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Quadratic form associated to a linear operator in an Hilbert space
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Let $A$ be a linear operator on $H$ Hilbert space (with a scalar product $(cdot,cdot)$ ), and let's define $r_A =sup_(Ax, x)$. We want to show that $r_A=|A|$, where we endow the space $L(H)$ of the linear continuous operator on H with the uniform norm. From Cauchy-Schwartz we easily obtain $r_A leq |A|$. I'm stuck with the other inequality; on the (rather dubious) note it should follow somehow from the polarization equality $(x, y)= frac 1 4 [|x+y|^2-|x-y|^2]$, but I can't manage to get it. Any help is appreciated.
functional-analysis operator-theory hilbert-spaces quadratic-forms spectral-radius
edited Jul 16 at 23:11
mechanodroid
22.3k52041
asked Jul 16 at 22:20
Giuseppe Bargagnati
984414
add a comment |Â
up vote
2
down vote
favorite
Let $A$ be a linear operator on $H$ Hilbert space (with a scalar product $(cdot,cdot)$ ), and let's define $r_A =sup_(Ax, x)$. We want to show that $r_A=|A|$, where we endow the space $L(H)$ of the linear continuous operator on H with the uniform norm. From Cauchy-Schwartz we easily obtain $r_A leq |A|$. I'm stuck with the other inequality; on the (rather dubious) note it should follow somehow from the polarization equality $(x, y)= frac 1 4 [|x+y|^2-|x-y|^2]$, but I can't manage to get it. Any help is appreciated.
functional-analysis operator-theory hilbert-spaces quadratic-forms spectral-radius
edited Jul 16 at 23:11
mechanodroid
22.3k52041
asked Jul 16 at 22:20
Giuseppe Bargagnati
984414
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $A$ be a linear operator on $H$ Hilbert space (with a scalar product $(cdot,cdot)$ ), and let's define $r_A =sup_(Ax, x)$. We want to show that $r_A=|A|$, where we endow the space $L(H)$ of the linear continuous operator on H with the uniform norm. From Cauchy-Schwartz we easily obtain $r_A leq |A|$. I'm stuck with the other inequality; on the (rather dubious) note it should follow somehow from the polarization equality $(x, y)= frac 1 4 [|x+y|^2-|x-y|^2]$, but I can't manage to get it. Any help is appreciated.
functional-analysis operator-theory hilbert-spaces quadratic-forms spectral-radius
edited Jul 16 at 23:11
mechanodroid
22.3k52041
asked Jul 16 at 22:20
Giuseppe Bargagnati
984414
Let $A$ be a linear operator on $H$ Hilbert space (with a scalar product $(cdot,cdot)$ ), and let's define $r_A =sup_(Ax, x)$. We want to show that $r_A=|A|$, where we endow the space $L(H)$ of the linear continuous operator on H with the uniform norm. From Cauchy-Schwartz we easily obtain $r_A leq |A|$. I'm stuck with the other inequality; on the (rather dubious) note it should follow somehow from the polarization equality $(x, y)= frac 1 4 [|x+y|^2-|x-y|^2]$, but I can't manage to get it. Any help is appreciated.
functional-analysis operator-theory hilbert-spaces quadratic-forms spectral-radius
edited Jul 16 at 23:11
mechanodroid
22.3k52041
asked Jul 16 at 22:20
Giuseppe Bargagnati
984414
edited Jul 16 at 23:11
mechanodroid
22.3k52041
edited Jul 16 at 23:11
mechanodroid
22.3k52041
edited Jul 16 at 23:11
mechanodroid
22.3k52041
22.3k52041
asked Jul 16 at 22:20
Giuseppe Bargagnati
984414
asked Jul 16 at 22:20
Giuseppe Bargagnati
984414
asked Jul 16 at 22:20
Giuseppe Bargagnati
984414
984414
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
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accepted
This isn't true in general. Consider $A = beginbmatrix 0 & 1 \ -1 & 0endbmatrix$. We have
$$leftlanglebeginbmatrix 0 & 1 \ -1 & 0endbmatrixbeginbmatrix x \ y endbmatrix, beginbmatrix x \ yendbmatrix rightrangle = leftlangle beginbmatrix y \ -xendbmatrix, beginbmatrix x \ yendbmatrix rightrangle = 0$$
so $r_A = 0$ but $|A| ne 0$.
However, this is true if $A$ is assumed to be self-adjoint.
Moreover, it is even true for normal operators if the Hilbert space is complex.
answered Jul 16 at 23:10
mechanodroid
22.3k52041
If $A$ is self-adjoint, does it follow from the equality $||A||=||A^2^n||^frac 1 2^n$ and from the fact that $r_A= lim _n to infty||A^n||^frac 1 n$?
– Giuseppe Bargagnati
Jul 17 at 12:28
(And that I know that the last sequence admits limit).
– Giuseppe Bargagnati
Jul 17 at 12:33
1
@GiuseppeBargagnati $lim_ntoinfty |A^n|^frac1n$ is the spectral radius of $A$, not the numerical radius $r_A$. But yes, for normal operators we have $|A^n| = |A|^n$ so the spectral radius of $A$ is equal to $|A|$.
– mechanodroid
Jul 17 at 14:19
Ok, I've some problems with the notation with the notes I'm studying, thank you very much.
– Giuseppe Bargagnati
Jul 17 at 14:22
add a comment |Â
up vote
1
down vote
I assume from your version of the polarization identity that we're working over a real Hilbert space.
Hint: Note that (if $A$ is self-adjoint)
$$
(Ax,y) = frac 14 left[ (x+y,A(x+y)) - (x-y,A(x-y))right]
$$
you can regard this as a generalization of the usual polarization identity. Thus, we have
$$
4sup_x (Ax,y) leq sup_x(A(x+y),(x+y)) - inf_x A(x-y,x-y)
$$
answered Jul 16 at 22:36
Omnomnomnom
121k784170
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
This isn't true in general. Consider $A = beginbmatrix 0 & 1 \ -1 & 0endbmatrix$. We have
$$leftlanglebeginbmatrix 0 & 1 \ -1 & 0endbmatrixbeginbmatrix x \ y endbmatrix, beginbmatrix x \ yendbmatrix rightrangle = leftlangle beginbmatrix y \ -xendbmatrix, beginbmatrix x \ yendbmatrix rightrangle = 0$$
so $r_A = 0$ but $|A| ne 0$.
However, this is true if $A$ is assumed to be self-adjoint.
Moreover, it is even true for normal operators if the Hilbert space is complex.
answered Jul 16 at 23:10
mechanodroid
22.3k52041
If $A$ is self-adjoint, does it follow from the equality $||A||=||A^2^n||^frac 1 2^n$ and from the fact that $r_A= lim _n to infty||A^n||^frac 1 n$?
– Giuseppe Bargagnati
Jul 17 at 12:28
(And that I know that the last sequence admits limit).
– Giuseppe Bargagnati
Jul 17 at 12:33
1
@GiuseppeBargagnati $lim_ntoinfty |A^n|^frac1n$ is the spectral radius of $A$, not the numerical radius $r_A$. But yes, for normal operators we have $|A^n| = |A|^n$ so the spectral radius of $A$ is equal to $|A|$.
– mechanodroid
Jul 17 at 14:19
Ok, I've some problems with the notation with the notes I'm studying, thank you very much.
– Giuseppe Bargagnati
Jul 17 at 14:22
add a comment |Â
up vote
3
down vote
accepted
This isn't true in general. Consider $A = beginbmatrix 0 & 1 \ -1 & 0endbmatrix$. We have
$$leftlanglebeginbmatrix 0 & 1 \ -1 & 0endbmatrixbeginbmatrix x \ y endbmatrix, beginbmatrix x \ yendbmatrix rightrangle = leftlangle beginbmatrix y \ -xendbmatrix, beginbmatrix x \ yendbmatrix rightrangle = 0$$
so $r_A = 0$ but $|A| ne 0$.
However, this is true if $A$ is assumed to be self-adjoint.
Moreover, it is even true for normal operators if the Hilbert space is complex.
answered Jul 16 at 23:10
mechanodroid
22.3k52041
If $A$ is self-adjoint, does it follow from the equality $||A||=||A^2^n||^frac 1 2^n$ and from the fact that $r_A= lim _n to infty||A^n||^frac 1 n$?
– Giuseppe Bargagnati
Jul 17 at 12:28
(And that I know that the last sequence admits limit).
– Giuseppe Bargagnati
Jul 17 at 12:33
1
@GiuseppeBargagnati $lim_ntoinfty |A^n|^frac1n$ is the spectral radius of $A$, not the numerical radius $r_A$. But yes, for normal operators we have $|A^n| = |A|^n$ so the spectral radius of $A$ is equal to $|A|$.
– mechanodroid
Jul 17 at 14:19
Ok, I've some problems with the notation with the notes I'm studying, thank you very much.
– Giuseppe Bargagnati
Jul 17 at 14:22
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
This isn't true in general. Consider $A = beginbmatrix 0 & 1 \ -1 & 0endbmatrix$. We have
$$leftlanglebeginbmatrix 0 & 1 \ -1 & 0endbmatrixbeginbmatrix x \ y endbmatrix, beginbmatrix x \ yendbmatrix rightrangle = leftlangle beginbmatrix y \ -xendbmatrix, beginbmatrix x \ yendbmatrix rightrangle = 0$$
so $r_A = 0$ but $|A| ne 0$.
However, this is true if $A$ is assumed to be self-adjoint.
Moreover, it is even true for normal operators if the Hilbert space is complex.
answered Jul 16 at 23:10
mechanodroid
22.3k52041
This isn't true in general. Consider $A = beginbmatrix 0 & 1 \ -1 & 0endbmatrix$. We have
$$leftlanglebeginbmatrix 0 & 1 \ -1 & 0endbmatrixbeginbmatrix x \ y endbmatrix, beginbmatrix x \ yendbmatrix rightrangle = leftlangle beginbmatrix y \ -xendbmatrix, beginbmatrix x \ yendbmatrix rightrangle = 0$$
so $r_A = 0$ but $|A| ne 0$.
However, this is true if $A$ is assumed to be self-adjoint.
Moreover, it is even true for normal operators if the Hilbert space is complex.
answered Jul 16 at 23:10
mechanodroid
22.3k52041
answered Jul 16 at 23:10
mechanodroid
22.3k52041
answered Jul 16 at 23:10
mechanodroid
22.3k52041
answered Jul 16 at 23:10
mechanodroid
22.3k52041
22.3k52041
If $A$ is self-adjoint, does it follow from the equality $||A||=||A^2^n||^frac 1 2^n$ and from the fact that $r_A= lim _n to infty||A^n||^frac 1 n$?
– Giuseppe Bargagnati
Jul 17 at 12:28
(And that I know that the last sequence admits limit).
– Giuseppe Bargagnati
Jul 17 at 12:33
1
@GiuseppeBargagnati $lim_ntoinfty |A^n|^frac1n$ is the spectral radius of $A$, not the numerical radius $r_A$. But yes, for normal operators we have $|A^n| = |A|^n$ so the spectral radius of $A$ is equal to $|A|$.
– mechanodroid
Jul 17 at 14:19
Ok, I've some problems with the notation with the notes I'm studying, thank you very much.
– Giuseppe Bargagnati
Jul 17 at 14:22
add a comment |Â
If $A$ is self-adjoint, does it follow from the equality $||A||=||A^2^n||^frac 1 2^n$ and from the fact that $r_A= lim _n to infty||A^n||^frac 1 n$?
– Giuseppe Bargagnati
Jul 17 at 12:28
(And that I know that the last sequence admits limit).
– Giuseppe Bargagnati
Jul 17 at 12:33
1
@GiuseppeBargagnati $lim_ntoinfty |A^n|^frac1n$ is the spectral radius of $A$, not the numerical radius $r_A$. But yes, for normal operators we have $|A^n| = |A|^n$ so the spectral radius of $A$ is equal to $|A|$.
– mechanodroid
Jul 17 at 14:19
Ok, I've some problems with the notation with the notes I'm studying, thank you very much.
– Giuseppe Bargagnati
Jul 17 at 14:22
If $A$ is self-adjoint, does it follow from the equality $||A||=||A^2^n||^frac 1 2^n$ and from the fact that $r_A= lim _n to infty||A^n||^frac 1 n$?
– Giuseppe Bargagnati
Jul 17 at 12:28
If $A$ is self-adjoint, does it follow from the equality $||A||=||A^2^n||^frac 1 2^n$ and from the fact that $r_A= lim _n to infty||A^n||^frac 1 n$?
– Giuseppe Bargagnati
Jul 17 at 12:28
(And that I know that the last sequence admits limit).
– Giuseppe Bargagnati
Jul 17 at 12:33
(And that I know that the last sequence admits limit).
– Giuseppe Bargagnati
Jul 17 at 12:33
1
1
@GiuseppeBargagnati $lim_ntoinfty |A^n|^frac1n$ is the spectral radius of $A$, not the numerical radius $r_A$. But yes, for normal operators we have $|A^n| = |A|^n$ so the spectral radius of $A$ is equal to $|A|$.
– mechanodroid
Jul 17 at 14:19
@GiuseppeBargagnati $lim_ntoinfty |A^n|^frac1n$ is the spectral radius of $A$, not the numerical radius $r_A$. But yes, for normal operators we have $|A^n| = |A|^n$ so the spectral radius of $A$ is equal to $|A|$.
– mechanodroid
Jul 17 at 14:19
Ok, I've some problems with the notation with the notes I'm studying, thank you very much.
– Giuseppe Bargagnati
Jul 17 at 14:22
Ok, I've some problems with the notation with the notes I'm studying, thank you very much.
– Giuseppe Bargagnati
Jul 17 at 14:22
add a comment |Â
up vote
1
down vote
I assume from your version of the polarization identity that we're working over a real Hilbert space.
Hint: Note that (if $A$ is self-adjoint)
$$
(Ax,y) = frac 14 left[ (x+y,A(x+y)) - (x-y,A(x-y))right]
$$
you can regard this as a generalization of the usual polarization identity. Thus, we have
$$
4sup_x (Ax,y) leq sup_x(A(x+y),(x+y)) - inf_x A(x-y,x-y)
$$
answered Jul 16 at 22:36
Omnomnomnom
121k784170
add a comment |Â
up vote
1
down vote
I assume from your version of the polarization identity that we're working over a real Hilbert space.
Hint: Note that (if $A$ is self-adjoint)
$$
(Ax,y) = frac 14 left[ (x+y,A(x+y)) - (x-y,A(x-y))right]
$$
you can regard this as a generalization of the usual polarization identity. Thus, we have
$$
4sup_x (Ax,y) leq sup_x(A(x+y),(x+y)) - inf_x A(x-y,x-y)
$$
answered Jul 16 at 22:36
Omnomnomnom
121k784170
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I assume from your version of the polarization identity that we're working over a real Hilbert space.
Hint: Note that (if $A$ is self-adjoint)
$$
(Ax,y) = frac 14 left[ (x+y,A(x+y)) - (x-y,A(x-y))right]
$$
you can regard this as a generalization of the usual polarization identity. Thus, we have
$$
4sup_x (Ax,y) leq sup_x(A(x+y),(x+y)) - inf_x A(x-y,x-y)
$$
answered Jul 16 at 22:36
Omnomnomnom
121k784170
I assume from your version of the polarization identity that we're working over a real Hilbert space.
Hint: Note that (if $A$ is self-adjoint)
$$
(Ax,y) = frac 14 left[ (x+y,A(x+y)) - (x-y,A(x-y))right]
$$
you can regard this as a generalization of the usual polarization identity. Thus, we have
$$
4sup_x (Ax,y) leq sup_x(A(x+y),(x+y)) - inf_x A(x-y,x-y)
$$
answered Jul 16 at 22:36
Omnomnomnom
121k784170
edited Jul 17 at 1:18
answered Jul 16 at 22:36
Omnomnomnom
121k784170
answered Jul 16 at 22:36
Omnomnomnom
121k784170
answered Jul 16 at 22:36
Omnomnomnom
121k784170
121k784170
add a comment |Â
add a comment |Â
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