Finding the norm of a matrix given an orthonormal basis
$begingroup$
So I am working on practice problems out of a textbook I am working through over break. One of the problems asks to show that given a complex $ntimes n$ matrix $X$ and any orthonormal basis ${u_1,u_2,...,u_n}$ of $mathbb{C}^n$ that the Hilbert-Schmidt norm of the matrix given as
$$| X |=bigg(sum_{j,k=1}^n|X_{jk}|^2bigg)^{1/2}$$
can be expressed as
$$| X |^2=sum_{j,k=1}^n|langle u_j,Xu_krangle|^2$$
I figured that, if we consider the coordinate vectors w.r.t. our basis, we essentially have
$$langle u_j,Xu_krangle rightarrowlangle e_j,Xe_krangle$$
where it follows that $$langle e_j,Xe_krangle=X_{jk}$$
However, I am rusty on my linear algebra and I know this is not fully correct. I was thinking that, in order to use the coordinate vectors, I need to change $X$ to be w.r.t. our orthonormal basis but I keep confusing myself in the process.
linear-algebra norm inner-product-space
asked Jan 2 at 22:25
Ian B.Ian B.
898
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add a comment |
$begingroup$
So I am working on practice problems out of a textbook I am working through over break. One of the problems asks to show that given a complex $ntimes n$ matrix $X$ and any orthonormal basis ${u_1,u_2,...,u_n}$ of $mathbb{C}^n$ that the Hilbert-Schmidt norm of the matrix given as
$$| X |=bigg(sum_{j,k=1}^n|X_{jk}|^2bigg)^{1/2}$$
can be expressed as
$$| X |^2=sum_{j,k=1}^n|langle u_j,Xu_krangle|^2$$
I figured that, if we consider the coordinate vectors w.r.t. our basis, we essentially have
$$langle u_j,Xu_krangle rightarrowlangle e_j,Xe_krangle$$
where it follows that $$langle e_j,Xe_krangle=X_{jk}$$
However, I am rusty on my linear algebra and I know this is not fully correct. I was thinking that, in order to use the coordinate vectors, I need to change $X$ to be w.r.t. our orthonormal basis but I keep confusing myself in the process.
linear-algebra norm inner-product-space
asked Jan 2 at 22:25
Ian B.Ian B.
898
$endgroup$
add a comment |
$begingroup$
So I am working on practice problems out of a textbook I am working through over break. One of the problems asks to show that given a complex $ntimes n$ matrix $X$ and any orthonormal basis ${u_1,u_2,...,u_n}$ of $mathbb{C}^n$ that the Hilbert-Schmidt norm of the matrix given as
$$| X |=bigg(sum_{j,k=1}^n|X_{jk}|^2bigg)^{1/2}$$
can be expressed as
$$| X |^2=sum_{j,k=1}^n|langle u_j,Xu_krangle|^2$$
I figured that, if we consider the coordinate vectors w.r.t. our basis, we essentially have
$$langle u_j,Xu_krangle rightarrowlangle e_j,Xe_krangle$$
where it follows that $$langle e_j,Xe_krangle=X_{jk}$$
However, I am rusty on my linear algebra and I know this is not fully correct. I was thinking that, in order to use the coordinate vectors, I need to change $X$ to be w.r.t. our orthonormal basis but I keep confusing myself in the process.
linear-algebra norm inner-product-space
asked Jan 2 at 22:25
Ian B.Ian B.
898
$endgroup$
So I am working on practice problems out of a textbook I am working through over break. One of the problems asks to show that given a complex $ntimes n$ matrix $X$ and any orthonormal basis ${u_1,u_2,...,u_n}$ of $mathbb{C}^n$ that the Hilbert-Schmidt norm of the matrix given as
$$| X |=bigg(sum_{j,k=1}^n|X_{jk}|^2bigg)^{1/2}$$
can be expressed as
$$| X |^2=sum_{j,k=1}^n|langle u_j,Xu_krangle|^2$$
I figured that, if we consider the coordinate vectors w.r.t. our basis, we essentially have
$$langle u_j,Xu_krangle rightarrowlangle e_j,Xe_krangle$$
where it follows that $$langle e_j,Xe_krangle=X_{jk}$$
However, I am rusty on my linear algebra and I know this is not fully correct. I was thinking that, in order to use the coordinate vectors, I need to change $X$ to be w.r.t. our orthonormal basis but I keep confusing myself in the process.
linear-algebra norm inner-product-space
linear-algebra norm inner-product-space
asked Jan 2 at 22:25
Ian B.Ian B.
898
asked Jan 2 at 22:25
Ian B.Ian B.
898
asked Jan 2 at 22:25
Ian B.Ian B.
898
asked Jan 2 at 22:25
Ian B.Ian B.
898
asked Jan 2 at 22:25
Ian B.Ian B.
898
898
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
To complete you may use two things:
- To figure out that
$$
|X|^2=bigg(sum_{j,k=1}^n|X_{jk}|^2bigg)=operatorname{trace}X^*X
$$
($X^*$ is the Hermitian conjugate of $X$) - To know the cyclic trace property
$$
operatorname{trace}(AB)=operatorname{trace}(BA).
$$
Then everything comes nicely: for any unitary matrix $U$ (the columns are an orthonormal basis $u_1ldots,u_n$) we have $UU^*=I$, so
begin{align}
|X|^2&=operatorname{trace}X^*X=operatorname{trace}X^*XUU^*=\
&=operatorname{trace}U^*X^*XU=operatorname{trace}U^*X^*UU^*XU=|U^*XU|^2.
end{align}
Here any element of $U^*XU$ is
$$
u_j^*Xu_k=langle u_j,Xu_krangle.
$$
answered Jan 2 at 22:36
A.Γ.A.Γ.
22.7k32656
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To complete you may use two things:
- To figure out that
$$
|X|^2=bigg(sum_{j,k=1}^n|X_{jk}|^2bigg)=operatorname{trace}X^*X
$$
($X^*$ is the Hermitian conjugate of $X$) - To know the cyclic trace property
$$
operatorname{trace}(AB)=operatorname{trace}(BA).
$$
Then everything comes nicely: for any unitary matrix $U$ (the columns are an orthonormal basis $u_1ldots,u_n$) we have $UU^*=I$, so
begin{align}
|X|^2&=operatorname{trace}X^*X=operatorname{trace}X^*XUU^*=\
&=operatorname{trace}U^*X^*XU=operatorname{trace}U^*X^*UU^*XU=|U^*XU|^2.
end{align}
Here any element of $U^*XU$ is
$$
u_j^*Xu_k=langle u_j,Xu_krangle.
$$
answered Jan 2 at 22:36
A.Γ.A.Γ.
22.7k32656
$endgroup$
add a comment |
$begingroup$
To complete you may use two things:
- To figure out that
$$
|X|^2=bigg(sum_{j,k=1}^n|X_{jk}|^2bigg)=operatorname{trace}X^*X
$$
($X^*$ is the Hermitian conjugate of $X$) - To know the cyclic trace property
$$
operatorname{trace}(AB)=operatorname{trace}(BA).
$$
Then everything comes nicely: for any unitary matrix $U$ (the columns are an orthonormal basis $u_1ldots,u_n$) we have $UU^*=I$, so
begin{align}
|X|^2&=operatorname{trace}X^*X=operatorname{trace}X^*XUU^*=\
&=operatorname{trace}U^*X^*XU=operatorname{trace}U^*X^*UU^*XU=|U^*XU|^2.
end{align}
Here any element of $U^*XU$ is
$$
u_j^*Xu_k=langle u_j,Xu_krangle.
$$
answered Jan 2 at 22:36
A.Γ.A.Γ.
22.7k32656
$endgroup$
add a comment |
$begingroup$
To complete you may use two things:
- To figure out that
$$
|X|^2=bigg(sum_{j,k=1}^n|X_{jk}|^2bigg)=operatorname{trace}X^*X
$$
($X^*$ is the Hermitian conjugate of $X$) - To know the cyclic trace property
$$
operatorname{trace}(AB)=operatorname{trace}(BA).
$$
Then everything comes nicely: for any unitary matrix $U$ (the columns are an orthonormal basis $u_1ldots,u_n$) we have $UU^*=I$, so
begin{align}
|X|^2&=operatorname{trace}X^*X=operatorname{trace}X^*XUU^*=\
&=operatorname{trace}U^*X^*XU=operatorname{trace}U^*X^*UU^*XU=|U^*XU|^2.
end{align}
Here any element of $U^*XU$ is
$$
u_j^*Xu_k=langle u_j,Xu_krangle.
$$
answered Jan 2 at 22:36
A.Γ.A.Γ.
22.7k32656
$endgroup$
To complete you may use two things:
- To figure out that
$$
|X|^2=bigg(sum_{j,k=1}^n|X_{jk}|^2bigg)=operatorname{trace}X^*X
$$
($X^*$ is the Hermitian conjugate of $X$) - To know the cyclic trace property
$$
operatorname{trace}(AB)=operatorname{trace}(BA).
$$
Then everything comes nicely: for any unitary matrix $U$ (the columns are an orthonormal basis $u_1ldots,u_n$) we have $UU^*=I$, so
begin{align}
|X|^2&=operatorname{trace}X^*X=operatorname{trace}X^*XUU^*=\
&=operatorname{trace}U^*X^*XU=operatorname{trace}U^*X^*UU^*XU=|U^*XU|^2.
end{align}
Here any element of $U^*XU$ is
$$
u_j^*Xu_k=langle u_j,Xu_krangle.
$$
answered Jan 2 at 22:36
A.Γ.A.Γ.
22.7k32656
edited Jan 2 at 22:47
answered Jan 2 at 22:36
A.Γ.A.Γ.
22.7k32656
answered Jan 2 at 22:36
A.Γ.A.Γ.
22.7k32656
answered Jan 2 at 22:36
A.Γ.A.Γ.
22.7k32656
22.7k32656
add a comment |
add a comment |
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