Why is the polarization identity important, intuitively?
$begingroup$
The polarization identity states, roughly, that a norm satisfying the parallelogram law induces a vector space inner product (and vice versa). This has many nice applications, such as a simple characterization of unitary operators (which is useful in quantum mechanics), recovering operators from their associated quadratic forms, proving things about expectations of stochastic variables, or showing that the limit of a sequence of quadratic forms is a quadratic form.
What I don't see fully is why, when facing some problem, it would occur to someone to attempt to polarize something. What does polarization mean, philosophically, and why and when would I expect it to make an appearance?
Any other illustrative consequences of the polarization identity, or useful references are also welcome.
linear-algebra functional-analysis quadratic-forms motivation
asked Dec 11 '18 at 7:55
Kevin L.Kevin L.
162
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add a comment |
$begingroup$
The polarization identity states, roughly, that a norm satisfying the parallelogram law induces a vector space inner product (and vice versa). This has many nice applications, such as a simple characterization of unitary operators (which is useful in quantum mechanics), recovering operators from their associated quadratic forms, proving things about expectations of stochastic variables, or showing that the limit of a sequence of quadratic forms is a quadratic form.
What I don't see fully is why, when facing some problem, it would occur to someone to attempt to polarize something. What does polarization mean, philosophically, and why and when would I expect it to make an appearance?
Any other illustrative consequences of the polarization identity, or useful references are also welcome.
linear-algebra functional-analysis quadratic-forms motivation
asked Dec 11 '18 at 7:55
Kevin L.Kevin L.
162
$endgroup$
add a comment |
$begingroup$
The polarization identity states, roughly, that a norm satisfying the parallelogram law induces a vector space inner product (and vice versa). This has many nice applications, such as a simple characterization of unitary operators (which is useful in quantum mechanics), recovering operators from their associated quadratic forms, proving things about expectations of stochastic variables, or showing that the limit of a sequence of quadratic forms is a quadratic form.
What I don't see fully is why, when facing some problem, it would occur to someone to attempt to polarize something. What does polarization mean, philosophically, and why and when would I expect it to make an appearance?
Any other illustrative consequences of the polarization identity, or useful references are also welcome.
linear-algebra functional-analysis quadratic-forms motivation
asked Dec 11 '18 at 7:55
Kevin L.Kevin L.
162
$endgroup$
The polarization identity states, roughly, that a norm satisfying the parallelogram law induces a vector space inner product (and vice versa). This has many nice applications, such as a simple characterization of unitary operators (which is useful in quantum mechanics), recovering operators from their associated quadratic forms, proving things about expectations of stochastic variables, or showing that the limit of a sequence of quadratic forms is a quadratic form.
What I don't see fully is why, when facing some problem, it would occur to someone to attempt to polarize something. What does polarization mean, philosophically, and why and when would I expect it to make an appearance?
Any other illustrative consequences of the polarization identity, or useful references are also welcome.
linear-algebra functional-analysis quadratic-forms motivation
linear-algebra functional-analysis quadratic-forms motivation
asked Dec 11 '18 at 7:55
Kevin L.Kevin L.
162
asked Dec 11 '18 at 7:55
Kevin L.Kevin L.
162
asked Dec 11 '18 at 7:55
Kevin L.Kevin L.
162
asked Dec 11 '18 at 7:55
Kevin L.Kevin L.
162
asked Dec 11 '18 at 7:55
Kevin L.Kevin L.
162
162
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The polarization identity tells you that the geometry of a (complex) Hibert space is captured in its norm through the parallelogram law. This was precisely John von Neumann's goal in formulating his result that an inner product could be defined from a norm, assuming that the norm satisfies the parallelogram law. And that allowed simpler characterizations of various operators by considering only their quadratic forms, instead of their full forms. By the way, it was von Neumann who first defined an abstract inner product, even though Hilbert often gets that credit. Hilbert was von Neumann's advisor. von Neumman did many interesting things in his career, including proving the Spectral Theorem for selfadjoint operators, and inventing the CPU, just to name a couple of things.
Another interesting application is to selfadjoint operators. Suppose $X$ is a complex inner product space, and further suppose that $A : X rightarrow X$ is such that $langle Ax,xrangle$ is real for all $xin X$. Then, $langle Ax,xrangle=overline{langle Ax,xrangle}=langle x,Axrangle$, which also gives
begin{align}
langle Ax,yrangle &= frac{1}{4}sum_{n=0}^{3}i^nlangle A(x+i^n y),x+i^nyrangle \
&= frac{1}{4}sum_{n=0}^{3}i^nlangle x+i^ny,A(x+i^ny)rangle \
&= langle x,Ayrangle.
end{align}
From this, it also follows that
begin{align}
langle A(alpha x+beta y),zrangle
&=langle alpha x+beta y,Azrangle \
&=alpha langle x,Azrangle+betalangle y,Azrangle \
&=alpha langle Ax,zrangle+betalangle Ay,zrangle \
&=langle alpha Ax+beta Ay,zrangle.
end{align}
Hence $A$ is linear because the above holds for all $z$. So real Quantum measurements give rise to linear selfadjoint operators.
answered Dec 11 '18 at 20:58
DisintegratingByPartsDisintegratingByParts
59.3k42580
$endgroup$
1
$begingroup$
other "big" contributions of Neumann to history: "Von Neumann, four other scientists, and various military personnel were included in the target selection committee that was responsible for choosing the Japanese cities of Hiroshima and Nagasaki as the first targets of the atomic bomb. " Just to name a phew :)
$endgroup$
– Masacroso
Dec 12 '18 at 1:30
$begingroup$
@Masacroso "With the tongue we praise our Lord and Father, and with it we curse human beings, who have been made in God’s likeness. Out of the same mouth come praise and cursing. My brothers and sisters, this should not be." - James 3
$endgroup$
– DisintegratingByParts
Dec 12 '18 at 2:28
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
The polarization identity tells you that the geometry of a (complex) Hibert space is captured in its norm through the parallelogram law. This was precisely John von Neumann's goal in formulating his result that an inner product could be defined from a norm, assuming that the norm satisfies the parallelogram law. And that allowed simpler characterizations of various operators by considering only their quadratic forms, instead of their full forms. By the way, it was von Neumann who first defined an abstract inner product, even though Hilbert often gets that credit. Hilbert was von Neumann's advisor. von Neumman did many interesting things in his career, including proving the Spectral Theorem for selfadjoint operators, and inventing the CPU, just to name a couple of things.
Another interesting application is to selfadjoint operators. Suppose $X$ is a complex inner product space, and further suppose that $A : X rightarrow X$ is such that $langle Ax,xrangle$ is real for all $xin X$. Then, $langle Ax,xrangle=overline{langle Ax,xrangle}=langle x,Axrangle$, which also gives
begin{align}
langle Ax,yrangle &= frac{1}{4}sum_{n=0}^{3}i^nlangle A(x+i^n y),x+i^nyrangle \
&= frac{1}{4}sum_{n=0}^{3}i^nlangle x+i^ny,A(x+i^ny)rangle \
&= langle x,Ayrangle.
end{align}
From this, it also follows that
begin{align}
langle A(alpha x+beta y),zrangle
&=langle alpha x+beta y,Azrangle \
&=alpha langle x,Azrangle+betalangle y,Azrangle \
&=alpha langle Ax,zrangle+betalangle Ay,zrangle \
&=langle alpha Ax+beta Ay,zrangle.
end{align}
Hence $A$ is linear because the above holds for all $z$. So real Quantum measurements give rise to linear selfadjoint operators.
answered Dec 11 '18 at 20:58
DisintegratingByPartsDisintegratingByParts
59.3k42580
$endgroup$
1
$begingroup$
other "big" contributions of Neumann to history: "Von Neumann, four other scientists, and various military personnel were included in the target selection committee that was responsible for choosing the Japanese cities of Hiroshima and Nagasaki as the first targets of the atomic bomb. " Just to name a phew :)
$endgroup$
– Masacroso
Dec 12 '18 at 1:30
$begingroup$
@Masacroso "With the tongue we praise our Lord and Father, and with it we curse human beings, who have been made in God’s likeness. Out of the same mouth come praise and cursing. My brothers and sisters, this should not be." - James 3
$endgroup$
– DisintegratingByParts
Dec 12 '18 at 2:28
add a comment |
$begingroup$
The polarization identity tells you that the geometry of a (complex) Hibert space is captured in its norm through the parallelogram law. This was precisely John von Neumann's goal in formulating his result that an inner product could be defined from a norm, assuming that the norm satisfies the parallelogram law. And that allowed simpler characterizations of various operators by considering only their quadratic forms, instead of their full forms. By the way, it was von Neumann who first defined an abstract inner product, even though Hilbert often gets that credit. Hilbert was von Neumann's advisor. von Neumman did many interesting things in his career, including proving the Spectral Theorem for selfadjoint operators, and inventing the CPU, just to name a couple of things.
Another interesting application is to selfadjoint operators. Suppose $X$ is a complex inner product space, and further suppose that $A : X rightarrow X$ is such that $langle Ax,xrangle$ is real for all $xin X$. Then, $langle Ax,xrangle=overline{langle Ax,xrangle}=langle x,Axrangle$, which also gives
begin{align}
langle Ax,yrangle &= frac{1}{4}sum_{n=0}^{3}i^nlangle A(x+i^n y),x+i^nyrangle \
&= frac{1}{4}sum_{n=0}^{3}i^nlangle x+i^ny,A(x+i^ny)rangle \
&= langle x,Ayrangle.
end{align}
From this, it also follows that
begin{align}
langle A(alpha x+beta y),zrangle
&=langle alpha x+beta y,Azrangle \
&=alpha langle x,Azrangle+betalangle y,Azrangle \
&=alpha langle Ax,zrangle+betalangle Ay,zrangle \
&=langle alpha Ax+beta Ay,zrangle.
end{align}
Hence $A$ is linear because the above holds for all $z$. So real Quantum measurements give rise to linear selfadjoint operators.
answered Dec 11 '18 at 20:58
DisintegratingByPartsDisintegratingByParts
59.3k42580
$endgroup$
1
$begingroup$
other "big" contributions of Neumann to history: "Von Neumann, four other scientists, and various military personnel were included in the target selection committee that was responsible for choosing the Japanese cities of Hiroshima and Nagasaki as the first targets of the atomic bomb. " Just to name a phew :)
$endgroup$
– Masacroso
Dec 12 '18 at 1:30
$begingroup$
@Masacroso "With the tongue we praise our Lord and Father, and with it we curse human beings, who have been made in God’s likeness. Out of the same mouth come praise and cursing. My brothers and sisters, this should not be." - James 3
$endgroup$
– DisintegratingByParts
Dec 12 '18 at 2:28
add a comment |
$begingroup$
The polarization identity tells you that the geometry of a (complex) Hibert space is captured in its norm through the parallelogram law. This was precisely John von Neumann's goal in formulating his result that an inner product could be defined from a norm, assuming that the norm satisfies the parallelogram law. And that allowed simpler characterizations of various operators by considering only their quadratic forms, instead of their full forms. By the way, it was von Neumann who first defined an abstract inner product, even though Hilbert often gets that credit. Hilbert was von Neumann's advisor. von Neumman did many interesting things in his career, including proving the Spectral Theorem for selfadjoint operators, and inventing the CPU, just to name a couple of things.
Another interesting application is to selfadjoint operators. Suppose $X$ is a complex inner product space, and further suppose that $A : X rightarrow X$ is such that $langle Ax,xrangle$ is real for all $xin X$. Then, $langle Ax,xrangle=overline{langle Ax,xrangle}=langle x,Axrangle$, which also gives
begin{align}
langle Ax,yrangle &= frac{1}{4}sum_{n=0}^{3}i^nlangle A(x+i^n y),x+i^nyrangle \
&= frac{1}{4}sum_{n=0}^{3}i^nlangle x+i^ny,A(x+i^ny)rangle \
&= langle x,Ayrangle.
end{align}
From this, it also follows that
begin{align}
langle A(alpha x+beta y),zrangle
&=langle alpha x+beta y,Azrangle \
&=alpha langle x,Azrangle+betalangle y,Azrangle \
&=alpha langle Ax,zrangle+betalangle Ay,zrangle \
&=langle alpha Ax+beta Ay,zrangle.
end{align}
Hence $A$ is linear because the above holds for all $z$. So real Quantum measurements give rise to linear selfadjoint operators.
answered Dec 11 '18 at 20:58
DisintegratingByPartsDisintegratingByParts
59.3k42580
$endgroup$
The polarization identity tells you that the geometry of a (complex) Hibert space is captured in its norm through the parallelogram law. This was precisely John von Neumann's goal in formulating his result that an inner product could be defined from a norm, assuming that the norm satisfies the parallelogram law. And that allowed simpler characterizations of various operators by considering only their quadratic forms, instead of their full forms. By the way, it was von Neumann who first defined an abstract inner product, even though Hilbert often gets that credit. Hilbert was von Neumann's advisor. von Neumman did many interesting things in his career, including proving the Spectral Theorem for selfadjoint operators, and inventing the CPU, just to name a couple of things.
Another interesting application is to selfadjoint operators. Suppose $X$ is a complex inner product space, and further suppose that $A : X rightarrow X$ is such that $langle Ax,xrangle$ is real for all $xin X$. Then, $langle Ax,xrangle=overline{langle Ax,xrangle}=langle x,Axrangle$, which also gives
begin{align}
langle Ax,yrangle &= frac{1}{4}sum_{n=0}^{3}i^nlangle A(x+i^n y),x+i^nyrangle \
&= frac{1}{4}sum_{n=0}^{3}i^nlangle x+i^ny,A(x+i^ny)rangle \
&= langle x,Ayrangle.
end{align}
From this, it also follows that
begin{align}
langle A(alpha x+beta y),zrangle
&=langle alpha x+beta y,Azrangle \
&=alpha langle x,Azrangle+betalangle y,Azrangle \
&=alpha langle Ax,zrangle+betalangle Ay,zrangle \
&=langle alpha Ax+beta Ay,zrangle.
end{align}
Hence $A$ is linear because the above holds for all $z$. So real Quantum measurements give rise to linear selfadjoint operators.
answered Dec 11 '18 at 20:58
DisintegratingByPartsDisintegratingByParts
59.3k42580
answered Dec 11 '18 at 20:58
DisintegratingByPartsDisintegratingByParts
59.3k42580
answered Dec 11 '18 at 20:58
DisintegratingByPartsDisintegratingByParts
59.3k42580
answered Dec 11 '18 at 20:58
DisintegratingByPartsDisintegratingByParts
59.3k42580
59.3k42580
1
$begingroup$
other "big" contributions of Neumann to history: "Von Neumann, four other scientists, and various military personnel were included in the target selection committee that was responsible for choosing the Japanese cities of Hiroshima and Nagasaki as the first targets of the atomic bomb. " Just to name a phew :)
$endgroup$
– Masacroso
Dec 12 '18 at 1:30
$begingroup$
@Masacroso "With the tongue we praise our Lord and Father, and with it we curse human beings, who have been made in God’s likeness. Out of the same mouth come praise and cursing. My brothers and sisters, this should not be." - James 3
$endgroup$
– DisintegratingByParts
Dec 12 '18 at 2:28
add a comment |
1
$begingroup$
other "big" contributions of Neumann to history: "Von Neumann, four other scientists, and various military personnel were included in the target selection committee that was responsible for choosing the Japanese cities of Hiroshima and Nagasaki as the first targets of the atomic bomb. " Just to name a phew :)
$endgroup$
– Masacroso
Dec 12 '18 at 1:30
$begingroup$
@Masacroso "With the tongue we praise our Lord and Father, and with it we curse human beings, who have been made in God’s likeness. Out of the same mouth come praise and cursing. My brothers and sisters, this should not be." - James 3
$endgroup$
– DisintegratingByParts
Dec 12 '18 at 2:28
1
1
$begingroup$
other "big" contributions of Neumann to history: "Von Neumann, four other scientists, and various military personnel were included in the target selection committee that was responsible for choosing the Japanese cities of Hiroshima and Nagasaki as the first targets of the atomic bomb. " Just to name a phew :)
$endgroup$
– Masacroso
Dec 12 '18 at 1:30
$begingroup$
other "big" contributions of Neumann to history: "Von Neumann, four other scientists, and various military personnel were included in the target selection committee that was responsible for choosing the Japanese cities of Hiroshima and Nagasaki as the first targets of the atomic bomb. " Just to name a phew :)
$endgroup$
– Masacroso
Dec 12 '18 at 1:30
$begingroup$
@Masacroso "With the tongue we praise our Lord and Father, and with it we curse human beings, who have been made in God’s likeness. Out of the same mouth come praise and cursing. My brothers and sisters, this should not be." - James 3
$endgroup$
– DisintegratingByParts
Dec 12 '18 at 2:28
$begingroup$
@Masacroso "With the tongue we praise our Lord and Father, and with it we curse human beings, who have been made in God’s likeness. Out of the same mouth come praise and cursing. My brothers and sisters, this should not be." - James 3
$endgroup$
– DisintegratingByParts
Dec 12 '18 at 2:28
add a comment |
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