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$begingroup$
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
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
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
edited 2 days ago
Rócherz
2,8912821
asked Apr 1 '13 at 17:30
dineshdileepdineshdileep
5,98611935
$endgroup$
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
linear-algebra matrices convex-optimization quadratic-forms block-matrices
edited 2 days ago
Rócherz
2,8912821
asked Apr 1 '13 at 17:30
dineshdileepdineshdileep
5,98611935
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
edited 2 days ago
Rócherz
2,8912821
2,8912821
asked Apr 1 '13 at 17:30
dineshdileepdineshdileep
5,98611935
asked Apr 1 '13 at 17:30
dineshdileepdineshdileep
5,98611935
asked Apr 1 '13 at 17:30
dineshdileepdineshdileep
5,98611935
5,98611935
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $(bar{h}_jh_i^T)otimes(g_jg_i^H)$.
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
Your Answer
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $(bar{h}_jh_i^T)otimes(g_jg_i^H)$.
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 $(bar{h}_jh_i^T)otimes(g_jg_i^H)$.
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 $(bar{h}_jh_i^T)otimes(g_jg_i^H)$.
edited 2 days ago
Rócherz
2,8912821
answered Apr 1 '13 at 18:13
user1551user1551
73.5k566129
$endgroup$
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 $(bar{h}_jh_i^T)otimes(g_jg_i^H)$.
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
edited 2 days ago
Rócherz
2,8912821
2,8912821
answered Apr 1 '13 at 18:13
user1551user1551
73.5k566129
answered Apr 1 '13 at 18:13
user1551user1551
73.5k566129
answered Apr 1 '13 at 18:13
user1551user1551
73.5k566129
73.5k566129
$begingroup$
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
$begingroup$
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
$begingroup$
you are always reliable!!!
$endgroup$
– dineshdileep
Apr 2 '13 at 8:50
$begingroup$
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
$begingroup$
@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
$begingroup$
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
$begingroup$
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
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
$begingroup$
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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