Mixing
Froebenius norm is unitarily invariant.
$begingroup$
I'm considering the norm defined on matrices by
$$|A|_F = sqrt{sum_{i,j}|a_{ij}|^2}$$
I want to show that it is unitarily invariant, so that for unitary $U$ we have that
$$|UA|_F = |A|_F = |AU|_F$$
however I have trouble doing it directly. Writing $|UA|_F$ directly I find by Cauchy-Schwarz that
$$|UA|_F = sqrt{sum_{i,j}left|sum_{k=1}^{n}u_{ik}a_{kj}right|^2}= sqrt{sum_{i,j}|langle U_i,overline{A_j}rangle|^2}leq sqrt{sum_{i,j}|A_j|^2}$$
where $U_i$ denotes the $i$th row of $U$ and $A_j$ the $j$th column of $A$. However this estimate is to crude and will not equal $|A|_F$. I would like to prove this without refering to trace or singular values and would appreciate a hint, rather than a full solution, on how to tackle this problem.
EDIT: Completion of the proof based on the answer from $A.Gamma$:
Since the rows of $U$ constitute an orthonormal basis for $mathbb{C}^n$ we find by Parsevals theorem that
$$|UA_j|_2^2 = sum_{i=1}^{n}left|sum_{k=1}^{n}u_{ik}a_{kj}right|^2 = sum_{i=1}^{n}|langle U_i,overline{A_j}rangle|^2 = |overline{A_j}|_2^2 = |A_j|_2^2$$
linear-algebra matrices norm
asked Jan 4 at 12:03
Olof RubinOlof Rubin
1,131316
$endgroup$
add a comment |
$begingroup$
I'm considering the norm defined on matrices by
$$|A|_F = sqrt{sum_{i,j}|a_{ij}|^2}$$
I want to show that it is unitarily invariant, so that for unitary $U$ we have that
$$|UA|_F = |A|_F = |AU|_F$$
however I have trouble doing it directly. Writing $|UA|_F$ directly I find by Cauchy-Schwarz that
$$|UA|_F = sqrt{sum_{i,j}left|sum_{k=1}^{n}u_{ik}a_{kj}right|^2}= sqrt{sum_{i,j}|langle U_i,overline{A_j}rangle|^2}leq sqrt{sum_{i,j}|A_j|^2}$$
where $U_i$ denotes the $i$th row of $U$ and $A_j$ the $j$th column of $A$. However this estimate is to crude and will not equal $|A|_F$. I would like to prove this without refering to trace or singular values and would appreciate a hint, rather than a full solution, on how to tackle this problem.
EDIT: Completion of the proof based on the answer from $A.Gamma$:
Since the rows of $U$ constitute an orthonormal basis for $mathbb{C}^n$ we find by Parsevals theorem that
$$|UA_j|_2^2 = sum_{i=1}^{n}left|sum_{k=1}^{n}u_{ik}a_{kj}right|^2 = sum_{i=1}^{n}|langle U_i,overline{A_j}rangle|^2 = |overline{A_j}|_2^2 = |A_j|_2^2$$
linear-algebra matrices norm
asked Jan 4 at 12:03
Olof RubinOlof Rubin
1,131316
$endgroup$
add a comment |
$begingroup$
I'm considering the norm defined on matrices by
$$|A|_F = sqrt{sum_{i,j}|a_{ij}|^2}$$
I want to show that it is unitarily invariant, so that for unitary $U$ we have that
$$|UA|_F = |A|_F = |AU|_F$$
however I have trouble doing it directly. Writing $|UA|_F$ directly I find by Cauchy-Schwarz that
$$|UA|_F = sqrt{sum_{i,j}left|sum_{k=1}^{n}u_{ik}a_{kj}right|^2}= sqrt{sum_{i,j}|langle U_i,overline{A_j}rangle|^2}leq sqrt{sum_{i,j}|A_j|^2}$$
where $U_i$ denotes the $i$th row of $U$ and $A_j$ the $j$th column of $A$. However this estimate is to crude and will not equal $|A|_F$. I would like to prove this without refering to trace or singular values and would appreciate a hint, rather than a full solution, on how to tackle this problem.
EDIT: Completion of the proof based on the answer from $A.Gamma$:
Since the rows of $U$ constitute an orthonormal basis for $mathbb{C}^n$ we find by Parsevals theorem that
$$|UA_j|_2^2 = sum_{i=1}^{n}left|sum_{k=1}^{n}u_{ik}a_{kj}right|^2 = sum_{i=1}^{n}|langle U_i,overline{A_j}rangle|^2 = |overline{A_j}|_2^2 = |A_j|_2^2$$
linear-algebra matrices norm
asked Jan 4 at 12:03
Olof RubinOlof Rubin
1,131316
$endgroup$
I'm considering the norm defined on matrices by
$$|A|_F = sqrt{sum_{i,j}|a_{ij}|^2}$$
I want to show that it is unitarily invariant, so that for unitary $U$ we have that
$$|UA|_F = |A|_F = |AU|_F$$
however I have trouble doing it directly. Writing $|UA|_F$ directly I find by Cauchy-Schwarz that
$$|UA|_F = sqrt{sum_{i,j}left|sum_{k=1}^{n}u_{ik}a_{kj}right|^2}= sqrt{sum_{i,j}|langle U_i,overline{A_j}rangle|^2}leq sqrt{sum_{i,j}|A_j|^2}$$
where $U_i$ denotes the $i$th row of $U$ and $A_j$ the $j$th column of $A$. However this estimate is to crude and will not equal $|A|_F$. I would like to prove this without refering to trace or singular values and would appreciate a hint, rather than a full solution, on how to tackle this problem.
EDIT: Completion of the proof based on the answer from $A.Gamma$:
Since the rows of $U$ constitute an orthonormal basis for $mathbb{C}^n$ we find by Parsevals theorem that
$$|UA_j|_2^2 = sum_{i=1}^{n}left|sum_{k=1}^{n}u_{ik}a_{kj}right|^2 = sum_{i=1}^{n}|langle U_i,overline{A_j}rangle|^2 = |overline{A_j}|_2^2 = |A_j|_2^2$$
linear-algebra matrices norm
linear-algebra matrices norm
asked Jan 4 at 12:03
Olof RubinOlof Rubin
1,131316
asked Jan 4 at 12:03
Olof RubinOlof Rubin
1,131316
edited Jan 4 at 12:39
Olof Rubin
asked Jan 4 at 12:03
Olof RubinOlof Rubin
1,131316
asked Jan 4 at 12:03
Olof RubinOlof Rubin
1,131316
asked Jan 4 at 12:03
Olof RubinOlof Rubin
1,131316
1,131316
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since
$$
UA=[UA_1 UA_2 ldots UA_n]
$$
you need to prove that
$$
|UA|_F^2=sum_{j=1}^n|UA_j|_2^2stackrel{?}{=}sum_{j=1}^n|A_j|_2^2=|A|_F^2.
$$
It suffice to prove that $|UA_j|_2^2=|A_j|_2^2$.
P.S. For $AU$ use conjugation.
answered Jan 4 at 12:13
A.Γ.A.Γ.
22.7k32656
$endgroup$
$begingroup$
I see we can use that the rows of $U$ constitute an ONB for $mathbb{C}^n$, I completed my proof above using your recommendation. Thank you. I realise now that part of my confusion came from the fact that unitary matrices don't need to have norm $1$.
$endgroup$
– Olof Rubin
Jan 4 at 12:40
2
$begingroup$
@OlofRubin Alternatively, one can use that a unitary matrix is isometric, i.e. $|Ux|_2=|x|_2$. Proof: $$|Ux|_2^2=(Ux)^*Ux=x^*underbrace{U^*U}_{=I}x=x^*x=|x|_2^2.$$
$endgroup$
– A.Γ.
Jan 4 at 12:46
$begingroup$
Ah yes that is more direct.
$endgroup$
– Olof Rubin
Jan 4 at 13:26
add a comment |
$begingroup$
Quick and dirty:
$$|UA|_F^2 = operatorname{Tr}((UA)^*(UA)) = operatorname{Tr}(A^*U^*UA) = operatorname{Tr}(A^*A) = |A|_F^2$$
and then since $U^*$ is also unitary
$$|AU|_F^2 = |(U^*A^*)^*|_F^2 = |U^*A^*|_F^2 = |A^*|_F^2 = |A|_F^2$$
Alternative argument:
Note that $A^*A ge 0$ so there exists an orthonormal basis ${u_1, ldots, u_n}$ for $mathbb{C}^n$ such that $A^*A u_i = lambda_i u_i$ for some $lambda ge 0$.
We have
$$sum_{i=1}^n |Au_i|_2^2 = sum_{i=1}^n langle Au_i, Au_irangle = sum_{i=1}^n langle A^*Au_i, u_irangle = sum_{i=1}^n lambda_i =operatorname{Tr}(A^*A) = |A|_F^2$$
because the trace is the sum of eigenvalues.
The interesting part is that the sum $sum_{i=1}^n |Au_i|_2^2$ is actually independent of the choice of the orthonormal basis ${u_1, ldots, u_n}$. Indeed, if ${v_1, ldots, v_n}$ is some other orthonormal basis for $mathbb{C}^n$, we have
begin{align}
sum_{i=1}^n |Au_i|_2^2 &= sum_{i=1}^n langle A^*Au_i, u_irangle\
&= sum_{i=1}^n leftlangle sum_{j=1}^nlangle u_i,v_jrangle A^*A v_j , sum_{k=1}^nlangle u_i,v_krangle v_krightrangle\
&= sum_{j=1}^n sum_{k=1}^n left(sum_{i=1}^nlangle u_i,v_jrangle langle v_k,u_irangleright)langle A^*A v_j,v_krangle\
&= sum_{j=1}^n sum_{k=1}^n langle v_j,v_kranglelangle A^*A v_j,v_krangle\
&= sum_{j=1}^n langle A^*A v_j,v_jrangle\
&= sum_{j=1}^n |Av_j|_2^2
end{align}
Now, if $U$ is unitary, for any orthonormal basis ${u_1, ldots, u_n}$ we have that ${Uu_1, ldots, Uu_n}$ is also an orthonormal basis so:
$$|AU|_F^2 = sum_{i=1}^n |A(Ue_i)|^2 = |A|_F^2$$
answered Jan 4 at 12:20
mechanodroidmechanodroid
27.1k62446
$endgroup$
$begingroup$
That's true, but the OP doesn't want to refer to trace.
$endgroup$
– A.Γ.
Jan 4 at 12:26
$begingroup$
@A.Γ. True, I missed that. Your argument is great. I have added an alternative argument which may be helpful, but it still uses the trace.
$endgroup$
– mechanodroid
Jan 4 at 12:30
add a comment |
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2 Answers
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2 Answers
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active
oldest
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$begingroup$
Since
$$
UA=[UA_1 UA_2 ldots UA_n]
$$
you need to prove that
$$
|UA|_F^2=sum_{j=1}^n|UA_j|_2^2stackrel{?}{=}sum_{j=1}^n|A_j|_2^2=|A|_F^2.
$$
It suffice to prove that $|UA_j|_2^2=|A_j|_2^2$.
P.S. For $AU$ use conjugation.
answered Jan 4 at 12:13
A.Γ.A.Γ.
22.7k32656
$endgroup$
$begingroup$
I see we can use that the rows of $U$ constitute an ONB for $mathbb{C}^n$, I completed my proof above using your recommendation. Thank you. I realise now that part of my confusion came from the fact that unitary matrices don't need to have norm $1$.
$endgroup$
– Olof Rubin
Jan 4 at 12:40
2
$begingroup$
@OlofRubin Alternatively, one can use that a unitary matrix is isometric, i.e. $|Ux|_2=|x|_2$. Proof: $$|Ux|_2^2=(Ux)^*Ux=x^*underbrace{U^*U}_{=I}x=x^*x=|x|_2^2.$$
$endgroup$
– A.Γ.
Jan 4 at 12:46
$begingroup$
Ah yes that is more direct.
$endgroup$
– Olof Rubin
Jan 4 at 13:26
add a comment |
$begingroup$
Since
$$
UA=[UA_1 UA_2 ldots UA_n]
$$
you need to prove that
$$
|UA|_F^2=sum_{j=1}^n|UA_j|_2^2stackrel{?}{=}sum_{j=1}^n|A_j|_2^2=|A|_F^2.
$$
It suffice to prove that $|UA_j|_2^2=|A_j|_2^2$.
P.S. For $AU$ use conjugation.
answered Jan 4 at 12:13
A.Γ.A.Γ.
22.7k32656
$endgroup$
$begingroup$
I see we can use that the rows of $U$ constitute an ONB for $mathbb{C}^n$, I completed my proof above using your recommendation. Thank you. I realise now that part of my confusion came from the fact that unitary matrices don't need to have norm $1$.
$endgroup$
– Olof Rubin
Jan 4 at 12:40
2
$begingroup$
@OlofRubin Alternatively, one can use that a unitary matrix is isometric, i.e. $|Ux|_2=|x|_2$. Proof: $$|Ux|_2^2=(Ux)^*Ux=x^*underbrace{U^*U}_{=I}x=x^*x=|x|_2^2.$$
$endgroup$
– A.Γ.
Jan 4 at 12:46
$begingroup$
Ah yes that is more direct.
$endgroup$
– Olof Rubin
Jan 4 at 13:26
add a comment |
$begingroup$
Since
$$
UA=[UA_1 UA_2 ldots UA_n]
$$
you need to prove that
$$
|UA|_F^2=sum_{j=1}^n|UA_j|_2^2stackrel{?}{=}sum_{j=1}^n|A_j|_2^2=|A|_F^2.
$$
It suffice to prove that $|UA_j|_2^2=|A_j|_2^2$.
P.S. For $AU$ use conjugation.
answered Jan 4 at 12:13
A.Γ.A.Γ.
22.7k32656
$endgroup$
Since
$$
UA=[UA_1 UA_2 ldots UA_n]
$$
you need to prove that
$$
|UA|_F^2=sum_{j=1}^n|UA_j|_2^2stackrel{?}{=}sum_{j=1}^n|A_j|_2^2=|A|_F^2.
$$
It suffice to prove that $|UA_j|_2^2=|A_j|_2^2$.
P.S. For $AU$ use conjugation.
answered Jan 4 at 12:13
A.Γ.A.Γ.
22.7k32656
answered Jan 4 at 12:13
A.Γ.A.Γ.
22.7k32656
answered Jan 4 at 12:13
A.Γ.A.Γ.
22.7k32656
answered Jan 4 at 12:13
A.Γ.A.Γ.
22.7k32656
22.7k32656
$begingroup$
I see we can use that the rows of $U$ constitute an ONB for $mathbb{C}^n$, I completed my proof above using your recommendation. Thank you. I realise now that part of my confusion came from the fact that unitary matrices don't need to have norm $1$.
$endgroup$
– Olof Rubin
Jan 4 at 12:40
2
$begingroup$
@OlofRubin Alternatively, one can use that a unitary matrix is isometric, i.e. $|Ux|_2=|x|_2$. Proof: $$|Ux|_2^2=(Ux)^*Ux=x^*underbrace{U^*U}_{=I}x=x^*x=|x|_2^2.$$
$endgroup$
– A.Γ.
Jan 4 at 12:46
$begingroup$
Ah yes that is more direct.
$endgroup$
– Olof Rubin
Jan 4 at 13:26
add a comment |
$begingroup$
I see we can use that the rows of $U$ constitute an ONB for $mathbb{C}^n$, I completed my proof above using your recommendation. Thank you. I realise now that part of my confusion came from the fact that unitary matrices don't need to have norm $1$.
$endgroup$
– Olof Rubin
Jan 4 at 12:40
2
$begingroup$
@OlofRubin Alternatively, one can use that a unitary matrix is isometric, i.e. $|Ux|_2=|x|_2$. Proof: $$|Ux|_2^2=(Ux)^*Ux=x^*underbrace{U^*U}_{=I}x=x^*x=|x|_2^2.$$
$endgroup$
– A.Γ.
Jan 4 at 12:46
$begingroup$
Ah yes that is more direct.
$endgroup$
– Olof Rubin
Jan 4 at 13:26
$begingroup$
I see we can use that the rows of $U$ constitute an ONB for $mathbb{C}^n$, I completed my proof above using your recommendation. Thank you. I realise now that part of my confusion came from the fact that unitary matrices don't need to have norm $1$.
$endgroup$
– Olof Rubin
Jan 4 at 12:40
$begingroup$
I see we can use that the rows of $U$ constitute an ONB for $mathbb{C}^n$, I completed my proof above using your recommendation. Thank you. I realise now that part of my confusion came from the fact that unitary matrices don't need to have norm $1$.
$endgroup$
– Olof Rubin
Jan 4 at 12:40
2
2
$begingroup$
@OlofRubin Alternatively, one can use that a unitary matrix is isometric, i.e. $|Ux|_2=|x|_2$. Proof: $$|Ux|_2^2=(Ux)^*Ux=x^*underbrace{U^*U}_{=I}x=x^*x=|x|_2^2.$$
$endgroup$
– A.Γ.
Jan 4 at 12:46
$begingroup$
@OlofRubin Alternatively, one can use that a unitary matrix is isometric, i.e. $|Ux|_2=|x|_2$. Proof: $$|Ux|_2^2=(Ux)^*Ux=x^*underbrace{U^*U}_{=I}x=x^*x=|x|_2^2.$$
$endgroup$
– A.Γ.
Jan 4 at 12:46
$begingroup$
Ah yes that is more direct.
$endgroup$
– Olof Rubin
Jan 4 at 13:26
$begingroup$
Ah yes that is more direct.
$endgroup$
– Olof Rubin
Jan 4 at 13:26
add a comment |
$begingroup$
Quick and dirty:
$$|UA|_F^2 = operatorname{Tr}((UA)^*(UA)) = operatorname{Tr}(A^*U^*UA) = operatorname{Tr}(A^*A) = |A|_F^2$$
and then since $U^*$ is also unitary
$$|AU|_F^2 = |(U^*A^*)^*|_F^2 = |U^*A^*|_F^2 = |A^*|_F^2 = |A|_F^2$$
Alternative argument:
Note that $A^*A ge 0$ so there exists an orthonormal basis ${u_1, ldots, u_n}$ for $mathbb{C}^n$ such that $A^*A u_i = lambda_i u_i$ for some $lambda ge 0$.
We have
$$sum_{i=1}^n |Au_i|_2^2 = sum_{i=1}^n langle Au_i, Au_irangle = sum_{i=1}^n langle A^*Au_i, u_irangle = sum_{i=1}^n lambda_i =operatorname{Tr}(A^*A) = |A|_F^2$$
because the trace is the sum of eigenvalues.
The interesting part is that the sum $sum_{i=1}^n |Au_i|_2^2$ is actually independent of the choice of the orthonormal basis ${u_1, ldots, u_n}$. Indeed, if ${v_1, ldots, v_n}$ is some other orthonormal basis for $mathbb{C}^n$, we have
begin{align}
sum_{i=1}^n |Au_i|_2^2 &= sum_{i=1}^n langle A^*Au_i, u_irangle\
&= sum_{i=1}^n leftlangle sum_{j=1}^nlangle u_i,v_jrangle A^*A v_j , sum_{k=1}^nlangle u_i,v_krangle v_krightrangle\
&= sum_{j=1}^n sum_{k=1}^n left(sum_{i=1}^nlangle u_i,v_jrangle langle v_k,u_irangleright)langle A^*A v_j,v_krangle\
&= sum_{j=1}^n sum_{k=1}^n langle v_j,v_kranglelangle A^*A v_j,v_krangle\
&= sum_{j=1}^n langle A^*A v_j,v_jrangle\
&= sum_{j=1}^n |Av_j|_2^2
end{align}
Now, if $U$ is unitary, for any orthonormal basis ${u_1, ldots, u_n}$ we have that ${Uu_1, ldots, Uu_n}$ is also an orthonormal basis so:
$$|AU|_F^2 = sum_{i=1}^n |A(Ue_i)|^2 = |A|_F^2$$
answered Jan 4 at 12:20
mechanodroidmechanodroid
27.1k62446
$endgroup$
$begingroup$
That's true, but the OP doesn't want to refer to trace.
$endgroup$
– A.Γ.
Jan 4 at 12:26
$begingroup$
@A.Γ. True, I missed that. Your argument is great. I have added an alternative argument which may be helpful, but it still uses the trace.
$endgroup$
– mechanodroid
Jan 4 at 12:30
add a comment |
$begingroup$
Quick and dirty:
$$|UA|_F^2 = operatorname{Tr}((UA)^*(UA)) = operatorname{Tr}(A^*U^*UA) = operatorname{Tr}(A^*A) = |A|_F^2$$
and then since $U^*$ is also unitary
$$|AU|_F^2 = |(U^*A^*)^*|_F^2 = |U^*A^*|_F^2 = |A^*|_F^2 = |A|_F^2$$
Alternative argument:
Note that $A^*A ge 0$ so there exists an orthonormal basis ${u_1, ldots, u_n}$ for $mathbb{C}^n$ such that $A^*A u_i = lambda_i u_i$ for some $lambda ge 0$.
We have
$$sum_{i=1}^n |Au_i|_2^2 = sum_{i=1}^n langle Au_i, Au_irangle = sum_{i=1}^n langle A^*Au_i, u_irangle = sum_{i=1}^n lambda_i =operatorname{Tr}(A^*A) = |A|_F^2$$
because the trace is the sum of eigenvalues.
The interesting part is that the sum $sum_{i=1}^n |Au_i|_2^2$ is actually independent of the choice of the orthonormal basis ${u_1, ldots, u_n}$. Indeed, if ${v_1, ldots, v_n}$ is some other orthonormal basis for $mathbb{C}^n$, we have
begin{align}
sum_{i=1}^n |Au_i|_2^2 &= sum_{i=1}^n langle A^*Au_i, u_irangle\
&= sum_{i=1}^n leftlangle sum_{j=1}^nlangle u_i,v_jrangle A^*A v_j , sum_{k=1}^nlangle u_i,v_krangle v_krightrangle\
&= sum_{j=1}^n sum_{k=1}^n left(sum_{i=1}^nlangle u_i,v_jrangle langle v_k,u_irangleright)langle A^*A v_j,v_krangle\
&= sum_{j=1}^n sum_{k=1}^n langle v_j,v_kranglelangle A^*A v_j,v_krangle\
&= sum_{j=1}^n langle A^*A v_j,v_jrangle\
&= sum_{j=1}^n |Av_j|_2^2
end{align}
Now, if $U$ is unitary, for any orthonormal basis ${u_1, ldots, u_n}$ we have that ${Uu_1, ldots, Uu_n}$ is also an orthonormal basis so:
$$|AU|_F^2 = sum_{i=1}^n |A(Ue_i)|^2 = |A|_F^2$$
answered Jan 4 at 12:20
mechanodroidmechanodroid
27.1k62446
$endgroup$
$begingroup$
That's true, but the OP doesn't want to refer to trace.
$endgroup$
– A.Γ.
Jan 4 at 12:26
$begingroup$
@A.Γ. True, I missed that. Your argument is great. I have added an alternative argument which may be helpful, but it still uses the trace.
$endgroup$
– mechanodroid
Jan 4 at 12:30
add a comment |
$begingroup$
Quick and dirty:
$$|UA|_F^2 = operatorname{Tr}((UA)^*(UA)) = operatorname{Tr}(A^*U^*UA) = operatorname{Tr}(A^*A) = |A|_F^2$$
and then since $U^*$ is also unitary
$$|AU|_F^2 = |(U^*A^*)^*|_F^2 = |U^*A^*|_F^2 = |A^*|_F^2 = |A|_F^2$$
Alternative argument:
Note that $A^*A ge 0$ so there exists an orthonormal basis ${u_1, ldots, u_n}$ for $mathbb{C}^n$ such that $A^*A u_i = lambda_i u_i$ for some $lambda ge 0$.
We have
$$sum_{i=1}^n |Au_i|_2^2 = sum_{i=1}^n langle Au_i, Au_irangle = sum_{i=1}^n langle A^*Au_i, u_irangle = sum_{i=1}^n lambda_i =operatorname{Tr}(A^*A) = |A|_F^2$$
because the trace is the sum of eigenvalues.
The interesting part is that the sum $sum_{i=1}^n |Au_i|_2^2$ is actually independent of the choice of the orthonormal basis ${u_1, ldots, u_n}$. Indeed, if ${v_1, ldots, v_n}$ is some other orthonormal basis for $mathbb{C}^n$, we have
begin{align}
sum_{i=1}^n |Au_i|_2^2 &= sum_{i=1}^n langle A^*Au_i, u_irangle\
&= sum_{i=1}^n leftlangle sum_{j=1}^nlangle u_i,v_jrangle A^*A v_j , sum_{k=1}^nlangle u_i,v_krangle v_krightrangle\
&= sum_{j=1}^n sum_{k=1}^n left(sum_{i=1}^nlangle u_i,v_jrangle langle v_k,u_irangleright)langle A^*A v_j,v_krangle\
&= sum_{j=1}^n sum_{k=1}^n langle v_j,v_kranglelangle A^*A v_j,v_krangle\
&= sum_{j=1}^n langle A^*A v_j,v_jrangle\
&= sum_{j=1}^n |Av_j|_2^2
end{align}
Now, if $U$ is unitary, for any orthonormal basis ${u_1, ldots, u_n}$ we have that ${Uu_1, ldots, Uu_n}$ is also an orthonormal basis so:
$$|AU|_F^2 = sum_{i=1}^n |A(Ue_i)|^2 = |A|_F^2$$
answered Jan 4 at 12:20
mechanodroidmechanodroid
27.1k62446
$endgroup$
Quick and dirty:
$$|UA|_F^2 = operatorname{Tr}((UA)^*(UA)) = operatorname{Tr}(A^*U^*UA) = operatorname{Tr}(A^*A) = |A|_F^2$$
and then since $U^*$ is also unitary
$$|AU|_F^2 = |(U^*A^*)^*|_F^2 = |U^*A^*|_F^2 = |A^*|_F^2 = |A|_F^2$$
Alternative argument:
Note that $A^*A ge 0$ so there exists an orthonormal basis ${u_1, ldots, u_n}$ for $mathbb{C}^n$ such that $A^*A u_i = lambda_i u_i$ for some $lambda ge 0$.
We have
$$sum_{i=1}^n |Au_i|_2^2 = sum_{i=1}^n langle Au_i, Au_irangle = sum_{i=1}^n langle A^*Au_i, u_irangle = sum_{i=1}^n lambda_i =operatorname{Tr}(A^*A) = |A|_F^2$$
because the trace is the sum of eigenvalues.
The interesting part is that the sum $sum_{i=1}^n |Au_i|_2^2$ is actually independent of the choice of the orthonormal basis ${u_1, ldots, u_n}$. Indeed, if ${v_1, ldots, v_n}$ is some other orthonormal basis for $mathbb{C}^n$, we have
begin{align}
sum_{i=1}^n |Au_i|_2^2 &= sum_{i=1}^n langle A^*Au_i, u_irangle\
&= sum_{i=1}^n leftlangle sum_{j=1}^nlangle u_i,v_jrangle A^*A v_j , sum_{k=1}^nlangle u_i,v_krangle v_krightrangle\
&= sum_{j=1}^n sum_{k=1}^n left(sum_{i=1}^nlangle u_i,v_jrangle langle v_k,u_irangleright)langle A^*A v_j,v_krangle\
&= sum_{j=1}^n sum_{k=1}^n langle v_j,v_kranglelangle A^*A v_j,v_krangle\
&= sum_{j=1}^n langle A^*A v_j,v_jrangle\
&= sum_{j=1}^n |Av_j|_2^2
end{align}
Now, if $U$ is unitary, for any orthonormal basis ${u_1, ldots, u_n}$ we have that ${Uu_1, ldots, Uu_n}$ is also an orthonormal basis so:
$$|AU|_F^2 = sum_{i=1}^n |A(Ue_i)|^2 = |A|_F^2$$
answered Jan 4 at 12:20
mechanodroidmechanodroid
27.1k62446
edited Jan 4 at 12:28
answered Jan 4 at 12:20
mechanodroidmechanodroid
27.1k62446
answered Jan 4 at 12:20
mechanodroidmechanodroid
27.1k62446
answered Jan 4 at 12:20
mechanodroidmechanodroid
27.1k62446
27.1k62446
$begingroup$
That's true, but the OP doesn't want to refer to trace.
$endgroup$
– A.Γ.
Jan 4 at 12:26
$begingroup$
@A.Γ. True, I missed that. Your argument is great. I have added an alternative argument which may be helpful, but it still uses the trace.
$endgroup$
– mechanodroid
Jan 4 at 12:30
add a comment |
$begingroup$
That's true, but the OP doesn't want to refer to trace.
$endgroup$
– A.Γ.
Jan 4 at 12:26
$begingroup$
@A.Γ. True, I missed that. Your argument is great. I have added an alternative argument which may be helpful, but it still uses the trace.
$endgroup$
– mechanodroid
Jan 4 at 12:30
$begingroup$
That's true, but the OP doesn't want to refer to trace.
$endgroup$
– A.Γ.
Jan 4 at 12:26
$begingroup$
That's true, but the OP doesn't want to refer to trace.
$endgroup$
– A.Γ.
Jan 4 at 12:26
$begingroup$
@A.Γ. True, I missed that. Your argument is great. I have added an alternative argument which may be helpful, but it still uses the trace.
$endgroup$
– mechanodroid
Jan 4 at 12:30
$begingroup$
@A.Γ. True, I missed that. Your argument is great. I have added an alternative argument which may be helpful, but it still uses the trace.
$endgroup$
– mechanodroid
Jan 4 at 12:30
add a comment |
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