Express matrix in more compact form
$begingroup$
I have the following sum:
$$
S = sum_{j=1}^n sum_{k=1}^n a_{jk} x_j^T B x_k
$$
where $x_j$ is a vector of length $m$, and $B$ is an $m times m$ positive definite matrix, and $a_{jk}$ are positive scalars. I want to rewrite this sum as a symmetric bilinear form
$$
S = x^T C x
$$
where $x$ is a vector of length $nm$:
$$
x = begin{pmatrix}
x_1 \
x_2 \
vdots \
x_n
end{pmatrix}
$$
and $C$ is an $nm times nm$ matrix. By writing out all the terms, I found that:
$$
C = begin{pmatrix}
a_{11} B & a_{12} B & dots & a_{1n} B \
a_{21} B & a_{22} B & dots & a_{2n} B \
vdots & vdots & vdots & vdots \
a_{n1} B & a_{n2} B & dots & a_{nn} B
end{pmatrix}
$$
My question is: is there a way to express this matrix in a more elegant, compact form, using the matrix $A_{ij} = a_{ij}$ and $B$, without having to write out all the entries like above?
linear-algebra matrix-equations
asked Dec 10 '18 at 19:15
vibevibe
1648
$endgroup$
add a comment |
$begingroup$
I have the following sum:
$$
S = sum_{j=1}^n sum_{k=1}^n a_{jk} x_j^T B x_k
$$
where $x_j$ is a vector of length $m$, and $B$ is an $m times m$ positive definite matrix, and $a_{jk}$ are positive scalars. I want to rewrite this sum as a symmetric bilinear form
$$
S = x^T C x
$$
where $x$ is a vector of length $nm$:
$$
x = begin{pmatrix}
x_1 \
x_2 \
vdots \
x_n
end{pmatrix}
$$
and $C$ is an $nm times nm$ matrix. By writing out all the terms, I found that:
$$
C = begin{pmatrix}
a_{11} B & a_{12} B & dots & a_{1n} B \
a_{21} B & a_{22} B & dots & a_{2n} B \
vdots & vdots & vdots & vdots \
a_{n1} B & a_{n2} B & dots & a_{nn} B
end{pmatrix}
$$
My question is: is there a way to express this matrix in a more elegant, compact form, using the matrix $A_{ij} = a_{ij}$ and $B$, without having to write out all the entries like above?
linear-algebra matrix-equations
asked Dec 10 '18 at 19:15
vibevibe
1648
$endgroup$
add a comment |
$begingroup$
I have the following sum:
$$
S = sum_{j=1}^n sum_{k=1}^n a_{jk} x_j^T B x_k
$$
where $x_j$ is a vector of length $m$, and $B$ is an $m times m$ positive definite matrix, and $a_{jk}$ are positive scalars. I want to rewrite this sum as a symmetric bilinear form
$$
S = x^T C x
$$
where $x$ is a vector of length $nm$:
$$
x = begin{pmatrix}
x_1 \
x_2 \
vdots \
x_n
end{pmatrix}
$$
and $C$ is an $nm times nm$ matrix. By writing out all the terms, I found that:
$$
C = begin{pmatrix}
a_{11} B & a_{12} B & dots & a_{1n} B \
a_{21} B & a_{22} B & dots & a_{2n} B \
vdots & vdots & vdots & vdots \
a_{n1} B & a_{n2} B & dots & a_{nn} B
end{pmatrix}
$$
My question is: is there a way to express this matrix in a more elegant, compact form, using the matrix $A_{ij} = a_{ij}$ and $B$, without having to write out all the entries like above?
linear-algebra matrix-equations
asked Dec 10 '18 at 19:15
vibevibe
1648
$endgroup$
I have the following sum:
$$
S = sum_{j=1}^n sum_{k=1}^n a_{jk} x_j^T B x_k
$$
where $x_j$ is a vector of length $m$, and $B$ is an $m times m$ positive definite matrix, and $a_{jk}$ are positive scalars. I want to rewrite this sum as a symmetric bilinear form
$$
S = x^T C x
$$
where $x$ is a vector of length $nm$:
$$
x = begin{pmatrix}
x_1 \
x_2 \
vdots \
x_n
end{pmatrix}
$$
and $C$ is an $nm times nm$ matrix. By writing out all the terms, I found that:
$$
C = begin{pmatrix}
a_{11} B & a_{12} B & dots & a_{1n} B \
a_{21} B & a_{22} B & dots & a_{2n} B \
vdots & vdots & vdots & vdots \
a_{n1} B & a_{n2} B & dots & a_{nn} B
end{pmatrix}
$$
My question is: is there a way to express this matrix in a more elegant, compact form, using the matrix $A_{ij} = a_{ij}$ and $B$, without having to write out all the entries like above?
linear-algebra matrix-equations
linear-algebra matrix-equations
asked Dec 10 '18 at 19:15
vibevibe
1648
asked Dec 10 '18 at 19:15
vibevibe
1648
asked Dec 10 '18 at 19:15
vibevibe
1648
asked Dec 10 '18 at 19:15
vibevibe
1648
asked Dec 10 '18 at 19:15
vibevibe
1648
1648
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Your matrix $C$ is the Kronecker product $C = A otimes B$.
answered Dec 10 '18 at 19:17
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Your matrix $C$ is the Kronecker product $C = A otimes B$.
answered Dec 10 '18 at 19:17
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
add a comment |
$begingroup$
Your matrix $C$ is the Kronecker product $C = A otimes B$.
answered Dec 10 '18 at 19:17
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
add a comment |
$begingroup$
Your matrix $C$ is the Kronecker product $C = A otimes B$.
answered Dec 10 '18 at 19:17
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
Your matrix $C$ is the Kronecker product $C = A otimes B$.
answered Dec 10 '18 at 19:17
OmnomnomnomOmnomnomnom
128k791184
answered Dec 10 '18 at 19:17
OmnomnomnomOmnomnomnom
128k791184
answered Dec 10 '18 at 19:17
OmnomnomnomOmnomnomnom
128k791184
answered Dec 10 '18 at 19:17
OmnomnomnomOmnomnomnom
128k791184
128k791184
add a comment |
add a comment |
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