What does the term in the right mean?
$begingroup$
Reading an article about clustering, I came across this equality:
It's basically a sum over $ mu_{i,j}(1-cos(x_i, p_j))$ (where $ cos(x_i, p_j) $ is the cosinus between two vectors in $mathbb{R}^n$). And then suddenly some magic happens and the formula in the right appears. Can somebody explain?
Note: I understand that the parenthesis was expanded (each term got multiplied by $ mu_{i,j}$). But what is that huge dot product and why does it have this form?
vectors products
asked Nov 24 '18 at 16:56
Hello LiliHello Lili
1104
$endgroup$
add a comment |
$begingroup$
Reading an article about clustering, I came across this equality:
It's basically a sum over $ mu_{i,j}(1-cos(x_i, p_j))$ (where $ cos(x_i, p_j) $ is the cosinus between two vectors in $mathbb{R}^n$). And then suddenly some magic happens and the formula in the right appears. Can somebody explain?
Note: I understand that the parenthesis was expanded (each term got multiplied by $ mu_{i,j}$). But what is that huge dot product and why does it have this form?
vectors products
asked Nov 24 '18 at 16:56
Hello LiliHello Lili
1104
$endgroup$
1
$begingroup$
Refer to the inner products on $mathbb R^n$. Use the linearity w.r.t. each entry.
$endgroup$
– xbh
Nov 24 '18 at 17:00
$begingroup$
@xbh I changed R to $mathbb{R}$. I don't understand what you mentioned about the linearity. Oh, ok I'm stupid, you were actually answering my question, not referring to formating. I suck at math.
$endgroup$
– Hello Lili
Nov 24 '18 at 17:03
1
$begingroup$
$(lambda a+mu b|c) = lambda (a|c) + mu(b|c)$, where $(a|b) = langle a,b rangle$ [and this is just a preference of notations about $(|)$ and $langle,rangle$].
$endgroup$
– xbh
Nov 24 '18 at 17:05
add a comment |
$begingroup$
Reading an article about clustering, I came across this equality:
It's basically a sum over $ mu_{i,j}(1-cos(x_i, p_j))$ (where $ cos(x_i, p_j) $ is the cosinus between two vectors in $mathbb{R}^n$). And then suddenly some magic happens and the formula in the right appears. Can somebody explain?
Note: I understand that the parenthesis was expanded (each term got multiplied by $ mu_{i,j}$). But what is that huge dot product and why does it have this form?
vectors products
asked Nov 24 '18 at 16:56
Hello LiliHello Lili
1104
$endgroup$
Reading an article about clustering, I came across this equality:
It's basically a sum over $ mu_{i,j}(1-cos(x_i, p_j))$ (where $ cos(x_i, p_j) $ is the cosinus between two vectors in $mathbb{R}^n$). And then suddenly some magic happens and the formula in the right appears. Can somebody explain?
Note: I understand that the parenthesis was expanded (each term got multiplied by $ mu_{i,j}$). But what is that huge dot product and why does it have this form?
vectors products
vectors products
asked Nov 24 '18 at 16:56
Hello LiliHello Lili
1104
asked Nov 24 '18 at 16:56
Hello LiliHello Lili
1104
edited Nov 24 '18 at 17:02
Hello Lili
asked Nov 24 '18 at 16:56
Hello LiliHello Lili
1104
asked Nov 24 '18 at 16:56
Hello LiliHello Lili
1104
asked Nov 24 '18 at 16:56
Hello LiliHello Lili
1104
1104
1
$begingroup$
Refer to the inner products on $mathbb R^n$. Use the linearity w.r.t. each entry.
$endgroup$
– xbh
Nov 24 '18 at 17:00
$begingroup$
@xbh I changed R to $mathbb{R}$. I don't understand what you mentioned about the linearity. Oh, ok I'm stupid, you were actually answering my question, not referring to formating. I suck at math.
$endgroup$
– Hello Lili
Nov 24 '18 at 17:03
1
$begingroup$
$(lambda a+mu b|c) = lambda (a|c) + mu(b|c)$, where $(a|b) = langle a,b rangle$ [and this is just a preference of notations about $(|)$ and $langle,rangle$].
$endgroup$
– xbh
Nov 24 '18 at 17:05
add a comment |
1
$begingroup$
Refer to the inner products on $mathbb R^n$. Use the linearity w.r.t. each entry.
$endgroup$
– xbh
Nov 24 '18 at 17:00
$begingroup$
@xbh I changed R to $mathbb{R}$. I don't understand what you mentioned about the linearity. Oh, ok I'm stupid, you were actually answering my question, not referring to formating. I suck at math.
$endgroup$
– Hello Lili
Nov 24 '18 at 17:03
1
$begingroup$
$(lambda a+mu b|c) = lambda (a|c) + mu(b|c)$, where $(a|b) = langle a,b rangle$ [and this is just a preference of notations about $(|)$ and $langle,rangle$].
$endgroup$
– xbh
Nov 24 '18 at 17:05
1
1
$begingroup$
Refer to the inner products on $mathbb R^n$. Use the linearity w.r.t. each entry.
$endgroup$
– xbh
Nov 24 '18 at 17:00
$begingroup$
Refer to the inner products on $mathbb R^n$. Use the linearity w.r.t. each entry.
$endgroup$
– xbh
Nov 24 '18 at 17:00
$begingroup$
@xbh I changed R to $mathbb{R}$. I don't understand what you mentioned about the linearity. Oh, ok I'm stupid, you were actually answering my question, not referring to formating. I suck at math.
$endgroup$
– Hello Lili
Nov 24 '18 at 17:03
$begingroup$
@xbh I changed R to $mathbb{R}$. I don't understand what you mentioned about the linearity. Oh, ok I'm stupid, you were actually answering my question, not referring to formating. I suck at math.
$endgroup$
– Hello Lili
Nov 24 '18 at 17:03
1
1
$begingroup$
$(lambda a+mu b|c) = lambda (a|c) + mu(b|c)$, where $(a|b) = langle a,b rangle$ [and this is just a preference of notations about $(|)$ and $langle,rangle$].
$endgroup$
– xbh
Nov 24 '18 at 17:05
$begingroup$
$(lambda a+mu b|c) = lambda (a|c) + mu(b|c)$, where $(a|b) = langle a,b rangle$ [and this is just a preference of notations about $(|)$ and $langle,rangle$].
$endgroup$
– xbh
Nov 24 '18 at 17:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Remember that the inner product is bilinear: for three vectors $u,v,w$ and scalars $alpha,beta$,
$$
langle alpha u, beta vrangle = alphabetalangle u, vrangle, qquad langle u, v+wrangle = langle u, vrangle+langle u, wrangle
$$
What happened above is exactly what follows when applying these rules: for every $i,j$
$$
frac{langle x_i, p_jrangle}{lVert x_irVertlVert p_jrVert} = leftlangle frac{x_i}{lVert x_irVert}, frac{p_j}{lVert p_jrVert}rightrangle tag{1}
$$
since $lVert x_irVert,lVert p_jrVertinmathbb{R}$ are scalars (the brackets became "bigger" just for style considerations, like parentheses would be drawn bigger when what's inside looks "large").
Then, each $mu_{ij}$ is also a scalar, and you have a sum, so
$$
sum_{i} mu_{ij}leftlangle frac{x_i}{lVert x_irVert}, frac{p_j}{lVert p_jrVert}rightrangle
= sum_{i} leftlangle mu_{ij}frac{x_i}{lVert x_irVert}, frac{p_j}{lVert p_jrVert}rightrangle
= leftlangle sum_{i}mu_{ij}frac{x_i}{lVert x_irVert}, frac{p_j}{lVert p_jrVert}rightrangle tag{2}
$$
Note that it does make sense: $sum_{i}mu_{ij}frac{x_i}{lVert x_irVert}$ is a vector, and so is $frac{p_j}{lVert p_jrVert}$. So it's indeed the inner product between two vectors.
answered Nov 24 '18 at 17:04
Clement C.Clement C.
49.8k33886
$endgroup$
$begingroup$
This answers my question. Thank you! I will accept this answer in 3 minutes.
$endgroup$
– Hello Lili
Nov 24 '18 at 17:07
$begingroup$
@HelloLili Glad this helped!
$endgroup$
– Clement C.
Nov 24 '18 at 17:08
add a comment |
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$begingroup$
Remember that the inner product is bilinear: for three vectors $u,v,w$ and scalars $alpha,beta$,
$$
langle alpha u, beta vrangle = alphabetalangle u, vrangle, qquad langle u, v+wrangle = langle u, vrangle+langle u, wrangle
$$
What happened above is exactly what follows when applying these rules: for every $i,j$
$$
frac{langle x_i, p_jrangle}{lVert x_irVertlVert p_jrVert} = leftlangle frac{x_i}{lVert x_irVert}, frac{p_j}{lVert p_jrVert}rightrangle tag{1}
$$
since $lVert x_irVert,lVert p_jrVertinmathbb{R}$ are scalars (the brackets became "bigger" just for style considerations, like parentheses would be drawn bigger when what's inside looks "large").
Then, each $mu_{ij}$ is also a scalar, and you have a sum, so
$$
sum_{i} mu_{ij}leftlangle frac{x_i}{lVert x_irVert}, frac{p_j}{lVert p_jrVert}rightrangle
= sum_{i} leftlangle mu_{ij}frac{x_i}{lVert x_irVert}, frac{p_j}{lVert p_jrVert}rightrangle
= leftlangle sum_{i}mu_{ij}frac{x_i}{lVert x_irVert}, frac{p_j}{lVert p_jrVert}rightrangle tag{2}
$$
Note that it does make sense: $sum_{i}mu_{ij}frac{x_i}{lVert x_irVert}$ is a vector, and so is $frac{p_j}{lVert p_jrVert}$. So it's indeed the inner product between two vectors.
answered Nov 24 '18 at 17:04
Clement C.Clement C.
49.8k33886
$endgroup$
$begingroup$
This answers my question. Thank you! I will accept this answer in 3 minutes.
$endgroup$
– Hello Lili
Nov 24 '18 at 17:07
$begingroup$
@HelloLili Glad this helped!
$endgroup$
– Clement C.
Nov 24 '18 at 17:08
add a comment |
$begingroup$
Remember that the inner product is bilinear: for three vectors $u,v,w$ and scalars $alpha,beta$,
$$
langle alpha u, beta vrangle = alphabetalangle u, vrangle, qquad langle u, v+wrangle = langle u, vrangle+langle u, wrangle
$$
What happened above is exactly what follows when applying these rules: for every $i,j$
$$
frac{langle x_i, p_jrangle}{lVert x_irVertlVert p_jrVert} = leftlangle frac{x_i}{lVert x_irVert}, frac{p_j}{lVert p_jrVert}rightrangle tag{1}
$$
since $lVert x_irVert,lVert p_jrVertinmathbb{R}$ are scalars (the brackets became "bigger" just for style considerations, like parentheses would be drawn bigger when what's inside looks "large").
Then, each $mu_{ij}$ is also a scalar, and you have a sum, so
$$
sum_{i} mu_{ij}leftlangle frac{x_i}{lVert x_irVert}, frac{p_j}{lVert p_jrVert}rightrangle
= sum_{i} leftlangle mu_{ij}frac{x_i}{lVert x_irVert}, frac{p_j}{lVert p_jrVert}rightrangle
= leftlangle sum_{i}mu_{ij}frac{x_i}{lVert x_irVert}, frac{p_j}{lVert p_jrVert}rightrangle tag{2}
$$
Note that it does make sense: $sum_{i}mu_{ij}frac{x_i}{lVert x_irVert}$ is a vector, and so is $frac{p_j}{lVert p_jrVert}$. So it's indeed the inner product between two vectors.
answered Nov 24 '18 at 17:04
Clement C.Clement C.
49.8k33886
$endgroup$
$begingroup$
This answers my question. Thank you! I will accept this answer in 3 minutes.
$endgroup$
– Hello Lili
Nov 24 '18 at 17:07
$begingroup$
@HelloLili Glad this helped!
$endgroup$
– Clement C.
Nov 24 '18 at 17:08
add a comment |
$begingroup$
Remember that the inner product is bilinear: for three vectors $u,v,w$ and scalars $alpha,beta$,
$$
langle alpha u, beta vrangle = alphabetalangle u, vrangle, qquad langle u, v+wrangle = langle u, vrangle+langle u, wrangle
$$
What happened above is exactly what follows when applying these rules: for every $i,j$
$$
frac{langle x_i, p_jrangle}{lVert x_irVertlVert p_jrVert} = leftlangle frac{x_i}{lVert x_irVert}, frac{p_j}{lVert p_jrVert}rightrangle tag{1}
$$
since $lVert x_irVert,lVert p_jrVertinmathbb{R}$ are scalars (the brackets became "bigger" just for style considerations, like parentheses would be drawn bigger when what's inside looks "large").
Then, each $mu_{ij}$ is also a scalar, and you have a sum, so
$$
sum_{i} mu_{ij}leftlangle frac{x_i}{lVert x_irVert}, frac{p_j}{lVert p_jrVert}rightrangle
= sum_{i} leftlangle mu_{ij}frac{x_i}{lVert x_irVert}, frac{p_j}{lVert p_jrVert}rightrangle
= leftlangle sum_{i}mu_{ij}frac{x_i}{lVert x_irVert}, frac{p_j}{lVert p_jrVert}rightrangle tag{2}
$$
Note that it does make sense: $sum_{i}mu_{ij}frac{x_i}{lVert x_irVert}$ is a vector, and so is $frac{p_j}{lVert p_jrVert}$. So it's indeed the inner product between two vectors.
answered Nov 24 '18 at 17:04
Clement C.Clement C.
49.8k33886
$endgroup$
Remember that the inner product is bilinear: for three vectors $u,v,w$ and scalars $alpha,beta$,
$$
langle alpha u, beta vrangle = alphabetalangle u, vrangle, qquad langle u, v+wrangle = langle u, vrangle+langle u, wrangle
$$
What happened above is exactly what follows when applying these rules: for every $i,j$
$$
frac{langle x_i, p_jrangle}{lVert x_irVertlVert p_jrVert} = leftlangle frac{x_i}{lVert x_irVert}, frac{p_j}{lVert p_jrVert}rightrangle tag{1}
$$
since $lVert x_irVert,lVert p_jrVertinmathbb{R}$ are scalars (the brackets became "bigger" just for style considerations, like parentheses would be drawn bigger when what's inside looks "large").
Then, each $mu_{ij}$ is also a scalar, and you have a sum, so
$$
sum_{i} mu_{ij}leftlangle frac{x_i}{lVert x_irVert}, frac{p_j}{lVert p_jrVert}rightrangle
= sum_{i} leftlangle mu_{ij}frac{x_i}{lVert x_irVert}, frac{p_j}{lVert p_jrVert}rightrangle
= leftlangle sum_{i}mu_{ij}frac{x_i}{lVert x_irVert}, frac{p_j}{lVert p_jrVert}rightrangle tag{2}
$$
Note that it does make sense: $sum_{i}mu_{ij}frac{x_i}{lVert x_irVert}$ is a vector, and so is $frac{p_j}{lVert p_jrVert}$. So it's indeed the inner product between two vectors.
answered Nov 24 '18 at 17:04
Clement C.Clement C.
49.8k33886
answered Nov 24 '18 at 17:04
Clement C.Clement C.
49.8k33886
answered Nov 24 '18 at 17:04
Clement C.Clement C.
49.8k33886
answered Nov 24 '18 at 17:04
Clement C.Clement C.
49.8k33886
49.8k33886
$begingroup$
This answers my question. Thank you! I will accept this answer in 3 minutes.
$endgroup$
– Hello Lili
Nov 24 '18 at 17:07
$begingroup$
@HelloLili Glad this helped!
$endgroup$
– Clement C.
Nov 24 '18 at 17:08
add a comment |
$begingroup$
This answers my question. Thank you! I will accept this answer in 3 minutes.
$endgroup$
– Hello Lili
Nov 24 '18 at 17:07
$begingroup$
@HelloLili Glad this helped!
$endgroup$
– Clement C.
Nov 24 '18 at 17:08
$begingroup$
This answers my question. Thank you! I will accept this answer in 3 minutes.
$endgroup$
– Hello Lili
Nov 24 '18 at 17:07
$begingroup$
This answers my question. Thank you! I will accept this answer in 3 minutes.
$endgroup$
– Hello Lili
Nov 24 '18 at 17:07
$begingroup$
@HelloLili Glad this helped!
$endgroup$
– Clement C.
Nov 24 '18 at 17:08
$begingroup$
@HelloLili Glad this helped!
$endgroup$
– Clement C.
Nov 24 '18 at 17:08
add a comment |
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$begingroup$
Refer to the inner products on $mathbb R^n$. Use the linearity w.r.t. each entry.
$endgroup$
– xbh
Nov 24 '18 at 17:00
$begingroup$
@xbh I changed R to $mathbb{R}$. I don't understand what you mentioned about the linearity. Oh, ok I'm stupid, you were actually answering my question, not referring to formating. I suck at math.
$endgroup$
– Hello Lili
Nov 24 '18 at 17:03
1
$begingroup$
$(lambda a+mu b|c) = lambda (a|c) + mu(b|c)$, where $(a|b) = langle a,b rangle$ [and this is just a preference of notations about $(|)$ and $langle,rangle$].
$endgroup$
– xbh
Nov 24 '18 at 17:05