Suppose $U_1,dots,U_k$ and $V_1,dots,V_k$ are $ntimes n$ unitary matrices. Show that $|U_1cdots U_k-V_1cdots V_k|leqsum_i=1^k|U_i-V_i|$Is there a condition for the following consequence?Orthogonal Inner Product Proofprove change of basis matrix is unitarya matrix metricMatrix of non-degenerate product invertible?Prove that there is a $uin V$, such that $<u,v_i>$ is greater than zero, for every $i in 1,..,m$.Find shortest vectors $u_1,v_1,cdots,u_N,v_N$ such that $langle u_i,v_jrangle=1$ if $ile j$ and $langle u_i,v_jrangle=0$ if $i>j$Inner product of dual basisColumn Spaces and SubsetsInduced inner product on tensor powers.
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Suppose $U_1,dots,U_k$ and $V_1,dots,V_k$ are $ntimes n$ unitary matrices. Show that $|U_1cdots U_k-V_1cdots V_k|leqsum_i=1^k|U_i-V_i|$
Is there a condition for the following consequence?Orthogonal Inner Product Proofprove change of basis matrix is unitarya matrix metricMatrix of non-degenerate product invertible?Prove that there is a $uin V$, such that $<u,v_i>$ is greater than zero, for every $i in 1,..,m$.Find shortest vectors $u_1,v_1,cdots,u_N,v_N$ such that $langle u_i,v_jrangle=1$ if $ile j$ and $langle u_i,v_jrangle=0$ if $i>j$Inner product of dual basisColumn Spaces and SubsetsInduced inner product on tensor powers.
$begingroup$
Let $V,W$ be complex inner product spaces. Suppose $T: V to W$ is a linear map, then we define
$$|T|:=supv$$ where $|v_V|:=sqrtlangle v,vrangle$ and $|Tv|_W:=sqrtlangle Tv,Tvrangle$.
Question: Suppose $U_1,ldots,U_k$ and $V_1,ldots,V_k$ are $n times n$ unitary matrices. Show that
$$|U_1cdots U_k-V_1cdots V_k| leq sum_i=1^k|U_i-V_i|$$
I have tried to use triangle inequality for norms and induction but failed. Can anyone give some hints? Thank you!
linear-algebra matrices functional-analysis norm
edited 2 hours ago
Asaf Karagila♦
306k33438769
asked 5 hours ago
bbwbbw
52239
$endgroup$
add a comment |
$begingroup$
Let $V,W$ be complex inner product spaces. Suppose $T: V to W$ is a linear map, then we define
$$|T|:=supv$$ where $|v_V|:=sqrtlangle v,vrangle$ and $|Tv|_W:=sqrtlangle Tv,Tvrangle$.
Question: Suppose $U_1,ldots,U_k$ and $V_1,ldots,V_k$ are $n times n$ unitary matrices. Show that
$$|U_1cdots U_k-V_1cdots V_k| leq sum_i=1^k|U_i-V_i|$$
I have tried to use triangle inequality for norms and induction but failed. Can anyone give some hints? Thank you!
linear-algebra matrices functional-analysis norm
edited 2 hours ago
Asaf Karagila♦
306k33438769
asked 5 hours ago
bbwbbw
52239
$endgroup$
add a comment |
$begingroup$
Let $V,W$ be complex inner product spaces. Suppose $T: V to W$ is a linear map, then we define
$$|T|:=supv$$ where $|v_V|:=sqrtlangle v,vrangle$ and $|Tv|_W:=sqrtlangle Tv,Tvrangle$.
Question: Suppose $U_1,ldots,U_k$ and $V_1,ldots,V_k$ are $n times n$ unitary matrices. Show that
$$|U_1cdots U_k-V_1cdots V_k| leq sum_i=1^k|U_i-V_i|$$
I have tried to use triangle inequality for norms and induction but failed. Can anyone give some hints? Thank you!
linear-algebra matrices functional-analysis norm
edited 2 hours ago
Asaf Karagila♦
306k33438769
asked 5 hours ago
bbwbbw
52239
$endgroup$
Let $V,W$ be complex inner product spaces. Suppose $T: V to W$ is a linear map, then we define
$$|T|:=supv$$ where $|v_V|:=sqrtlangle v,vrangle$ and $|Tv|_W:=sqrtlangle Tv,Tvrangle$.
Question: Suppose $U_1,ldots,U_k$ and $V_1,ldots,V_k$ are $n times n$ unitary matrices. Show that
$$|U_1cdots U_k-V_1cdots V_k| leq sum_i=1^k|U_i-V_i|$$
I have tried to use triangle inequality for norms and induction but failed. Can anyone give some hints? Thank you!
linear-algebra matrices functional-analysis norm
linear-algebra matrices functional-analysis norm
edited 2 hours ago
Asaf Karagila♦
306k33438769
asked 5 hours ago
bbwbbw
52239
edited 2 hours ago
Asaf Karagila♦
306k33438769
asked 5 hours ago
bbwbbw
52239
edited 2 hours ago
Asaf Karagila♦
306k33438769
edited 2 hours ago
Asaf Karagila♦
306k33438769
edited 2 hours ago
Asaf Karagila♦
306k33438769
306k33438769
asked 5 hours ago
bbwbbw
52239
asked 5 hours ago
bbwbbw
52239
asked 5 hours ago
bbwbbw
52239
52239
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
$$
beginarrayll
& ||U_1 U_2 - V_1 V_2||\ \
= & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
= &
|| ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
endarray
$$
(The last inequality is the triangle inequality.) Now, since $U_2$ is unitary, we have $||U_2|| = 1$, so
$$
|| ( U_1 - V_1) U_2 || leq || U_1 - V_1 ||.
$$
A similar bound obtains for $||V_1 (U_2 - V_2) ||$.
This should give you enough "building blocks".:)
answered 5 hours ago
avsavs
3,434513
$endgroup$
$begingroup$
Thank you so much!
$endgroup$
– bbw
4 hours ago
1
$begingroup$
You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
$endgroup$
– avs
4 hours ago
add a comment |
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1 Answer
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$begingroup$
For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
$$
beginarrayll
& ||U_1 U_2 - V_1 V_2||\ \
= & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
= &
|| ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
endarray
$$
(The last inequality is the triangle inequality.) Now, since $U_2$ is unitary, we have $||U_2|| = 1$, so
$$
|| ( U_1 - V_1) U_2 || leq || U_1 - V_1 ||.
$$
A similar bound obtains for $||V_1 (U_2 - V_2) ||$.
This should give you enough "building blocks".:)
answered 5 hours ago
avsavs
3,434513
$endgroup$
$begingroup$
Thank you so much!
$endgroup$
– bbw
4 hours ago
1
$begingroup$
You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
$endgroup$
– avs
4 hours ago
add a comment |
$begingroup$
For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
$$
beginarrayll
& ||U_1 U_2 - V_1 V_2||\ \
= & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
= &
|| ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
endarray
$$
(The last inequality is the triangle inequality.) Now, since $U_2$ is unitary, we have $||U_2|| = 1$, so
$$
|| ( U_1 - V_1) U_2 || leq || U_1 - V_1 ||.
$$
A similar bound obtains for $||V_1 (U_2 - V_2) ||$.
This should give you enough "building blocks".:)
answered 5 hours ago
avsavs
3,434513
$endgroup$
$begingroup$
Thank you so much!
$endgroup$
– bbw
4 hours ago
1
$begingroup$
You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
$endgroup$
– avs
4 hours ago
add a comment |
$begingroup$
For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
$$
beginarrayll
& ||U_1 U_2 - V_1 V_2||\ \
= & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
= &
|| ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
endarray
$$
(The last inequality is the triangle inequality.) Now, since $U_2$ is unitary, we have $||U_2|| = 1$, so
$$
|| ( U_1 - V_1) U_2 || leq || U_1 - V_1 ||.
$$
A similar bound obtains for $||V_1 (U_2 - V_2) ||$.
This should give you enough "building blocks".:)
answered 5 hours ago
avsavs
3,434513
$endgroup$
For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
$$
beginarrayll
& ||U_1 U_2 - V_1 V_2||\ \
= & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
= &
|| ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
endarray
$$
(The last inequality is the triangle inequality.) Now, since $U_2$ is unitary, we have $||U_2|| = 1$, so
$$
|| ( U_1 - V_1) U_2 || leq || U_1 - V_1 ||.
$$
A similar bound obtains for $||V_1 (U_2 - V_2) ||$.
This should give you enough "building blocks".:)
answered 5 hours ago
avsavs
3,434513
edited 4 hours ago
answered 5 hours ago
avsavs
3,434513
answered 5 hours ago
avsavs
3,434513
answered 5 hours ago
avsavs
3,434513
3,434513
$begingroup$
Thank you so much!
$endgroup$
– bbw
4 hours ago
1
$begingroup$
You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
$endgroup$
– avs
4 hours ago
add a comment |
$begingroup$
Thank you so much!
$endgroup$
– bbw
4 hours ago
1
$begingroup$
You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
$endgroup$
– avs
4 hours ago
$begingroup$
Thank you so much!
$endgroup$
– bbw
4 hours ago
$begingroup$
Thank you so much!
$endgroup$
– bbw
4 hours ago
1
1
$begingroup$
You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
$endgroup$
– avs
4 hours ago
$begingroup$
You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
$endgroup$
– avs
4 hours ago
add a comment |
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