Mixing
Show $||U_i-V_i||leqepsilon forall i=1,…,mLongrightarrow||U_m…U_1-V_m…V_1||leq mepsilon$ (spectral...
$begingroup$
Let ${U_i},{V_i}$ be sets of $m$ unitary operators with $||U_i-V_i||leqepsilon forall i=1,...,m$.
Then $||U_m...U_1-V_m...V_1||leq mepsilon$ with $||cdot||$ being the spectral norm.
Can someone help me show this conclusion? I'm going through some lecture notes and they omit the proof because it is "easy."
matrices norm spectral-norm
asked Jan 19 at 2:01
DanDan
447
$endgroup$
add a comment |
$begingroup$
Let ${U_i},{V_i}$ be sets of $m$ unitary operators with $||U_i-V_i||leqepsilon forall i=1,...,m$.
Then $||U_m...U_1-V_m...V_1||leq mepsilon$ with $||cdot||$ being the spectral norm.
Can someone help me show this conclusion? I'm going through some lecture notes and they omit the proof because it is "easy."
matrices norm spectral-norm
asked Jan 19 at 2:01
DanDan
447
$endgroup$
add a comment |
$begingroup$
Let ${U_i},{V_i}$ be sets of $m$ unitary operators with $||U_i-V_i||leqepsilon forall i=1,...,m$.
Then $||U_m...U_1-V_m...V_1||leq mepsilon$ with $||cdot||$ being the spectral norm.
Can someone help me show this conclusion? I'm going through some lecture notes and they omit the proof because it is "easy."
matrices norm spectral-norm
asked Jan 19 at 2:01
DanDan
447
$endgroup$
Let ${U_i},{V_i}$ be sets of $m$ unitary operators with $||U_i-V_i||leqepsilon forall i=1,...,m$.
Then $||U_m...U_1-V_m...V_1||leq mepsilon$ with $||cdot||$ being the spectral norm.
Can someone help me show this conclusion? I'm going through some lecture notes and they omit the proof because it is "easy."
matrices norm spectral-norm
matrices norm spectral-norm
asked Jan 19 at 2:01
DanDan
447
asked Jan 19 at 2:01
DanDan
447
asked Jan 19 at 2:01
DanDan
447
asked Jan 19 at 2:01
DanDan
447
asked Jan 19 at 2:01
DanDan
447
447
add a comment |
add a comment |
1 Answer
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$begingroup$
We have
$$begin{split}
||U_m...U_1-V_m...V_1|| &= left | sum_{i=1}^m U_m...U_iV_{i-1}...V_1-U_m...U_{i+1}V_i...V_1 right|\
&leq sum_{i=1}^m |U_m...U_iV_{i-1}...V_1-U_m...U_{i+1}V_i...V_1| \
&leq sum_{i=1}^m | U_m...U_{i+1}(U_i-V_i)V_{i-1}...V_1|\
&leq sum_{i=1}^m |U_m...U_{i+1}||U_i-V_i||V_{i-1}...V_1|\
end{split}$$
where the last inequality comes from the fact that with the spectral norm, $|AB|leq|A||B|$.
Also, any unitary transform $T$ has spectral norm $1$, so
$|U_m...U_{i+1}|=|V_{i-1}...V_1|=1$.
Finally, one can conclude that
$$||U_m...U_1-V_m...V_1||
leq sum_{i=1}^m | U_i-V_i|leq mvarepsilon
$$
answered Jan 19 at 3:25
Stefan LafonStefan Lafon
2,24019
$endgroup$
$begingroup$
Fantastic! I'd never seen that first step (the equality) done before!
$endgroup$
– Dan
Jan 19 at 7:27
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
We have
$$begin{split}
||U_m...U_1-V_m...V_1|| &= left | sum_{i=1}^m U_m...U_iV_{i-1}...V_1-U_m...U_{i+1}V_i...V_1 right|\
&leq sum_{i=1}^m |U_m...U_iV_{i-1}...V_1-U_m...U_{i+1}V_i...V_1| \
&leq sum_{i=1}^m | U_m...U_{i+1}(U_i-V_i)V_{i-1}...V_1|\
&leq sum_{i=1}^m |U_m...U_{i+1}||U_i-V_i||V_{i-1}...V_1|\
end{split}$$
where the last inequality comes from the fact that with the spectral norm, $|AB|leq|A||B|$.
Also, any unitary transform $T$ has spectral norm $1$, so
$|U_m...U_{i+1}|=|V_{i-1}...V_1|=1$.
Finally, one can conclude that
$$||U_m...U_1-V_m...V_1||
leq sum_{i=1}^m | U_i-V_i|leq mvarepsilon
$$
answered Jan 19 at 3:25
Stefan LafonStefan Lafon
2,24019
$endgroup$
$begingroup$
Fantastic! I'd never seen that first step (the equality) done before!
$endgroup$
– Dan
Jan 19 at 7:27
add a comment |
$begingroup$
We have
$$begin{split}
||U_m...U_1-V_m...V_1|| &= left | sum_{i=1}^m U_m...U_iV_{i-1}...V_1-U_m...U_{i+1}V_i...V_1 right|\
&leq sum_{i=1}^m |U_m...U_iV_{i-1}...V_1-U_m...U_{i+1}V_i...V_1| \
&leq sum_{i=1}^m | U_m...U_{i+1}(U_i-V_i)V_{i-1}...V_1|\
&leq sum_{i=1}^m |U_m...U_{i+1}||U_i-V_i||V_{i-1}...V_1|\
end{split}$$
where the last inequality comes from the fact that with the spectral norm, $|AB|leq|A||B|$.
Also, any unitary transform $T$ has spectral norm $1$, so
$|U_m...U_{i+1}|=|V_{i-1}...V_1|=1$.
Finally, one can conclude that
$$||U_m...U_1-V_m...V_1||
leq sum_{i=1}^m | U_i-V_i|leq mvarepsilon
$$
answered Jan 19 at 3:25
Stefan LafonStefan Lafon
2,24019
$endgroup$
$begingroup$
Fantastic! I'd never seen that first step (the equality) done before!
$endgroup$
– Dan
Jan 19 at 7:27
add a comment |
$begingroup$
We have
$$begin{split}
||U_m...U_1-V_m...V_1|| &= left | sum_{i=1}^m U_m...U_iV_{i-1}...V_1-U_m...U_{i+1}V_i...V_1 right|\
&leq sum_{i=1}^m |U_m...U_iV_{i-1}...V_1-U_m...U_{i+1}V_i...V_1| \
&leq sum_{i=1}^m | U_m...U_{i+1}(U_i-V_i)V_{i-1}...V_1|\
&leq sum_{i=1}^m |U_m...U_{i+1}||U_i-V_i||V_{i-1}...V_1|\
end{split}$$
where the last inequality comes from the fact that with the spectral norm, $|AB|leq|A||B|$.
Also, any unitary transform $T$ has spectral norm $1$, so
$|U_m...U_{i+1}|=|V_{i-1}...V_1|=1$.
Finally, one can conclude that
$$||U_m...U_1-V_m...V_1||
leq sum_{i=1}^m | U_i-V_i|leq mvarepsilon
$$
answered Jan 19 at 3:25
Stefan LafonStefan Lafon
2,24019
$endgroup$
We have
$$begin{split}
||U_m...U_1-V_m...V_1|| &= left | sum_{i=1}^m U_m...U_iV_{i-1}...V_1-U_m...U_{i+1}V_i...V_1 right|\
&leq sum_{i=1}^m |U_m...U_iV_{i-1}...V_1-U_m...U_{i+1}V_i...V_1| \
&leq sum_{i=1}^m | U_m...U_{i+1}(U_i-V_i)V_{i-1}...V_1|\
&leq sum_{i=1}^m |U_m...U_{i+1}||U_i-V_i||V_{i-1}...V_1|\
end{split}$$
where the last inequality comes from the fact that with the spectral norm, $|AB|leq|A||B|$.
Also, any unitary transform $T$ has spectral norm $1$, so
$|U_m...U_{i+1}|=|V_{i-1}...V_1|=1$.
Finally, one can conclude that
$$||U_m...U_1-V_m...V_1||
leq sum_{i=1}^m | U_i-V_i|leq mvarepsilon
$$
answered Jan 19 at 3:25
Stefan LafonStefan Lafon
2,24019
answered Jan 19 at 3:25
Stefan LafonStefan Lafon
2,24019
answered Jan 19 at 3:25
Stefan LafonStefan Lafon
2,24019
answered Jan 19 at 3:25
Stefan LafonStefan Lafon
2,24019
2,24019
$begingroup$
Fantastic! I'd never seen that first step (the equality) done before!
$endgroup$
– Dan
Jan 19 at 7:27
add a comment |
$begingroup$
Fantastic! I'd never seen that first step (the equality) done before!
$endgroup$
– Dan
Jan 19 at 7:27
$begingroup$
Fantastic! I'd never seen that first step (the equality) done before!
$endgroup$
– Dan
Jan 19 at 7:27
$begingroup$
Fantastic! I'd never seen that first step (the equality) done before!
$endgroup$
– Dan
Jan 19 at 7:27
add a comment |
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