
$begingroup$
I think the equivalence is true only for positive operators. That is it is better to write:'' Let $M$ be a positive operator. Then $M$ is invertible if and only if there is $epsilon>0$ such that $$sigma(M)subset [epsilon, infty).$$ '' Do you agree with me? Thanks
$endgroup$
– Schüler
Feb 9 at 16:17
$begingroup$
I think the equivalence is true only for positive operators. That is it is better to write:'' Let $M$ be a positive operator. Then $M$ is invertible if and only if there is $epsilon>0$ such that $$sigma(M)subset [epsilon, infty).$$ '' Do you agree with me? Thanks
$endgroup$
– Schüler
Feb 9 at 16:17
|
show 3 more comments
$begingroup$
$N(x)=N(y)$ implies $M(x)=N^2(x)=N^2(y)=M(y)$ since $M$ is injective, we deduce that $N$ is injective.
For every $y$, $y=M(x)=N(N(x))$ implies that $N$ is surjective and open (open mapping theorem)
This implies that the inverse of $N$ is continuous $N$ and $N$ is invertible.
answered Jan 5 at 14:06
Tsemo AristideTsemo Aristide
59.5k11446
$endgroup$
add a comment |
$begingroup$
$N(x)=N(y)$ implies $M(x)=N^2(x)=N^2(y)=M(y)$ since $M$ is injective, we deduce that $N$ is injective.
For every $y$, $y=M(x)=N(N(x))$ implies that $N$ is surjective and open (open mapping theorem)
This implies that the inverse of $N$ is continuous $N$ and $N$ is invertible.
answered Jan 5 at 14:06
Tsemo AristideTsemo Aristide
59.5k11446
$endgroup$
add a comment |
$begingroup$
$N(x)=N(y)$ implies $M(x)=N^2(x)=N^2(y)=M(y)$ since $M$ is injective, we deduce that $N$ is injective.
For every $y$, $y=M(x)=N(N(x))$ implies that $N$ is surjective and open (open mapping theorem)
This implies that the inverse of $N$ is continuous $N$ and $N$ is invertible.
answered Jan 5 at 14:06
Tsemo AristideTsemo Aristide
59.5k11446
$endgroup$
$N(x)=N(y)$ implies $M(x)=N^2(x)=N^2(y)=M(y)$ since $M$ is injective, we deduce that $N$ is injective.
For every $y$, $y=M(x)=N(N(x))$ implies that $N$ is surjective and open (open mapping theorem)
This implies that the inverse of $N$ is continuous $N$ and $N$ is invertible.
answered Jan 5 at 14:06
Tsemo AristideTsemo Aristide
59.5k11446
answered Jan 5 at 14:06
Tsemo AristideTsemo Aristide
59.5k11446
answered Jan 5 at 14:06
Tsemo AristideTsemo Aristide
59.5k11446
answered Jan 5 at 14:06
Tsemo AristideTsemo Aristide
59.5k11446
59.5k11446
add a comment |
add a comment |
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