Does bilinear models on vectors mean dot or outer product?
$begingroup$
If I have 2 vectors $x$ and $y$ where $x in mathcal{R}^{m}$ and $y in mathcal{R}^{n}$.
Does bilinear model mean?
$f(x,y) = x^TWy$ where $W in mathcal{R}^{m*n}$
which result in a scalar
or
$f(x,y) = W(x⊗y^T)$ where ⊗ is the outer
product and $W in mathcal{R}^{m*n}$.
which result in a matrix
I checked 2 papers, the first one Low-rank Bilinear Pooling
in page 2 in equation 1 their bilinear model produce a scalar
while in Compact Bilinear Pooling in section 3.1 they said "Bilinear models take the outer product of two vectors"
linear-algebra matrices
$endgroup$
add a comment |
$begingroup$
If I have 2 vectors $x$ and $y$ where $x in mathcal{R}^{m}$ and $y in mathcal{R}^{n}$.
Does bilinear model mean?
$f(x,y) = x^TWy$ where $W in mathcal{R}^{m*n}$
which result in a scalar
or
$f(x,y) = W(x⊗y^T)$ where ⊗ is the outer
product and $W in mathcal{R}^{m*n}$.
which result in a matrix
I checked 2 papers, the first one Low-rank Bilinear Pooling
in page 2 in equation 1 their bilinear model produce a scalar
while in Compact Bilinear Pooling in section 3.1 they said "Bilinear models take the outer product of two vectors"
linear-algebra matrices
$endgroup$
add a comment |
$begingroup$
If I have 2 vectors $x$ and $y$ where $x in mathcal{R}^{m}$ and $y in mathcal{R}^{n}$.
Does bilinear model mean?
$f(x,y) = x^TWy$ where $W in mathcal{R}^{m*n}$
which result in a scalar
or
$f(x,y) = W(x⊗y^T)$ where ⊗ is the outer
product and $W in mathcal{R}^{m*n}$.
which result in a matrix
I checked 2 papers, the first one Low-rank Bilinear Pooling
in page 2 in equation 1 their bilinear model produce a scalar
while in Compact Bilinear Pooling in section 3.1 they said "Bilinear models take the outer product of two vectors"
linear-algebra matrices
$endgroup$
If I have 2 vectors $x$ and $y$ where $x in mathcal{R}^{m}$ and $y in mathcal{R}^{n}$.
Does bilinear model mean?
$f(x,y) = x^TWy$ where $W in mathcal{R}^{m*n}$
which result in a scalar
or
$f(x,y) = W(x⊗y^T)$ where ⊗ is the outer
product and $W in mathcal{R}^{m*n}$.
which result in a matrix
I checked 2 papers, the first one Low-rank Bilinear Pooling
in page 2 in equation 1 their bilinear model produce a scalar
while in Compact Bilinear Pooling in section 3.1 they said "Bilinear models take the outer product of two vectors"
linear-algebra matrices
linear-algebra matrices
asked Dec 19 '18 at 11:53
floydfloyd
1032
1032
add a comment |
add a comment |
1 Answer
1
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$begingroup$
"Bilinear" is simply an adjective that you can apply to any function of two vectors to indicate that it is linear in each argument. So the linear map
$(x,y) rightarrow x^TWy$
and the tensor product
$(x,y) rightarrow xy^T$
can both be described as bilinear, even though their codomains are different.
$endgroup$
$begingroup$
What about the element-wise product between 2 vectors? The reason I mention it is because it's different than the dot and outer product because it's doesn't take into account all the interactions between the 2 vectors (like what the dot product does).
$endgroup$
– floyd
Jan 7 at 20:20
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
"Bilinear" is simply an adjective that you can apply to any function of two vectors to indicate that it is linear in each argument. So the linear map
$(x,y) rightarrow x^TWy$
and the tensor product
$(x,y) rightarrow xy^T$
can both be described as bilinear, even though their codomains are different.
$endgroup$
$begingroup$
What about the element-wise product between 2 vectors? The reason I mention it is because it's different than the dot and outer product because it's doesn't take into account all the interactions between the 2 vectors (like what the dot product does).
$endgroup$
– floyd
Jan 7 at 20:20
add a comment |
$begingroup$
"Bilinear" is simply an adjective that you can apply to any function of two vectors to indicate that it is linear in each argument. So the linear map
$(x,y) rightarrow x^TWy$
and the tensor product
$(x,y) rightarrow xy^T$
can both be described as bilinear, even though their codomains are different.
$endgroup$
$begingroup$
What about the element-wise product between 2 vectors? The reason I mention it is because it's different than the dot and outer product because it's doesn't take into account all the interactions between the 2 vectors (like what the dot product does).
$endgroup$
– floyd
Jan 7 at 20:20
add a comment |
$begingroup$
"Bilinear" is simply an adjective that you can apply to any function of two vectors to indicate that it is linear in each argument. So the linear map
$(x,y) rightarrow x^TWy$
and the tensor product
$(x,y) rightarrow xy^T$
can both be described as bilinear, even though their codomains are different.
$endgroup$
"Bilinear" is simply an adjective that you can apply to any function of two vectors to indicate that it is linear in each argument. So the linear map
$(x,y) rightarrow x^TWy$
and the tensor product
$(x,y) rightarrow xy^T$
can both be described as bilinear, even though their codomains are different.
answered Dec 19 '18 at 12:16
gandalf61gandalf61
9,199825
9,199825
$begingroup$
What about the element-wise product between 2 vectors? The reason I mention it is because it's different than the dot and outer product because it's doesn't take into account all the interactions between the 2 vectors (like what the dot product does).
$endgroup$
– floyd
Jan 7 at 20:20
add a comment |
$begingroup$
What about the element-wise product between 2 vectors? The reason I mention it is because it's different than the dot and outer product because it's doesn't take into account all the interactions between the 2 vectors (like what the dot product does).
$endgroup$
– floyd
Jan 7 at 20:20
$begingroup$
What about the element-wise product between 2 vectors? The reason I mention it is because it's different than the dot and outer product because it's doesn't take into account all the interactions between the 2 vectors (like what the dot product does).
$endgroup$
– floyd
Jan 7 at 20:20
$begingroup$
What about the element-wise product between 2 vectors? The reason I mention it is because it's different than the dot and outer product because it's doesn't take into account all the interactions between the 2 vectors (like what the dot product does).
$endgroup$
– floyd
Jan 7 at 20:20
add a comment |
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