prove/disprove if each two in $n$ operators can be diagonalizable simultaneously then all can be...
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I have an idea that for $n$ diagonalizable operators $A_1, A_2, ..., A_n in ell(V)$. if each $A_i, A_j$ can be diagonalizable simultaneously then all of them can be diagonalizable simultaneously.
if it is true then we can prove that for every permutation the answer of $A_{i_1}A_{i_2}...A_{i_n}$ is the same iff they are diagonalizable simultaneously.
can you prove or disprove that?
linear-algebra operator-theory diagonalization
asked Jan 14 at 6:33
Peyman mohseni kiasariPeyman mohseni kiasari
13711
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add a comment |
$begingroup$
I have an idea that for $n$ diagonalizable operators $A_1, A_2, ..., A_n in ell(V)$. if each $A_i, A_j$ can be diagonalizable simultaneously then all of them can be diagonalizable simultaneously.
if it is true then we can prove that for every permutation the answer of $A_{i_1}A_{i_2}...A_{i_n}$ is the same iff they are diagonalizable simultaneously.
can you prove or disprove that?
linear-algebra operator-theory diagonalization
asked Jan 14 at 6:33
Peyman mohseni kiasariPeyman mohseni kiasari
13711
$endgroup$
1
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If $A_i$ and $A_j$ are simultaneously diagonalisable, then they commute: $A_iA_j=A_jA_i$.
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– Lord Shark the Unknown
Jan 14 at 6:36
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@LordSharktheUnknown look at the "iff" part.
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– Peyman mohseni kiasari
Jan 14 at 6:37
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Are we talking about bounded operators on a Hilbert space? And by diagonalizable, do you mean unitarily equivalent to a multiplication operator?
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– MaoWao
Jan 15 at 12:40
add a comment |
$begingroup$
I have an idea that for $n$ diagonalizable operators $A_1, A_2, ..., A_n in ell(V)$. if each $A_i, A_j$ can be diagonalizable simultaneously then all of them can be diagonalizable simultaneously.
if it is true then we can prove that for every permutation the answer of $A_{i_1}A_{i_2}...A_{i_n}$ is the same iff they are diagonalizable simultaneously.
can you prove or disprove that?
linear-algebra operator-theory diagonalization
asked Jan 14 at 6:33
Peyman mohseni kiasariPeyman mohseni kiasari
13711
$endgroup$
I have an idea that for $n$ diagonalizable operators $A_1, A_2, ..., A_n in ell(V)$. if each $A_i, A_j$ can be diagonalizable simultaneously then all of them can be diagonalizable simultaneously.
if it is true then we can prove that for every permutation the answer of $A_{i_1}A_{i_2}...A_{i_n}$ is the same iff they are diagonalizable simultaneously.
can you prove or disprove that?
linear-algebra operator-theory diagonalization
linear-algebra operator-theory diagonalization
asked Jan 14 at 6:33
Peyman mohseni kiasariPeyman mohseni kiasari
13711
asked Jan 14 at 6:33
Peyman mohseni kiasariPeyman mohseni kiasari
13711
edited Jan 14 at 8:31
Peyman mohseni kiasari
asked Jan 14 at 6:33
Peyman mohseni kiasariPeyman mohseni kiasari
13711
asked Jan 14 at 6:33
Peyman mohseni kiasariPeyman mohseni kiasari
13711
asked Jan 14 at 6:33
Peyman mohseni kiasariPeyman mohseni kiasari
13711
13711
1
$begingroup$
If $A_i$ and $A_j$ are simultaneously diagonalisable, then they commute: $A_iA_j=A_jA_i$.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 6:36
$begingroup$
@LordSharktheUnknown look at the "iff" part.
$endgroup$
– Peyman mohseni kiasari
Jan 14 at 6:37
$begingroup$
Are we talking about bounded operators on a Hilbert space? And by diagonalizable, do you mean unitarily equivalent to a multiplication operator?
$endgroup$
– MaoWao
Jan 15 at 12:40
add a comment |
1
$begingroup$
If $A_i$ and $A_j$ are simultaneously diagonalisable, then they commute: $A_iA_j=A_jA_i$.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 6:36
$begingroup$
@LordSharktheUnknown look at the "iff" part.
$endgroup$
– Peyman mohseni kiasari
Jan 14 at 6:37
$begingroup$
Are we talking about bounded operators on a Hilbert space? And by diagonalizable, do you mean unitarily equivalent to a multiplication operator?
$endgroup$
– MaoWao
Jan 15 at 12:40
1
1
$begingroup$
If $A_i$ and $A_j$ are simultaneously diagonalisable, then they commute: $A_iA_j=A_jA_i$.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 6:36
$begingroup$
If $A_i$ and $A_j$ are simultaneously diagonalisable, then they commute: $A_iA_j=A_jA_i$.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 6:36
$begingroup$
@LordSharktheUnknown look at the "iff" part.
$endgroup$
– Peyman mohseni kiasari
Jan 14 at 6:37
$begingroup$
@LordSharktheUnknown look at the "iff" part.
$endgroup$
– Peyman mohseni kiasari
Jan 14 at 6:37
$begingroup$
Are we talking about bounded operators on a Hilbert space? And by diagonalizable, do you mean unitarily equivalent to a multiplication operator?
$endgroup$
– MaoWao
Jan 15 at 12:40
$begingroup$
Are we talking about bounded operators on a Hilbert space? And by diagonalizable, do you mean unitarily equivalent to a multiplication operator?
$endgroup$
– MaoWao
Jan 15 at 12:40
add a comment |
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$begingroup$
If $A_i$ and $A_j$ are simultaneously diagonalisable, then they commute: $A_iA_j=A_jA_i$.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 6:36
$begingroup$
@LordSharktheUnknown look at the "iff" part.
$endgroup$
– Peyman mohseni kiasari
Jan 14 at 6:37
$begingroup$
Are we talking about bounded operators on a Hilbert space? And by diagonalizable, do you mean unitarily equivalent to a multiplication operator?
$endgroup$
– MaoWao
Jan 15 at 12:40