Mixing
calculate the norm of the following operator on $mathbb{C}^2$
$begingroup$
It is well known that if $Ain mathcal{B}(F)$, then
$$|A|:=displaystylesup_{|x|=1}|Ax|.$$
and
$$omega(T):=displaystylesup_{|x|=1}|langle Ax, xrangle|,$$
where $F$ is a complex Hilbert space and $mathcal{B}(F)$ is the algebra of all bounded linear operators on $F$.
Consider the following operator on $mathbb{C}^2$
$$T=begin{pmatrix} 1 & 1 \ 0 & 1 end{pmatrix}.$$
I want to prove that
begin{align*}
omega(T)
&=supleft{||x|^2+yoverline{x}+|y|^2|,;;(x,y)in mathbb{C}^2,;text{and};|x|^2+|y|^2=1 right}=tfrac{3}{2},
end{align*}
and
begin{align*}
|T|
&=supleft{sqrt{|x+y|^2+|y|^2},;;(x_1,x_2)in mathbb{C}^2,;text{and};|x|^2+|y|^2=1 right}=frac{sqrt{5}+1}{2}.
end{align*}
Attempt:
Clearly, for all $(x,y)in mathbb{C}^2$ such that $|x|^2+|y|^2=1$, we have
$$||x|^2+yoverline{x}+|y|^2|leq frac12(|x|^2+|y|^2)+|x|^2+|y|^2=frac32.$$
Hence,
$$omega(T)leqfrac32.$$
To calculate $|T|$, I try to use the following result:
$$|T|=max{sqrt{lambda};;lambdain sigma(T^*T)},$$
where $sigma(A)$ denotes the spectrum of an operator $A$.
Clearly
$$T^*T=begin{pmatrix} 1 & 1 \ 1 & 2 end{pmatrix}.$$
Also
$$sigma(T^*T)={frac{sqrt{5}+3}{2},frac{3-sqrt{5}}{2}}.$$
Hence,
$$|T|^2=frac{sqrt{5}+3}{2}.$$
linear-algebra
asked Jan 21 at 20:07
SchülerSchüler
1,5391421
$endgroup$
add a comment |
$begingroup$
It is well known that if $Ain mathcal{B}(F)$, then
$$|A|:=displaystylesup_{|x|=1}|Ax|.$$
and
$$omega(T):=displaystylesup_{|x|=1}|langle Ax, xrangle|,$$
where $F$ is a complex Hilbert space and $mathcal{B}(F)$ is the algebra of all bounded linear operators on $F$.
Consider the following operator on $mathbb{C}^2$
$$T=begin{pmatrix} 1 & 1 \ 0 & 1 end{pmatrix}.$$
I want to prove that
begin{align*}
omega(T)
&=supleft{||x|^2+yoverline{x}+|y|^2|,;;(x,y)in mathbb{C}^2,;text{and};|x|^2+|y|^2=1 right}=tfrac{3}{2},
end{align*}
and
begin{align*}
|T|
&=supleft{sqrt{|x+y|^2+|y|^2},;;(x_1,x_2)in mathbb{C}^2,;text{and};|x|^2+|y|^2=1 right}=frac{sqrt{5}+1}{2}.
end{align*}
Attempt:
Clearly, for all $(x,y)in mathbb{C}^2$ such that $|x|^2+|y|^2=1$, we have
$$||x|^2+yoverline{x}+|y|^2|leq frac12(|x|^2+|y|^2)+|x|^2+|y|^2=frac32.$$
Hence,
$$omega(T)leqfrac32.$$
To calculate $|T|$, I try to use the following result:
$$|T|=max{sqrt{lambda};;lambdain sigma(T^*T)},$$
where $sigma(A)$ denotes the spectrum of an operator $A$.
Clearly
$$T^*T=begin{pmatrix} 1 & 1 \ 1 & 2 end{pmatrix}.$$
Also
$$sigma(T^*T)={frac{sqrt{5}+3}{2},frac{3-sqrt{5}}{2}}.$$
Hence,
$$|T|^2=frac{sqrt{5}+3}{2}.$$
linear-algebra
asked Jan 21 at 20:07
SchülerSchüler
1,5391421
$endgroup$
1
$begingroup$
Something is wrong here. The quantity $|x|^2+yoverline{x}+|y|^2$ need not be real, so that the supremum does not exist.
$endgroup$
– Ben W
Jan 21 at 20:13
$begingroup$
@BenW Sorry I have edited my question.
$endgroup$
– Schüler
Jan 21 at 20:17
$begingroup$
Np. If $x=y=(1+i)/4$ then $|x|^2+yoverline{x}+|y|^2=3/2$. That settles the first question. Can you do the second?
$endgroup$
– Ben W
Jan 21 at 20:43
$begingroup$
For the second, i try to solve it however i'm facing difficulties how can l start.
$endgroup$
– Schüler
Jan 21 at 20:47
$begingroup$
@BenW I think you should take $x=y=(1+i)/2$ Do you agree with me? Thanks
$endgroup$
– Schüler
Jan 24 at 6:13
add a comment |
$begingroup$
It is well known that if $Ain mathcal{B}(F)$, then
$$|A|:=displaystylesup_{|x|=1}|Ax|.$$
and
$$omega(T):=displaystylesup_{|x|=1}|langle Ax, xrangle|,$$
where $F$ is a complex Hilbert space and $mathcal{B}(F)$ is the algebra of all bounded linear operators on $F$.
Consider the following operator on $mathbb{C}^2$
$$T=begin{pmatrix} 1 & 1 \ 0 & 1 end{pmatrix}.$$
I want to prove that
begin{align*}
omega(T)
&=supleft{||x|^2+yoverline{x}+|y|^2|,;;(x,y)in mathbb{C}^2,;text{and};|x|^2+|y|^2=1 right}=tfrac{3}{2},
end{align*}
and
begin{align*}
|T|
&=supleft{sqrt{|x+y|^2+|y|^2},;;(x_1,x_2)in mathbb{C}^2,;text{and};|x|^2+|y|^2=1 right}=frac{sqrt{5}+1}{2}.
end{align*}
Attempt:
Clearly, for all $(x,y)in mathbb{C}^2$ such that $|x|^2+|y|^2=1$, we have
$$||x|^2+yoverline{x}+|y|^2|leq frac12(|x|^2+|y|^2)+|x|^2+|y|^2=frac32.$$
Hence,
$$omega(T)leqfrac32.$$
To calculate $|T|$, I try to use the following result:
$$|T|=max{sqrt{lambda};;lambdain sigma(T^*T)},$$
where $sigma(A)$ denotes the spectrum of an operator $A$.
Clearly
$$T^*T=begin{pmatrix} 1 & 1 \ 1 & 2 end{pmatrix}.$$
Also
$$sigma(T^*T)={frac{sqrt{5}+3}{2},frac{3-sqrt{5}}{2}}.$$
Hence,
$$|T|^2=frac{sqrt{5}+3}{2}.$$
linear-algebra
asked Jan 21 at 20:07
SchülerSchüler
1,5391421
$endgroup$
It is well known that if $Ain mathcal{B}(F)$, then
$$|A|:=displaystylesup_{|x|=1}|Ax|.$$
and
$$omega(T):=displaystylesup_{|x|=1}|langle Ax, xrangle|,$$
where $F$ is a complex Hilbert space and $mathcal{B}(F)$ is the algebra of all bounded linear operators on $F$.
Consider the following operator on $mathbb{C}^2$
$$T=begin{pmatrix} 1 & 1 \ 0 & 1 end{pmatrix}.$$
I want to prove that
begin{align*}
omega(T)
&=supleft{||x|^2+yoverline{x}+|y|^2|,;;(x,y)in mathbb{C}^2,;text{and};|x|^2+|y|^2=1 right}=tfrac{3}{2},
end{align*}
and
begin{align*}
|T|
&=supleft{sqrt{|x+y|^2+|y|^2},;;(x_1,x_2)in mathbb{C}^2,;text{and};|x|^2+|y|^2=1 right}=frac{sqrt{5}+1}{2}.
end{align*}
Attempt:
Clearly, for all $(x,y)in mathbb{C}^2$ such that $|x|^2+|y|^2=1$, we have
$$||x|^2+yoverline{x}+|y|^2|leq frac12(|x|^2+|y|^2)+|x|^2+|y|^2=frac32.$$
Hence,
$$omega(T)leqfrac32.$$
To calculate $|T|$, I try to use the following result:
$$|T|=max{sqrt{lambda};;lambdain sigma(T^*T)},$$
where $sigma(A)$ denotes the spectrum of an operator $A$.
Clearly
$$T^*T=begin{pmatrix} 1 & 1 \ 1 & 2 end{pmatrix}.$$
Also
$$sigma(T^*T)={frac{sqrt{5}+3}{2},frac{3-sqrt{5}}{2}}.$$
Hence,
$$|T|^2=frac{sqrt{5}+3}{2}.$$
linear-algebra
linear-algebra
asked Jan 21 at 20:07
SchülerSchüler
1,5391421
asked Jan 21 at 20:07
SchülerSchüler
1,5391421
edited Jan 23 at 19:00
Schüler
asked Jan 21 at 20:07
SchülerSchüler
1,5391421
asked Jan 21 at 20:07
SchülerSchüler
1,5391421
asked Jan 21 at 20:07
SchülerSchüler
1,5391421
1,5391421
1
$begingroup$
Something is wrong here. The quantity $|x|^2+yoverline{x}+|y|^2$ need not be real, so that the supremum does not exist.
$endgroup$
– Ben W
Jan 21 at 20:13
$begingroup$
@BenW Sorry I have edited my question.
$endgroup$
– Schüler
Jan 21 at 20:17
$begingroup$
Np. If $x=y=(1+i)/4$ then $|x|^2+yoverline{x}+|y|^2=3/2$. That settles the first question. Can you do the second?
$endgroup$
– Ben W
Jan 21 at 20:43
$begingroup$
For the second, i try to solve it however i'm facing difficulties how can l start.
$endgroup$
– Schüler
Jan 21 at 20:47
$begingroup$
@BenW I think you should take $x=y=(1+i)/2$ Do you agree with me? Thanks
$endgroup$
– Schüler
Jan 24 at 6:13
add a comment |
1
$begingroup$
Something is wrong here. The quantity $|x|^2+yoverline{x}+|y|^2$ need not be real, so that the supremum does not exist.
$endgroup$
– Ben W
Jan 21 at 20:13
$begingroup$
@BenW Sorry I have edited my question.
$endgroup$
– Schüler
Jan 21 at 20:17
$begingroup$
Np. If $x=y=(1+i)/4$ then $|x|^2+yoverline{x}+|y|^2=3/2$. That settles the first question. Can you do the second?
$endgroup$
– Ben W
Jan 21 at 20:43
$begingroup$
For the second, i try to solve it however i'm facing difficulties how can l start.
$endgroup$
– Schüler
Jan 21 at 20:47
$begingroup$
@BenW I think you should take $x=y=(1+i)/2$ Do you agree with me? Thanks
$endgroup$
– Schüler
Jan 24 at 6:13
1
1
$begingroup$
Something is wrong here. The quantity $|x|^2+yoverline{x}+|y|^2$ need not be real, so that the supremum does not exist.
$endgroup$
– Ben W
Jan 21 at 20:13
$begingroup$
Something is wrong here. The quantity $|x|^2+yoverline{x}+|y|^2$ need not be real, so that the supremum does not exist.
$endgroup$
– Ben W
Jan 21 at 20:13
$begingroup$
@BenW Sorry I have edited my question.
$endgroup$
– Schüler
Jan 21 at 20:17
$begingroup$
@BenW Sorry I have edited my question.
$endgroup$
– Schüler
Jan 21 at 20:17
$begingroup$
Np. If $x=y=(1+i)/4$ then $|x|^2+yoverline{x}+|y|^2=3/2$. That settles the first question. Can you do the second?
$endgroup$
– Ben W
Jan 21 at 20:43
$begingroup$
Np. If $x=y=(1+i)/4$ then $|x|^2+yoverline{x}+|y|^2=3/2$. That settles the first question. Can you do the second?
$endgroup$
– Ben W
Jan 21 at 20:43
$begingroup$
For the second, i try to solve it however i'm facing difficulties how can l start.
$endgroup$
– Schüler
Jan 21 at 20:47
$begingroup$
For the second, i try to solve it however i'm facing difficulties how can l start.
$endgroup$
– Schüler
Jan 21 at 20:47
$begingroup$
@BenW I think you should take $x=y=(1+i)/2$ Do you agree with me? Thanks
$endgroup$
– Schüler
Jan 24 at 6:13
$begingroup$
@BenW I think you should take $x=y=(1+i)/2$ Do you agree with me? Thanks
$endgroup$
– Schüler
Jan 24 at 6:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For the second, let's see why for all $z,winmathbb{C}$ with $|z|^2+|w|^2=1$ it is $sqrt{|z+w|^2+|w|^2}leq(sqrt{5}+1)/2$. Let $a=|z|, b=|y|inmathbb{R}$. It is obviously $sqrt{|z+w|^2+|w|^2}leqsqrt{1+2ab+b^2}$ while $a,b$ satisfy $a^2+b^2=1$. Now we can use Lagrange multipliers to solve a classical problem. I won't go into details, but it will work (at least a graph calculator says so).
Now let's see that there exist $z,winmathbb{C}$ such that $|z|^2+|w|^2=1$ and $sqrt{|z+w|^2+|w|^2}=(sqrt{5}+1)/2$. Rewrite this equivalently as $4(|z+w|^2+|w|^2)=6+2sqrt{5}$. This is equivalent to $2(|w|^2+2text{Re}(zoverline{w}))=1+sqrt{5}$. Now let $w$ be a real number, say $cin(0,1)$. Then we have that $2c^2+2ctext{Re}(z)=1+sqrt{5}$, hence $text{Re}(z)=frac{1+sqrt{5}-2c^2}{2c}$. Let for convenience $c$ be the solution to the equation $2t^2=1+sqrt{5}$ (it does indeed have a solution in $(0,1)$). Then $text{Re}(z)=0$. So now since $z$ is imaginary, it is $z=iy$ and $|z|=|y|$. solve $c^2+y^2=1$ for $y$ and note that it has a solution in $(0,1)$ too. the pair $(iy,c)inmathbb{C}^2$ makes this supremum a maximum.
answered Jan 21 at 21:51
JustDroppedInJustDroppedIn
2,153420
$endgroup$
$begingroup$
@Schüler are you familiar with the lagrange multipliers method?
$endgroup$
– JustDroppedIn
Jan 22 at 20:16
$begingroup$
@Schüler are you sure? I see that you're talking about spectra of operators in your edit, Lagrange multipliers is a standard calculus tool to maximize functions of several real variables.
$endgroup$
– JustDroppedIn
Jan 23 at 18:41
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3082342%2fcalculate-the-norm-of-the-following-operator-on-mathbbc2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the second, let's see why for all $z,winmathbb{C}$ with $|z|^2+|w|^2=1$ it is $sqrt{|z+w|^2+|w|^2}leq(sqrt{5}+1)/2$. Let $a=|z|, b=|y|inmathbb{R}$. It is obviously $sqrt{|z+w|^2+|w|^2}leqsqrt{1+2ab+b^2}$ while $a,b$ satisfy $a^2+b^2=1$. Now we can use Lagrange multipliers to solve a classical problem. I won't go into details, but it will work (at least a graph calculator says so).
Now let's see that there exist $z,winmathbb{C}$ such that $|z|^2+|w|^2=1$ and $sqrt{|z+w|^2+|w|^2}=(sqrt{5}+1)/2$. Rewrite this equivalently as $4(|z+w|^2+|w|^2)=6+2sqrt{5}$. This is equivalent to $2(|w|^2+2text{Re}(zoverline{w}))=1+sqrt{5}$. Now let $w$ be a real number, say $cin(0,1)$. Then we have that $2c^2+2ctext{Re}(z)=1+sqrt{5}$, hence $text{Re}(z)=frac{1+sqrt{5}-2c^2}{2c}$. Let for convenience $c$ be the solution to the equation $2t^2=1+sqrt{5}$ (it does indeed have a solution in $(0,1)$). Then $text{Re}(z)=0$. So now since $z$ is imaginary, it is $z=iy$ and $|z|=|y|$. solve $c^2+y^2=1$ for $y$ and note that it has a solution in $(0,1)$ too. the pair $(iy,c)inmathbb{C}^2$ makes this supremum a maximum.
answered Jan 21 at 21:51
JustDroppedInJustDroppedIn
2,153420
$endgroup$
$begingroup$
@Schüler are you familiar with the lagrange multipliers method?
$endgroup$
– JustDroppedIn
Jan 22 at 20:16
$begingroup$
@Schüler are you sure? I see that you're talking about spectra of operators in your edit, Lagrange multipliers is a standard calculus tool to maximize functions of several real variables.
$endgroup$
– JustDroppedIn
Jan 23 at 18:41
add a comment |
$begingroup$
For the second, let's see why for all $z,winmathbb{C}$ with $|z|^2+|w|^2=1$ it is $sqrt{|z+w|^2+|w|^2}leq(sqrt{5}+1)/2$. Let $a=|z|, b=|y|inmathbb{R}$. It is obviously $sqrt{|z+w|^2+|w|^2}leqsqrt{1+2ab+b^2}$ while $a,b$ satisfy $a^2+b^2=1$. Now we can use Lagrange multipliers to solve a classical problem. I won't go into details, but it will work (at least a graph calculator says so).
Now let's see that there exist $z,winmathbb{C}$ such that $|z|^2+|w|^2=1$ and $sqrt{|z+w|^2+|w|^2}=(sqrt{5}+1)/2$. Rewrite this equivalently as $4(|z+w|^2+|w|^2)=6+2sqrt{5}$. This is equivalent to $2(|w|^2+2text{Re}(zoverline{w}))=1+sqrt{5}$. Now let $w$ be a real number, say $cin(0,1)$. Then we have that $2c^2+2ctext{Re}(z)=1+sqrt{5}$, hence $text{Re}(z)=frac{1+sqrt{5}-2c^2}{2c}$. Let for convenience $c$ be the solution to the equation $2t^2=1+sqrt{5}$ (it does indeed have a solution in $(0,1)$). Then $text{Re}(z)=0$. So now since $z$ is imaginary, it is $z=iy$ and $|z|=|y|$. solve $c^2+y^2=1$ for $y$ and note that it has a solution in $(0,1)$ too. the pair $(iy,c)inmathbb{C}^2$ makes this supremum a maximum.
answered Jan 21 at 21:51
JustDroppedInJustDroppedIn
2,153420
$endgroup$
$begingroup$
@Schüler are you familiar with the lagrange multipliers method?
$endgroup$
– JustDroppedIn
Jan 22 at 20:16
$begingroup$
@Schüler are you sure? I see that you're talking about spectra of operators in your edit, Lagrange multipliers is a standard calculus tool to maximize functions of several real variables.
$endgroup$
– JustDroppedIn
Jan 23 at 18:41
add a comment |
$begingroup$
For the second, let's see why for all $z,winmathbb{C}$ with $|z|^2+|w|^2=1$ it is $sqrt{|z+w|^2+|w|^2}leq(sqrt{5}+1)/2$. Let $a=|z|, b=|y|inmathbb{R}$. It is obviously $sqrt{|z+w|^2+|w|^2}leqsqrt{1+2ab+b^2}$ while $a,b$ satisfy $a^2+b^2=1$. Now we can use Lagrange multipliers to solve a classical problem. I won't go into details, but it will work (at least a graph calculator says so).
Now let's see that there exist $z,winmathbb{C}$ such that $|z|^2+|w|^2=1$ and $sqrt{|z+w|^2+|w|^2}=(sqrt{5}+1)/2$. Rewrite this equivalently as $4(|z+w|^2+|w|^2)=6+2sqrt{5}$. This is equivalent to $2(|w|^2+2text{Re}(zoverline{w}))=1+sqrt{5}$. Now let $w$ be a real number, say $cin(0,1)$. Then we have that $2c^2+2ctext{Re}(z)=1+sqrt{5}$, hence $text{Re}(z)=frac{1+sqrt{5}-2c^2}{2c}$. Let for convenience $c$ be the solution to the equation $2t^2=1+sqrt{5}$ (it does indeed have a solution in $(0,1)$). Then $text{Re}(z)=0$. So now since $z$ is imaginary, it is $z=iy$ and $|z|=|y|$. solve $c^2+y^2=1$ for $y$ and note that it has a solution in $(0,1)$ too. the pair $(iy,c)inmathbb{C}^2$ makes this supremum a maximum.
answered Jan 21 at 21:51
JustDroppedInJustDroppedIn
2,153420
$endgroup$
For the second, let's see why for all $z,winmathbb{C}$ with $|z|^2+|w|^2=1$ it is $sqrt{|z+w|^2+|w|^2}leq(sqrt{5}+1)/2$. Let $a=|z|, b=|y|inmathbb{R}$. It is obviously $sqrt{|z+w|^2+|w|^2}leqsqrt{1+2ab+b^2}$ while $a,b$ satisfy $a^2+b^2=1$. Now we can use Lagrange multipliers to solve a classical problem. I won't go into details, but it will work (at least a graph calculator says so).
Now let's see that there exist $z,winmathbb{C}$ such that $|z|^2+|w|^2=1$ and $sqrt{|z+w|^2+|w|^2}=(sqrt{5}+1)/2$. Rewrite this equivalently as $4(|z+w|^2+|w|^2)=6+2sqrt{5}$. This is equivalent to $2(|w|^2+2text{Re}(zoverline{w}))=1+sqrt{5}$. Now let $w$ be a real number, say $cin(0,1)$. Then we have that $2c^2+2ctext{Re}(z)=1+sqrt{5}$, hence $text{Re}(z)=frac{1+sqrt{5}-2c^2}{2c}$. Let for convenience $c$ be the solution to the equation $2t^2=1+sqrt{5}$ (it does indeed have a solution in $(0,1)$). Then $text{Re}(z)=0$. So now since $z$ is imaginary, it is $z=iy$ and $|z|=|y|$. solve $c^2+y^2=1$ for $y$ and note that it has a solution in $(0,1)$ too. the pair $(iy,c)inmathbb{C}^2$ makes this supremum a maximum.
answered Jan 21 at 21:51
JustDroppedInJustDroppedIn
2,153420
answered Jan 21 at 21:51
JustDroppedInJustDroppedIn
2,153420
answered Jan 21 at 21:51
JustDroppedInJustDroppedIn
2,153420
answered Jan 21 at 21:51
JustDroppedInJustDroppedIn
2,153420
2,153420
$begingroup$
@Schüler are you familiar with the lagrange multipliers method?
$endgroup$
– JustDroppedIn
Jan 22 at 20:16
$begingroup$
@Schüler are you sure? I see that you're talking about spectra of operators in your edit, Lagrange multipliers is a standard calculus tool to maximize functions of several real variables.
$endgroup$
– JustDroppedIn
Jan 23 at 18:41
add a comment |
$begingroup$
@Schüler are you familiar with the lagrange multipliers method?
$endgroup$
– JustDroppedIn
Jan 22 at 20:16
$begingroup$
@Schüler are you sure? I see that you're talking about spectra of operators in your edit, Lagrange multipliers is a standard calculus tool to maximize functions of several real variables.
$endgroup$
– JustDroppedIn
Jan 23 at 18:41
$begingroup$
@Schüler are you familiar with the lagrange multipliers method?
$endgroup$
– JustDroppedIn
Jan 22 at 20:16
$begingroup$
@Schüler are you familiar with the lagrange multipliers method?
$endgroup$
– JustDroppedIn
Jan 22 at 20:16
$begingroup$
@Schüler are you sure? I see that you're talking about spectra of operators in your edit, Lagrange multipliers is a standard calculus tool to maximize functions of several real variables.
$endgroup$
– JustDroppedIn
Jan 23 at 18:41
$begingroup$
@Schüler are you sure? I see that you're talking about spectra of operators in your edit, Lagrange multipliers is a standard calculus tool to maximize functions of several real variables.
$endgroup$
– JustDroppedIn
Jan 23 at 18:41
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3082342%2fcalculate-the-norm-of-the-following-operator-on-mathbbc2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Something is wrong here. The quantity $|x|^2+yoverline{x}+|y|^2$ need not be real, so that the supremum does not exist.
$endgroup$
– Ben W
Jan 21 at 20:13
$begingroup$
@BenW Sorry I have edited my question.
$endgroup$
– Schüler
Jan 21 at 20:17
$begingroup$
Np. If $x=y=(1+i)/4$ then $|x|^2+yoverline{x}+|y|^2=3/2$. That settles the first question. Can you do the second?
$endgroup$
– Ben W
Jan 21 at 20:43
$begingroup$
For the second, i try to solve it however i'm facing difficulties how can l start.
$endgroup$
– Schüler
Jan 21 at 20:47
$begingroup$
@BenW I think you should take $x=y=(1+i)/2$ Do you agree with me? Thanks
$endgroup$
– Schüler
Jan 24 at 6:13