An inequality of matrix normIs there a condition for the following consequence?Orthogonal Inner Product...
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An inequality of matrix norm
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An inequality of matrix norm
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 $i in {{1,..,m}}$.Inner product of dual basisColumn Spaces and SubsetsProve matrix inequality in inner product spaceInduced 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
$$|T|:=sup{|Tv|_{W}:|v|_{V}=1}$$ where $|v_{V}|:=sqrt{langle v,vrangle}$ and $|Tv|_{W}:=sqrt{langle Tv,Tvrangle}$.
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 1 hour ago
Bernard
123k741116
asked 1 hour ago
bbwbbw
51739
$endgroup$
add a comment |
$begingroup$
Let $V,W$ be complex inner product spaces. Suppose $T: V to W$ is a linear map, then we define
$$|T|:=sup{|Tv|_{W}:|v|_{V}=1}$$ where $|v_{V}|:=sqrt{langle v,vrangle}$ and $|Tv|_{W}:=sqrt{langle Tv,Tvrangle}$.
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 1 hour ago
Bernard
123k741116
asked 1 hour ago
bbwbbw
51739
$endgroup$
add a comment |
$begingroup$
Let $V,W$ be complex inner product spaces. Suppose $T: V to W$ is a linear map, then we define
$$|T|:=sup{|Tv|_{W}:|v|_{V}=1}$$ where $|v_{V}|:=sqrt{langle v,vrangle}$ and $|Tv|_{W}:=sqrt{langle Tv,Tvrangle}$.
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 1 hour ago
Bernard
123k741116
asked 1 hour ago
bbwbbw
51739
$endgroup$
Let $V,W$ be complex inner product spaces. Suppose $T: V to W$ is a linear map, then we define
$$|T|:=sup{|Tv|_{W}:|v|_{V}=1}$$ where $|v_{V}|:=sqrt{langle v,vrangle}$ and $|Tv|_{W}:=sqrt{langle Tv,Tvrangle}$.
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 1 hour ago
Bernard
123k741116
asked 1 hour ago
bbwbbw
51739
edited 1 hour ago
Bernard
123k741116
asked 1 hour ago
bbwbbw
51739
edited 1 hour ago
Bernard
123k741116
edited 1 hour ago
Bernard
123k741116
edited 1 hour ago
Bernard
123k741116
123k741116
asked 1 hour ago
bbwbbw
51739
asked 1 hour ago
bbwbbw
51739
asked 1 hour ago
bbwbbw
51739
51739
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 $k = 2$:
$$
begin{array}{ll}
& ||U_1 U_2 - V_1 V_2||\ \
= & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
= &
|| ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
end{array}
$$
(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 1 hour ago
avsavs
3,424513
$endgroup$
$begingroup$
Thank you so much!
$endgroup$
– bbw
1 hour ago
1
$begingroup$
You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
$endgroup$
– avs
52 mins ago
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
$$
begin{array}{ll}
& ||U_1 U_2 - V_1 V_2||\ \
= & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
= &
|| ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
end{array}
$$
(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 1 hour ago
avsavs
3,424513
$endgroup$
$begingroup$
Thank you so much!
$endgroup$
– bbw
1 hour ago
1
$begingroup$
You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
$endgroup$
– avs
52 mins ago
add a comment |
$begingroup$
For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
$$
begin{array}{ll}
& ||U_1 U_2 - V_1 V_2||\ \
= & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
= &
|| ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
end{array}
$$
(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 1 hour ago
avsavs
3,424513
$endgroup$
$begingroup$
Thank you so much!
$endgroup$
– bbw
1 hour ago
1
$begingroup$
You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
$endgroup$
– avs
52 mins ago
add a comment |
$begingroup$
For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
$$
begin{array}{ll}
& ||U_1 U_2 - V_1 V_2||\ \
= & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
= &
|| ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
end{array}
$$
(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 1 hour ago
avsavs
3,424513
$endgroup$
For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
$$
begin{array}{ll}
& ||U_1 U_2 - V_1 V_2||\ \
= & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
= &
|| ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
end{array}
$$
(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 1 hour ago
avsavs
3,424513
edited 52 mins ago
answered 1 hour ago
avsavs
3,424513
answered 1 hour ago
avsavs
3,424513
answered 1 hour ago
avsavs
3,424513
3,424513
$begingroup$
Thank you so much!
$endgroup$
– bbw
1 hour ago
1
$begingroup$
You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
$endgroup$
– avs
52 mins ago
add a comment |
$begingroup$
Thank you so much!
$endgroup$
– bbw
1 hour ago
1
$begingroup$
You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
$endgroup$
– avs
52 mins ago
$begingroup$
Thank you so much!
$endgroup$
– bbw
1 hour ago
$begingroup$
Thank you so much!
$endgroup$
– bbw
1 hour ago
1
1
$begingroup$
You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
$endgroup$
– avs
52 mins ago
$begingroup$
You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
$endgroup$
– avs
52 mins ago
add a comment |
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