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0

$begingroup$

Consider the set of $N times N$ matrices ${W_i}_{i=1}^{i=L}$, set of $N times 1$ vectors ${g_i}_{i=1}^{i=L}$ and ${h_i}_{i=1}^{i=L}$. Now consider the following sum
begin{align}
S=sum_{i}sum_{j}g_i^HW_ih_ih_j^{H}W_j^{H}g_j
end{align}
where the summation runs through all $L$ for all $i,j$. Clearly, this equation is quadratic in the matrix variables ${W_i}_{i=1}^{i=L}$. Now define the column vector
begin{align}
w=begin{bmatrix} operatorname{vec}(W_1) \ operatorname{vec}(W_2) \ vdots \ operatorname{vec}(W_L) end{bmatrix}
end{align}
where for a matrix $A$, $operatorname{vec}(A)$ denotes the column vector containing the columns of $A$ starting from the first column. The question is, can we write
begin{align}
S=w^{H}Qw
end{align}
where $Q$ is a matrix which is a function of ${g_i}_{i=1}^{i=L}$ and ${h_i}_{i=1}^{i=L}$. If so, what is the structure of $Q$?

linear-algebra matrices convex-optimization quadratic-forms block-matrices

share|cite|improve this question

edited 2 days ago

Rócherz

2,8912821

asked Apr 1 '13 at 17:30

dineshdileepdineshdileep

5,98611935

$endgroup$

add a comment | 

0

$begingroup$

Consider the set of $N times N$ matrices ${W_i}_{i=1}^{i=L}$, set of $N times 1$ vectors ${g_i}_{i=1}^{i=L}$ and ${h_i}_{i=1}^{i=L}$. Now consider the following sum
begin{align}
S=sum_{i}sum_{j}g_i^HW_ih_ih_j^{H}W_j^{H}g_j
end{align}
where the summation runs through all $L$ for all $i,j$. Clearly, this equation is quadratic in the matrix variables ${W_i}_{i=1}^{i=L}$. Now define the column vector
begin{align}
w=begin{bmatrix} operatorname{vec}(W_1) \ operatorname{vec}(W_2) \ vdots \ operatorname{vec}(W_L) end{bmatrix}
end{align}
where for a matrix $A$, $operatorname{vec}(A)$ denotes the column vector containing the columns of $A$ starting from the first column. The question is, can we write
begin{align}
S=w^{H}Qw
end{align}
where $Q$ is a matrix which is a function of ${g_i}_{i=1}^{i=L}$ and ${h_i}_{i=1}^{i=L}$. If so, what is the structure of $Q$?

linear-algebra matrices convex-optimization quadratic-forms block-matrices

share|cite|improve this question

edited 2 days ago

Rócherz

2,8912821

asked Apr 1 '13 at 17:30

dineshdileepdineshdileep

5,98611935

$endgroup$

add a comment | 

$begingroup$

Consider the set of $N times N$ matrices ${W_i}_{i=1}^{i=L}$, set of $N times 1$ vectors ${g_i}_{i=1}^{i=L}$ and ${h_i}_{i=1}^{i=L}$. Now consider the following sum
begin{align}
S=sum_{i}sum_{j}g_i^HW_ih_ih_j^{H}W_j^{H}g_j
end{align}
where the summation runs through all $L$ for all $i,j$. Clearly, this equation is quadratic in the matrix variables ${W_i}_{i=1}^{i=L}$. Now define the column vector
begin{align}
w=begin{bmatrix} operatorname{vec}(W_1) \ operatorname{vec}(W_2) \ vdots \ operatorname{vec}(W_L) end{bmatrix}
end{align}
where for a matrix $A$, $operatorname{vec}(A)$ denotes the column vector containing the columns of $A$ starting from the first column. The question is, can we write
begin{align}
S=w^{H}Qw
end{align}
where $Q$ is a matrix which is a function of ${g_i}_{i=1}^{i=L}$ and ${h_i}_{i=1}^{i=L}$. If so, what is the structure of $Q$?

linear-algebra matrices convex-optimization quadratic-forms block-matrices

share|cite|improve this question

edited 2 days ago

Rócherz

2,8912821

asked Apr 1 '13 at 17:30

dineshdileepdineshdileep

5,98611935

$endgroup$

share|cite|improve this question

edited 2 days ago

Rócherz

2,8912821

asked Apr 1 '13 at 17:30

dineshdileepdineshdileep

5,98611935

share|cite|improve this question

edited 2 days ago

Rócherz

2,8912821

asked Apr 1 '13 at 17:30

dineshdileepdineshdileep

5,98611935

edited 2 days ago

Rócherz

2,8912821

edited 2 days ago

Rócherz

2,8912821

asked Apr 1 '13 at 17:30

dineshdileepdineshdileep

5,98611935

asked Apr 1 '13 at 17:30

dineshdileepdineshdileep

5,98611935

1 Answer
1

active

oldest

votes

2

$begingroup$

begin{align}
S&=sum_{i}sum_{j}g_i^HW_ih_ih_j^HW_j^Hg_j\
&=sum_{i}sum_{j}operatorname{trace}(g_jg_i^HW_ih_ih_j^HW_j^H)\
&=sum_{i}sum_{j}operatorname{vec}(W_j)^H operatorname{vec}(g_jg_i^HW_ih_ih_j^H)\
&qquadqquadqquad{because operatorname{trace}(AB^{H})=operatorname{trace}(A^{H}B)=operatorname{vec}(A)^{H}operatorname{vec}(B)}\
& \
&=sum_{i}sum_{j}operatorname{vec}(W_j)^H left((h_ih_j^H)^Totimes(g_jg_i^H)right)operatorname{vec}(W_i)\
&qquadqquadqquad{because operatorname{vec}(ABC)=(C^{T}otimes A)operatorname{vec}(B)}\
& \
&=sum_{i}sum_{j}operatorname{vec}(W_j)^H left((bar{h}_jh_i^T)otimes(g_jg_i^H)right)operatorname{vec}(W_i)\
&=w^H Qw,
end{align}
where $Q$ is a block matrix whose $(j,i)$-th (note: not $(i,j)$-th) subblock is $(bar{h}_jh_i^T)otimes(g_jg_i^H)$.

share|cite|improve this answer

edited 2 days ago

Rócherz

2,8912821

answered Apr 1 '13 at 18:13

user1551user1551

73.5k566129

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you are always reliable!!!
  $endgroup$
  – dineshdileep
  Apr 2 '13 at 8:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @dineshdileep What? Who knows if some mistakes are lurking around in some shadowy places. You should double-check the claimed result.
  $endgroup$
  – user1551
  Apr 2 '13 at 16:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I will definitely do that.
  $endgroup$
  – dineshdileep
  Apr 3 '13 at 4:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you explain (mathematically) the second last step, where you switch to the block matrix. (I almost have it intuitively, but would like to know a rigorous one).
  $endgroup$
  – dineshdileep
  Apr 3 '13 at 4:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Let $w_j=operatorname{vec}(W_j)$. Then $w^T=(w_1^T,ldots,w_n^T)$. Partition $Q$ into subblocks, each of size $ntimes n$. Call the $(j,i)$-th block $Q_{ji}$. Then by $w^HQw=sum_isum_j w_j^H Q_{ji} w_i$. So, by comparing coefficients, we should set $Q_{ji}=(bar{h}_jh_i^T)otimes(g_jg_i^H)$.
  $endgroup$
  – user1551
  Apr 3 '13 at 20:51
  

  
  
  
  

 | 
show 1 more comment

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2

$begingroup$

begin{align}
S&=sum_{i}sum_{j}g_i^HW_ih_ih_j^HW_j^Hg_j\
&=sum_{i}sum_{j}operatorname{trace}(g_jg_i^HW_ih_ih_j^HW_j^H)\
&=sum_{i}sum_{j}operatorname{vec}(W_j)^H operatorname{vec}(g_jg_i^HW_ih_ih_j^H)\
&qquadqquadqquad{because operatorname{trace}(AB^{H})=operatorname{trace}(A^{H}B)=operatorname{vec}(A)^{H}operatorname{vec}(B)}\
& \
&=sum_{i}sum_{j}operatorname{vec}(W_j)^H left((h_ih_j^H)^Totimes(g_jg_i^H)right)operatorname{vec}(W_i)\
&qquadqquadqquad{because operatorname{vec}(ABC)=(C^{T}otimes A)operatorname{vec}(B)}\
& \
&=sum_{i}sum_{j}operatorname{vec}(W_j)^H left((bar{h}_jh_i^T)otimes(g_jg_i^H)right)operatorname{vec}(W_i)\
&=w^H Qw,
end{align}
where $Q$ is a block matrix whose $(j,i)$-th (note: not $(i,j)$-th) subblock is $(bar{h}_jh_i^T)otimes(g_jg_i^H)$.

share|cite|improve this answer

edited 2 days ago

Rócherz

2,8912821

answered Apr 1 '13 at 18:13

user1551user1551

73.5k566129

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you are always reliable!!!
  $endgroup$
  – dineshdileep
  Apr 2 '13 at 8:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @dineshdileep What? Who knows if some mistakes are lurking around in some shadowy places. You should double-check the claimed result.
  $endgroup$
  – user1551
  Apr 2 '13 at 16:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I will definitely do that.
  $endgroup$
  – dineshdileep
  Apr 3 '13 at 4:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you explain (mathematically) the second last step, where you switch to the block matrix. (I almost have it intuitively, but would like to know a rigorous one).
  $endgroup$
  – dineshdileep
  Apr 3 '13 at 4:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Let $w_j=operatorname{vec}(W_j)$. Then $w^T=(w_1^T,ldots,w_n^T)$. Partition $Q$ into subblocks, each of size $ntimes n$. Call the $(j,i)$-th block $Q_{ji}$. Then by $w^HQw=sum_isum_j w_j^H Q_{ji} w_i$. So, by comparing coefficients, we should set $Q_{ji}=(bar{h}_jh_i^T)otimes(g_jg_i^H)$.
  $endgroup$
  – user1551
  Apr 3 '13 at 20:51
  

  
  
  
  

 | 
show 1 more comment

2

$begingroup$

begin{align}
S&=sum_{i}sum_{j}g_i^HW_ih_ih_j^HW_j^Hg_j\
&=sum_{i}sum_{j}operatorname{trace}(g_jg_i^HW_ih_ih_j^HW_j^H)\
&=sum_{i}sum_{j}operatorname{vec}(W_j)^H operatorname{vec}(g_jg_i^HW_ih_ih_j^H)\
&qquadqquadqquad{because operatorname{trace}(AB^{H})=operatorname{trace}(A^{H}B)=operatorname{vec}(A)^{H}operatorname{vec}(B)}\
& \
&=sum_{i}sum_{j}operatorname{vec}(W_j)^H left((h_ih_j^H)^Totimes(g_jg_i^H)right)operatorname{vec}(W_i)\
&qquadqquadqquad{because operatorname{vec}(ABC)=(C^{T}otimes A)operatorname{vec}(B)}\
& \
&=sum_{i}sum_{j}operatorname{vec}(W_j)^H left((bar{h}_jh_i^T)otimes(g_jg_i^H)right)operatorname{vec}(W_i)\
&=w^H Qw,
end{align}
where $Q$ is a block matrix whose $(j,i)$-th (note: not $(i,j)$-th) subblock is $(bar{h}_jh_i^T)otimes(g_jg_i^H)$.

share|cite|improve this answer

edited 2 days ago

Rócherz

2,8912821

answered Apr 1 '13 at 18:13

user1551user1551

73.5k566129

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you are always reliable!!!
  $endgroup$
  – dineshdileep
  Apr 2 '13 at 8:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @dineshdileep What? Who knows if some mistakes are lurking around in some shadowy places. You should double-check the claimed result.
  $endgroup$
  – user1551
  Apr 2 '13 at 16:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I will definitely do that.
  $endgroup$
  – dineshdileep
  Apr 3 '13 at 4:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you explain (mathematically) the second last step, where you switch to the block matrix. (I almost have it intuitively, but would like to know a rigorous one).
  $endgroup$
  – dineshdileep
  Apr 3 '13 at 4:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Let $w_j=operatorname{vec}(W_j)$. Then $w^T=(w_1^T,ldots,w_n^T)$. Partition $Q$ into subblocks, each of size $ntimes n$. Call the $(j,i)$-th block $Q_{ji}$. Then by $w^HQw=sum_isum_j w_j^H Q_{ji} w_i$. So, by comparing coefficients, we should set $Q_{ji}=(bar{h}_jh_i^T)otimes(g_jg_i^H)$.
  $endgroup$
  – user1551
  Apr 3 '13 at 20:51
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

begin{align}
S&=sum_{i}sum_{j}g_i^HW_ih_ih_j^HW_j^Hg_j\
&=sum_{i}sum_{j}operatorname{trace}(g_jg_i^HW_ih_ih_j^HW_j^H)\
&=sum_{i}sum_{j}operatorname{vec}(W_j)^H operatorname{vec}(g_jg_i^HW_ih_ih_j^H)\
&qquadqquadqquad{because operatorname{trace}(AB^{H})=operatorname{trace}(A^{H}B)=operatorname{vec}(A)^{H}operatorname{vec}(B)}\
& \
&=sum_{i}sum_{j}operatorname{vec}(W_j)^H left((h_ih_j^H)^Totimes(g_jg_i^H)right)operatorname{vec}(W_i)\
&qquadqquadqquad{because operatorname{vec}(ABC)=(C^{T}otimes A)operatorname{vec}(B)}\
& \
&=sum_{i}sum_{j}operatorname{vec}(W_j)^H left((bar{h}_jh_i^T)otimes(g_jg_i^H)right)operatorname{vec}(W_i)\
&=w^H Qw,
end{align}
where $Q$ is a block matrix whose $(j,i)$-th (note: not $(i,j)$-th) subblock is $(bar{h}_jh_i^T)otimes(g_jg_i^H)$.

share|cite|improve this answer

edited 2 days ago

Rócherz

2,8912821

answered Apr 1 '13 at 18:13

user1551user1551

73.5k566129

$endgroup$

share|cite|improve this answer

edited 2 days ago

Rócherz

2,8912821

answered Apr 1 '13 at 18:13

user1551user1551

73.5k566129

edited 2 days ago

Rócherz

2,8912821

edited 2 days ago

Rócherz

2,8912821

answered Apr 1 '13 at 18:13

user1551user1551

73.5k566129

answered Apr 1 '13 at 18:13

user1551user1551

73.5k566129