How to compute derivative/gradient of matrices? E.g. $w^T X^T Xw - 2w^T X^T t +t^Tt$?
How to compute derivative/gradient of matrices? E.g. $w^T X^T Xw - 2w^T X^T t +t^Tt$?
Intuitively $2w^TX^Tt$ looks like $2xy$ so the derivative would be $2X^Tt$.
But what about $w^TX^TXw$?
$X$ is $mathbb{R}^{n times (p+1)}$. $w,t$ are row/column vectors corresponding to "weights" and "predictors" in $w^T x=y$ sense.
The gradient is $frac{partial}{partial w}$.
matrices derivatives vector-analysis
asked Dec 7 at 15:21
mavavilj
2,67111032
add a comment |
How to compute derivative/gradient of matrices? E.g. $w^T X^T Xw - 2w^T X^T t +t^Tt$?
Intuitively $2w^TX^Tt$ looks like $2xy$ so the derivative would be $2X^Tt$.
But what about $w^TX^TXw$?
$X$ is $mathbb{R}^{n times (p+1)}$. $w,t$ are row/column vectors corresponding to "weights" and "predictors" in $w^T x=y$ sense.
The gradient is $frac{partial}{partial w}$.
matrices derivatives vector-analysis
asked Dec 7 at 15:21
mavavilj
2,67111032
add a comment |
How to compute derivative/gradient of matrices? E.g. $w^T X^T Xw - 2w^T X^T t +t^Tt$?
Intuitively $2w^TX^Tt$ looks like $2xy$ so the derivative would be $2X^Tt$.
But what about $w^TX^TXw$?
$X$ is $mathbb{R}^{n times (p+1)}$. $w,t$ are row/column vectors corresponding to "weights" and "predictors" in $w^T x=y$ sense.
The gradient is $frac{partial}{partial w}$.
matrices derivatives vector-analysis
asked Dec 7 at 15:21
mavavilj
2,67111032
How to compute derivative/gradient of matrices? E.g. $w^T X^T Xw - 2w^T X^T t +t^Tt$?
Intuitively $2w^TX^Tt$ looks like $2xy$ so the derivative would be $2X^Tt$.
But what about $w^TX^TXw$?
$X$ is $mathbb{R}^{n times (p+1)}$. $w,t$ are row/column vectors corresponding to "weights" and "predictors" in $w^T x=y$ sense.
The gradient is $frac{partial}{partial w}$.
matrices derivatives vector-analysis
matrices derivatives vector-analysis
asked Dec 7 at 15:21
mavavilj
2,67111032
asked Dec 7 at 15:21
mavavilj
2,67111032
edited Dec 7 at 15:28
asked Dec 7 at 15:21
mavavilj
2,67111032
asked Dec 7 at 15:21
mavavilj
2,67111032
asked Dec 7 at 15:21
mavavilj
2,67111032
2,67111032
add a comment |
add a comment |
2 Answers
2
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$w^TX^Tt = t^TXw$ is linear: its gradient is $t^TX$.
$w^T(X^TX)w$ is quadratic: its gradient is $2X^TXw$.
Edit: consider $f(w)=w^T A w$ for a symmetric matrix $A$. Then
$$
f(u+v)-f(u) = (u+v)^TA(u+v) - u^Tau = 2u^TAv + v^TAv = 2u^TAv + o(|v|),
$$
so $f'(u)v = 2u^TAv$.
answered Dec 7 at 15:33
Federico
4,459512
Yeah but how do you see that $w^T(X^TX)w$ becomes that without "opening" all the multiplications?
– mavavilj
Dec 7 at 15:36
@mavavilj I've added details
– Federico
Dec 7 at 15:52
add a comment |
Let $y=(Xw-t),,$ then the function is
$$eqalign{
phi &= y^Ty cr
dphi &= 2y^Tdy = 2y^T(X,dw) = 2(X^Ty)^Tdw cr
frac{partialphi}{partial w} &= 2X^Ty = 2X^T(Xw-t) cr
}$$
answered Dec 10 at 15:32
greg
7,5151721
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$w^TX^Tt = t^TXw$ is linear: its gradient is $t^TX$.
$w^T(X^TX)w$ is quadratic: its gradient is $2X^TXw$.
Edit: consider $f(w)=w^T A w$ for a symmetric matrix $A$. Then
$$
f(u+v)-f(u) = (u+v)^TA(u+v) - u^Tau = 2u^TAv + v^TAv = 2u^TAv + o(|v|),
$$
so $f'(u)v = 2u^TAv$.
answered Dec 7 at 15:33
Federico
4,459512
Yeah but how do you see that $w^T(X^TX)w$ becomes that without "opening" all the multiplications?
– mavavilj
Dec 7 at 15:36
@mavavilj I've added details
– Federico
Dec 7 at 15:52
add a comment |
$w^TX^Tt = t^TXw$ is linear: its gradient is $t^TX$.
$w^T(X^TX)w$ is quadratic: its gradient is $2X^TXw$.
Edit: consider $f(w)=w^T A w$ for a symmetric matrix $A$. Then
$$
f(u+v)-f(u) = (u+v)^TA(u+v) - u^Tau = 2u^TAv + v^TAv = 2u^TAv + o(|v|),
$$
so $f'(u)v = 2u^TAv$.
answered Dec 7 at 15:33
Federico
4,459512
Yeah but how do you see that $w^T(X^TX)w$ becomes that without "opening" all the multiplications?
– mavavilj
Dec 7 at 15:36
@mavavilj I've added details
– Federico
Dec 7 at 15:52
add a comment |
$w^TX^Tt = t^TXw$ is linear: its gradient is $t^TX$.
$w^T(X^TX)w$ is quadratic: its gradient is $2X^TXw$.
Edit: consider $f(w)=w^T A w$ for a symmetric matrix $A$. Then
$$
f(u+v)-f(u) = (u+v)^TA(u+v) - u^Tau = 2u^TAv + v^TAv = 2u^TAv + o(|v|),
$$
so $f'(u)v = 2u^TAv$.
answered Dec 7 at 15:33
Federico
4,459512
$w^TX^Tt = t^TXw$ is linear: its gradient is $t^TX$.
$w^T(X^TX)w$ is quadratic: its gradient is $2X^TXw$.
Edit: consider $f(w)=w^T A w$ for a symmetric matrix $A$. Then
$$
f(u+v)-f(u) = (u+v)^TA(u+v) - u^Tau = 2u^TAv + v^TAv = 2u^TAv + o(|v|),
$$
so $f'(u)v = 2u^TAv$.
answered Dec 7 at 15:33
Federico
4,459512
edited Dec 7 at 15:45
answered Dec 7 at 15:33
Federico
4,459512
answered Dec 7 at 15:33
Federico
4,459512
answered Dec 7 at 15:33
Federico
4,459512
4,459512
Yeah but how do you see that $w^T(X^TX)w$ becomes that without "opening" all the multiplications?
– mavavilj
Dec 7 at 15:36
@mavavilj I've added details
– Federico
Dec 7 at 15:52
add a comment |
Yeah but how do you see that $w^T(X^TX)w$ becomes that without "opening" all the multiplications?
– mavavilj
Dec 7 at 15:36
@mavavilj I've added details
– Federico
Dec 7 at 15:52
Yeah but how do you see that $w^T(X^TX)w$ becomes that without "opening" all the multiplications?
– mavavilj
Dec 7 at 15:36
Yeah but how do you see that $w^T(X^TX)w$ becomes that without "opening" all the multiplications?
– mavavilj
Dec 7 at 15:36
@mavavilj I've added details
– Federico
Dec 7 at 15:52
@mavavilj I've added details
– Federico
Dec 7 at 15:52
add a comment |
Let $y=(Xw-t),,$ then the function is
$$eqalign{
phi &= y^Ty cr
dphi &= 2y^Tdy = 2y^T(X,dw) = 2(X^Ty)^Tdw cr
frac{partialphi}{partial w} &= 2X^Ty = 2X^T(Xw-t) cr
}$$
answered Dec 10 at 15:32
greg
7,5151721
add a comment |
Let $y=(Xw-t),,$ then the function is
$$eqalign{
phi &= y^Ty cr
dphi &= 2y^Tdy = 2y^T(X,dw) = 2(X^Ty)^Tdw cr
frac{partialphi}{partial w} &= 2X^Ty = 2X^T(Xw-t) cr
}$$
answered Dec 10 at 15:32
greg
7,5151721
add a comment |
Let $y=(Xw-t),,$ then the function is
$$eqalign{
phi &= y^Ty cr
dphi &= 2y^Tdy = 2y^T(X,dw) = 2(X^Ty)^Tdw cr
frac{partialphi}{partial w} &= 2X^Ty = 2X^T(Xw-t) cr
}$$
answered Dec 10 at 15:32
greg
7,5151721
Let $y=(Xw-t),,$ then the function is
$$eqalign{
phi &= y^Ty cr
dphi &= 2y^Tdy = 2y^T(X,dw) = 2(X^Ty)^Tdw cr
frac{partialphi}{partial w} &= 2X^Ty = 2X^T(Xw-t) cr
}$$
answered Dec 10 at 15:32
greg
7,5151721
answered Dec 10 at 15:32
greg
7,5151721
answered Dec 10 at 15:32
greg
7,5151721
answered Dec 10 at 15:32
greg
7,5151721
7,5151721
add a comment |
add a comment |
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