Mixing
Suppose $A,B$ are positive operators with $AB=0$,what is the norm of $A+B$?
$begingroup$
Suppose $A,B$ are positive operators with $AB=0$,what is the norm of $A+B$?
operator-theory operator-algebras c-star-algebras
edited Feb 3 at 16:15
Asaf Karagila♦
308k33441775
asked Feb 3 at 11:52
mathrookiemathrookie
936512
$endgroup$
add a comment |
$begingroup$
Suppose $A,B$ are positive operators with $AB=0$,what is the norm of $A+B$?
operator-theory operator-algebras c-star-algebras
edited Feb 3 at 16:15
Asaf Karagila♦
308k33441775
asked Feb 3 at 11:52
mathrookiemathrookie
936512
$endgroup$
add a comment |
$begingroup$
Suppose $A,B$ are positive operators with $AB=0$,what is the norm of $A+B$?
operator-theory operator-algebras c-star-algebras
edited Feb 3 at 16:15
Asaf Karagila♦
308k33441775
asked Feb 3 at 11:52
mathrookiemathrookie
936512
$endgroup$
Suppose $A,B$ are positive operators with $AB=0$,what is the norm of $A+B$?
operator-theory operator-algebras c-star-algebras
operator-theory operator-algebras c-star-algebras
edited Feb 3 at 16:15
Asaf Karagila♦
308k33441775
asked Feb 3 at 11:52
mathrookiemathrookie
936512
edited Feb 3 at 16:15
Asaf Karagila♦
308k33441775
asked Feb 3 at 11:52
mathrookiemathrookie
936512
edited Feb 3 at 16:15
Asaf Karagila♦
308k33441775
edited Feb 3 at 16:15
Asaf Karagila♦
308k33441775
edited Feb 3 at 16:15
Asaf Karagila♦
308k33441775
308k33441775
asked Feb 3 at 11:52
mathrookiemathrookie
936512
asked Feb 3 at 11:52
mathrookiemathrookie
936512
asked Feb 3 at 11:52
mathrookiemathrookie
936512
936512
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Sorry that I misread the OP's question as confining to the finite dimensional case. Here's finite dimensional approach and general approach.
(Finite dimension approach) Observe that $O=(AB)^*=B^*A^*=BA$, hence $AB=BA =O$. Since $A$ and $B$ commute, they are simultaneously diagonalizable, i.e. there exists a unitary matrix $P$ such that
$$
A = PDP^*,quad B=PEP^*
$$ where $D,E$ are positive diagonal matrix. If we write $D=text{diag}(d_1,d_2,ldots,d_n)$ and $E=text{diag}(e_1,e_2,ldots,e_n)$, then we have $d_ie_i =0$ for every $i$ since $DE=P^*ABP=O$. Also, it holds that $|A| =max_i d_i$ and $|B|=max_i e_i$. This gives $$max_i (d_i+e_i) = max{max_i d_i,max_i e_i} = max {|A|,|B|},$$ hence $$|A+B|=max_i (d_i+e_i) =max{|A|,|B|}.$$
General approach: Observe that for every self-adjoint operator $A$, $|A^2|=|A^*A|=|A|^2$ holds. Inductively, we have $|A^{2^k}|=|A|^{2^k}$ for every $kge 1$. Now, assume without loss of generality that $|A|ge |B|$, then we have
$$
|(A+B)|^{2^k}=|(A+B)^{2^k}|=|A^{2^k}+B^{2^k}|le |A^{2^k}|+|B^{2^k}|le2|A|^{2^k}.
$$ Thus
$$
|A+B|le 2^{1/2^k}|A|xrightarrow{ktoinfty} |A|.
$$ Since $|A|le |A+B|$ is obvious from $displaystyle |Ax|le |(A+B)x| $, we get $$|A+B|=|A|=max{|A|,|B|}$$ as desired.
answered Feb 3 at 12:42
SongSong
18.6k21651
$endgroup$
$begingroup$
I don't understand the statement "since A,B commute,A,B are diagonalizable",why can we have this conclusion?
$endgroup$
– mathrookie
Feb 3 at 14:05
$begingroup$
@mathrookie You can check this eariler thread and other linked posts.
$endgroup$
– Song
Feb 3 at 14:07
$begingroup$
@mathrookie I've added more general argument.
$endgroup$
– Song
Feb 3 at 14:52
$begingroup$
The same proof as in the finite-dimensional case also works in the general case with the appropriate notion of diagonalization (i.e. spectral theorem).
$endgroup$
– MaoWao
Feb 3 at 20:59
add a comment |
$begingroup$
Here is another argument. Since we also have $BA=0$ (from $(AB)^*=0$), we have $ker Asubsetker B$ and viceversa, and the respective range projections are pairwise orthogonal. So we may choose an orthonormal basis so that
$$
A=begin{bmatrix} A_0&0&0\ 0&0&0\ 0&0&0end{bmatrix} , B=begin{bmatrix} 0&0&0\ 0&B_0&0\ 0&0&0end{bmatrix}.
$$
Then
$$
A+B=begin{bmatrix} A_0&0&0\ 0&B_0&0\ 0&0&0end{bmatrix},
$$
and we clearly have
$$
|A+B|=max{|A_0|,|B_0|}=max{|A|,|B|}.
$$
answered Feb 4 at 1:15
Martin ArgeramiMartin Argerami
130k1184185
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Sorry that I misread the OP's question as confining to the finite dimensional case. Here's finite dimensional approach and general approach.
(Finite dimension approach) Observe that $O=(AB)^*=B^*A^*=BA$, hence $AB=BA =O$. Since $A$ and $B$ commute, they are simultaneously diagonalizable, i.e. there exists a unitary matrix $P$ such that
$$
A = PDP^*,quad B=PEP^*
$$ where $D,E$ are positive diagonal matrix. If we write $D=text{diag}(d_1,d_2,ldots,d_n)$ and $E=text{diag}(e_1,e_2,ldots,e_n)$, then we have $d_ie_i =0$ for every $i$ since $DE=P^*ABP=O$. Also, it holds that $|A| =max_i d_i$ and $|B|=max_i e_i$. This gives $$max_i (d_i+e_i) = max{max_i d_i,max_i e_i} = max {|A|,|B|},$$ hence $$|A+B|=max_i (d_i+e_i) =max{|A|,|B|}.$$
General approach: Observe that for every self-adjoint operator $A$, $|A^2|=|A^*A|=|A|^2$ holds. Inductively, we have $|A^{2^k}|=|A|^{2^k}$ for every $kge 1$. Now, assume without loss of generality that $|A|ge |B|$, then we have
$$
|(A+B)|^{2^k}=|(A+B)^{2^k}|=|A^{2^k}+B^{2^k}|le |A^{2^k}|+|B^{2^k}|le2|A|^{2^k}.
$$ Thus
$$
|A+B|le 2^{1/2^k}|A|xrightarrow{ktoinfty} |A|.
$$ Since $|A|le |A+B|$ is obvious from $displaystyle |Ax|le |(A+B)x| $, we get $$|A+B|=|A|=max{|A|,|B|}$$ as desired.
answered Feb 3 at 12:42
SongSong
18.6k21651
$endgroup$
$begingroup$
I don't understand the statement "since A,B commute,A,B are diagonalizable",why can we have this conclusion?
$endgroup$
– mathrookie
Feb 3 at 14:05
$begingroup$
@mathrookie You can check this eariler thread and other linked posts.
$endgroup$
– Song
Feb 3 at 14:07
$begingroup$
@mathrookie I've added more general argument.
$endgroup$
– Song
Feb 3 at 14:52
$begingroup$
The same proof as in the finite-dimensional case also works in the general case with the appropriate notion of diagonalization (i.e. spectral theorem).
$endgroup$
– MaoWao
Feb 3 at 20:59
add a comment |
$begingroup$
Sorry that I misread the OP's question as confining to the finite dimensional case. Here's finite dimensional approach and general approach.
(Finite dimension approach) Observe that $O=(AB)^*=B^*A^*=BA$, hence $AB=BA =O$. Since $A$ and $B$ commute, they are simultaneously diagonalizable, i.e. there exists a unitary matrix $P$ such that
$$
A = PDP^*,quad B=PEP^*
$$ where $D,E$ are positive diagonal matrix. If we write $D=text{diag}(d_1,d_2,ldots,d_n)$ and $E=text{diag}(e_1,e_2,ldots,e_n)$, then we have $d_ie_i =0$ for every $i$ since $DE=P^*ABP=O$. Also, it holds that $|A| =max_i d_i$ and $|B|=max_i e_i$. This gives $$max_i (d_i+e_i) = max{max_i d_i,max_i e_i} = max {|A|,|B|},$$ hence $$|A+B|=max_i (d_i+e_i) =max{|A|,|B|}.$$
General approach: Observe that for every self-adjoint operator $A$, $|A^2|=|A^*A|=|A|^2$ holds. Inductively, we have $|A^{2^k}|=|A|^{2^k}$ for every $kge 1$. Now, assume without loss of generality that $|A|ge |B|$, then we have
$$
|(A+B)|^{2^k}=|(A+B)^{2^k}|=|A^{2^k}+B^{2^k}|le |A^{2^k}|+|B^{2^k}|le2|A|^{2^k}.
$$ Thus
$$
|A+B|le 2^{1/2^k}|A|xrightarrow{ktoinfty} |A|.
$$ Since $|A|le |A+B|$ is obvious from $displaystyle |Ax|le |(A+B)x| $, we get $$|A+B|=|A|=max{|A|,|B|}$$ as desired.
answered Feb 3 at 12:42
SongSong
18.6k21651
$endgroup$
$begingroup$
I don't understand the statement "since A,B commute,A,B are diagonalizable",why can we have this conclusion?
$endgroup$
– mathrookie
Feb 3 at 14:05
$begingroup$
@mathrookie You can check this eariler thread and other linked posts.
$endgroup$
– Song
Feb 3 at 14:07
$begingroup$
@mathrookie I've added more general argument.
$endgroup$
– Song
Feb 3 at 14:52
$begingroup$
The same proof as in the finite-dimensional case also works in the general case with the appropriate notion of diagonalization (i.e. spectral theorem).
$endgroup$
– MaoWao
Feb 3 at 20:59
add a comment |
$begingroup$
Sorry that I misread the OP's question as confining to the finite dimensional case. Here's finite dimensional approach and general approach.
(Finite dimension approach) Observe that $O=(AB)^*=B^*A^*=BA$, hence $AB=BA =O$. Since $A$ and $B$ commute, they are simultaneously diagonalizable, i.e. there exists a unitary matrix $P$ such that
$$
A = PDP^*,quad B=PEP^*
$$ where $D,E$ are positive diagonal matrix. If we write $D=text{diag}(d_1,d_2,ldots,d_n)$ and $E=text{diag}(e_1,e_2,ldots,e_n)$, then we have $d_ie_i =0$ for every $i$ since $DE=P^*ABP=O$. Also, it holds that $|A| =max_i d_i$ and $|B|=max_i e_i$. This gives $$max_i (d_i+e_i) = max{max_i d_i,max_i e_i} = max {|A|,|B|},$$ hence $$|A+B|=max_i (d_i+e_i) =max{|A|,|B|}.$$
General approach: Observe that for every self-adjoint operator $A$, $|A^2|=|A^*A|=|A|^2$ holds. Inductively, we have $|A^{2^k}|=|A|^{2^k}$ for every $kge 1$. Now, assume without loss of generality that $|A|ge |B|$, then we have
$$
|(A+B)|^{2^k}=|(A+B)^{2^k}|=|A^{2^k}+B^{2^k}|le |A^{2^k}|+|B^{2^k}|le2|A|^{2^k}.
$$ Thus
$$
|A+B|le 2^{1/2^k}|A|xrightarrow{ktoinfty} |A|.
$$ Since $|A|le |A+B|$ is obvious from $displaystyle |Ax|le |(A+B)x| $, we get $$|A+B|=|A|=max{|A|,|B|}$$ as desired.
answered Feb 3 at 12:42
SongSong
18.6k21651
$endgroup$
Sorry that I misread the OP's question as confining to the finite dimensional case. Here's finite dimensional approach and general approach.
(Finite dimension approach) Observe that $O=(AB)^*=B^*A^*=BA$, hence $AB=BA =O$. Since $A$ and $B$ commute, they are simultaneously diagonalizable, i.e. there exists a unitary matrix $P$ such that
$$
A = PDP^*,quad B=PEP^*
$$ where $D,E$ are positive diagonal matrix. If we write $D=text{diag}(d_1,d_2,ldots,d_n)$ and $E=text{diag}(e_1,e_2,ldots,e_n)$, then we have $d_ie_i =0$ for every $i$ since $DE=P^*ABP=O$. Also, it holds that $|A| =max_i d_i$ and $|B|=max_i e_i$. This gives $$max_i (d_i+e_i) = max{max_i d_i,max_i e_i} = max {|A|,|B|},$$ hence $$|A+B|=max_i (d_i+e_i) =max{|A|,|B|}.$$
General approach: Observe that for every self-adjoint operator $A$, $|A^2|=|A^*A|=|A|^2$ holds. Inductively, we have $|A^{2^k}|=|A|^{2^k}$ for every $kge 1$. Now, assume without loss of generality that $|A|ge |B|$, then we have
$$
|(A+B)|^{2^k}=|(A+B)^{2^k}|=|A^{2^k}+B^{2^k}|le |A^{2^k}|+|B^{2^k}|le2|A|^{2^k}.
$$ Thus
$$
|A+B|le 2^{1/2^k}|A|xrightarrow{ktoinfty} |A|.
$$ Since $|A|le |A+B|$ is obvious from $displaystyle |Ax|le |(A+B)x| $, we get $$|A+B|=|A|=max{|A|,|B|}$$ as desired.
answered Feb 3 at 12:42
SongSong
18.6k21651
edited Feb 3 at 14:51
answered Feb 3 at 12:42
SongSong
18.6k21651
answered Feb 3 at 12:42
SongSong
18.6k21651
answered Feb 3 at 12:42
SongSong
18.6k21651
18.6k21651
$begingroup$
I don't understand the statement "since A,B commute,A,B are diagonalizable",why can we have this conclusion?
$endgroup$
– mathrookie
Feb 3 at 14:05
$begingroup$
@mathrookie You can check this eariler thread and other linked posts.
$endgroup$
– Song
Feb 3 at 14:07
$begingroup$
@mathrookie I've added more general argument.
$endgroup$
– Song
Feb 3 at 14:52
$begingroup$
The same proof as in the finite-dimensional case also works in the general case with the appropriate notion of diagonalization (i.e. spectral theorem).
$endgroup$
– MaoWao
Feb 3 at 20:59
add a comment |
$begingroup$
I don't understand the statement "since A,B commute,A,B are diagonalizable",why can we have this conclusion?
$endgroup$
– mathrookie
Feb 3 at 14:05
$begingroup$
@mathrookie You can check this eariler thread and other linked posts.
$endgroup$
– Song
Feb 3 at 14:07
$begingroup$
@mathrookie I've added more general argument.
$endgroup$
– Song
Feb 3 at 14:52
$begingroup$
The same proof as in the finite-dimensional case also works in the general case with the appropriate notion of diagonalization (i.e. spectral theorem).
$endgroup$
– MaoWao
Feb 3 at 20:59
$begingroup$
I don't understand the statement "since A,B commute,A,B are diagonalizable",why can we have this conclusion?
$endgroup$
– mathrookie
Feb 3 at 14:05
$begingroup$
I don't understand the statement "since A,B commute,A,B are diagonalizable",why can we have this conclusion?
$endgroup$
– mathrookie
Feb 3 at 14:05
$begingroup$
@mathrookie You can check this eariler thread and other linked posts.
$endgroup$
– Song
Feb 3 at 14:07
$begingroup$
@mathrookie You can check this eariler thread and other linked posts.
$endgroup$
– Song
Feb 3 at 14:07
$begingroup$
@mathrookie I've added more general argument.
$endgroup$
– Song
Feb 3 at 14:52
$begingroup$
@mathrookie I've added more general argument.
$endgroup$
– Song
Feb 3 at 14:52
$begingroup$
The same proof as in the finite-dimensional case also works in the general case with the appropriate notion of diagonalization (i.e. spectral theorem).
$endgroup$
– MaoWao
Feb 3 at 20:59
$begingroup$
The same proof as in the finite-dimensional case also works in the general case with the appropriate notion of diagonalization (i.e. spectral theorem).
$endgroup$
– MaoWao
Feb 3 at 20:59
add a comment |
$begingroup$
Here is another argument. Since we also have $BA=0$ (from $(AB)^*=0$), we have $ker Asubsetker B$ and viceversa, and the respective range projections are pairwise orthogonal. So we may choose an orthonormal basis so that
$$
A=begin{bmatrix} A_0&0&0\ 0&0&0\ 0&0&0end{bmatrix} , B=begin{bmatrix} 0&0&0\ 0&B_0&0\ 0&0&0end{bmatrix}.
$$
Then
$$
A+B=begin{bmatrix} A_0&0&0\ 0&B_0&0\ 0&0&0end{bmatrix},
$$
and we clearly have
$$
|A+B|=max{|A_0|,|B_0|}=max{|A|,|B|}.
$$
answered Feb 4 at 1:15
Martin ArgeramiMartin Argerami
130k1184185
$endgroup$
add a comment |
$begingroup$
Here is another argument. Since we also have $BA=0$ (from $(AB)^*=0$), we have $ker Asubsetker B$ and viceversa, and the respective range projections are pairwise orthogonal. So we may choose an orthonormal basis so that
$$
A=begin{bmatrix} A_0&0&0\ 0&0&0\ 0&0&0end{bmatrix} , B=begin{bmatrix} 0&0&0\ 0&B_0&0\ 0&0&0end{bmatrix}.
$$
Then
$$
A+B=begin{bmatrix} A_0&0&0\ 0&B_0&0\ 0&0&0end{bmatrix},
$$
and we clearly have
$$
|A+B|=max{|A_0|,|B_0|}=max{|A|,|B|}.
$$
answered Feb 4 at 1:15
Martin ArgeramiMartin Argerami
130k1184185
$endgroup$
add a comment |
$begingroup$
Here is another argument. Since we also have $BA=0$ (from $(AB)^*=0$), we have $ker Asubsetker B$ and viceversa, and the respective range projections are pairwise orthogonal. So we may choose an orthonormal basis so that
$$
A=begin{bmatrix} A_0&0&0\ 0&0&0\ 0&0&0end{bmatrix} , B=begin{bmatrix} 0&0&0\ 0&B_0&0\ 0&0&0end{bmatrix}.
$$
Then
$$
A+B=begin{bmatrix} A_0&0&0\ 0&B_0&0\ 0&0&0end{bmatrix},
$$
and we clearly have
$$
|A+B|=max{|A_0|,|B_0|}=max{|A|,|B|}.
$$
answered Feb 4 at 1:15
Martin ArgeramiMartin Argerami
130k1184185
$endgroup$
Here is another argument. Since we also have $BA=0$ (from $(AB)^*=0$), we have $ker Asubsetker B$ and viceversa, and the respective range projections are pairwise orthogonal. So we may choose an orthonormal basis so that
$$
A=begin{bmatrix} A_0&0&0\ 0&0&0\ 0&0&0end{bmatrix} , B=begin{bmatrix} 0&0&0\ 0&B_0&0\ 0&0&0end{bmatrix}.
$$
Then
$$
A+B=begin{bmatrix} A_0&0&0\ 0&B_0&0\ 0&0&0end{bmatrix},
$$
and we clearly have
$$
|A+B|=max{|A_0|,|B_0|}=max{|A|,|B|}.
$$
answered Feb 4 at 1:15
Martin ArgeramiMartin Argerami
130k1184185
answered Feb 4 at 1:15
Martin ArgeramiMartin Argerami
130k1184185
answered Feb 4 at 1:15
Martin ArgeramiMartin Argerami
130k1184185
answered Feb 4 at 1:15
Martin ArgeramiMartin Argerami
130k1184185
130k1184185
add a comment |
add a comment |
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