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An inequality of matrix norm

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3

1

$begingroup$

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

share|cite|improve this question

edited 1 hour ago

Bernard

123k741116

asked 1 hour ago

bbwbbw

51739

$endgroup$

add a comment | 

3

1

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

edited 1 hour ago

Bernard

123k741116

asked 1 hour ago

bbwbbw

51739

$endgroup$

share|cite|improve this question

edited 1 hour ago

Bernard

123k741116

asked 1 hour ago

bbwbbw

51739

share|cite|improve this question

edited 1 hour ago

Bernard

123k741116

asked 1 hour ago

bbwbbw

51739

edited 1 hour ago

Bernard

123k741116

edited 1 hour ago

Bernard

123k741116

asked 1 hour ago

bbwbbw

51739

asked 1 hour ago

bbwbbw

51739

1 Answer
1

active

oldest

votes

5

$begingroup$

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".:)

share|cite|improve this answer

edited 2 mins ago

answered 56 mins ago

avsavs

3,414513

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much!
  $endgroup$
  – bbw
  31 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
  $endgroup$
  – avs
  2 mins ago
  

  
  
  
  

add a comment | 

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5

$begingroup$

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".:)

share|cite|improve this answer

edited 2 mins ago

answered 56 mins ago

avsavs

3,414513

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much!
  $endgroup$
  – bbw
  31 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
  $endgroup$
  – avs
  2 mins ago
  

  
  
  
  

add a comment | 

5

$begingroup$

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".:)

share|cite|improve this answer

edited 2 mins ago

answered 56 mins ago

avsavs

3,414513

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much!
  $endgroup$
  – bbw
  31 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
  $endgroup$
  – avs
  2 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".:)

share|cite|improve this answer

edited 2 mins ago

answered 56 mins ago

avsavs

3,414513

$endgroup$

share|cite|improve this answer

edited 2 mins ago

answered 56 mins ago

avsavs

3,414513

answered 56 mins ago

avsavs

3,414513

answered 56 mins ago

avsavs

3,414513