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Derivation of Linear Regression using Normal Equations
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I was going through Andrew Ng's course on ML and had a doubt regarding one of the steps while deriving the solution for linear regression using normal equations.
Normal equation: $theta=(X^TX)^{-1}X^TY$
While deriving, there's this step:
$frac{delta}{deltatheta}theta^TX^TXtheta = X^TXfrac{delta}{deltatheta}theta^Ttheta$
But isn't matrix multiplication commutative, for us to take out $X^TX$ from inside the derivative?
Thanks
matrix-calculus linear-regression
asked Jan 13 at 19:13
Rish1618Rish1618
31
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add a comment |
$begingroup$
I was going through Andrew Ng's course on ML and had a doubt regarding one of the steps while deriving the solution for linear regression using normal equations.
Normal equation: $theta=(X^TX)^{-1}X^TY$
While deriving, there's this step:
$frac{delta}{deltatheta}theta^TX^TXtheta = X^TXfrac{delta}{deltatheta}theta^Ttheta$
But isn't matrix multiplication commutative, for us to take out $X^TX$ from inside the derivative?
Thanks
matrix-calculus linear-regression
asked Jan 13 at 19:13
Rish1618Rish1618
31
$endgroup$
add a comment |
$begingroup$
I was going through Andrew Ng's course on ML and had a doubt regarding one of the steps while deriving the solution for linear regression using normal equations.
Normal equation: $theta=(X^TX)^{-1}X^TY$
While deriving, there's this step:
$frac{delta}{deltatheta}theta^TX^TXtheta = X^TXfrac{delta}{deltatheta}theta^Ttheta$
But isn't matrix multiplication commutative, for us to take out $X^TX$ from inside the derivative?
Thanks
matrix-calculus linear-regression
asked Jan 13 at 19:13
Rish1618Rish1618
31
$endgroup$
I was going through Andrew Ng's course on ML and had a doubt regarding one of the steps while deriving the solution for linear regression using normal equations.
Normal equation: $theta=(X^TX)^{-1}X^TY$
While deriving, there's this step:
$frac{delta}{deltatheta}theta^TX^TXtheta = X^TXfrac{delta}{deltatheta}theta^Ttheta$
But isn't matrix multiplication commutative, for us to take out $X^TX$ from inside the derivative?
Thanks
matrix-calculus linear-regression
matrix-calculus linear-regression
asked Jan 13 at 19:13
Rish1618Rish1618
31
asked Jan 13 at 19:13
Rish1618Rish1618
31
asked Jan 13 at 19:13
Rish1618Rish1618
31
asked Jan 13 at 19:13
Rish1618Rish1618
31
asked Jan 13 at 19:13
Rish1618Rish1618
31
31
add a comment |
add a comment |
2 Answers
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Given two symmetric $(A, B)$ consider these following the scalar functions and their gradients
$$eqalign{
alpha &= theta^TAtheta &implies frac{partialalpha}{partialtheta}=2Atheta cr
beta &= theta^TBtheta &implies frac{partialbeta}{partialtheta}=2Btheta cr
}$$
It's not terribly illuminating, but you can write the second gradient in terms of the first, i.e.
$$frac{partialbeta}{partialtheta} = BA^{-1}frac{partialalpha}{partialtheta}$$
For the purposes of your question, $A=I$ and $B=X^TX$.
answered Jan 13 at 21:30
greggreg
8,2751823
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Although that equality is true, it does not give insight into why it is true.
There are many ways to compute that gradient, but here is a direct approach that simply computes all the partial derivatives individually.
Let $A$ be a symmetric matrix. (In your context, $A= X^top X$.)
The partial derivative of $theta^top A theta = sum_i sum_j A_{ij} theta_i theta_j$ with respect to $theta_k$ is
$$frac{partial}{partial theta_k} theta^top A theta = sum_i sum_j A_{ij} frac{partial}{partial theta_k}(theta_i theta_j) = A_{kk} cdot 2 theta_k + sum_{i ne k} A_{ik} cdot theta_ i + sum_{j ne k} A_{kj} theta_j = 2sum_i A_{ki} theta_i = 2 (A theta)_k$$
Stacking the partial derivatives into a vector gives you the gradient, so
$$nabla_theta theta^top A theta = 2 A theta.$$
answered Jan 13 at 19:29
angryavianangryavian
41k23380
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2 Answers
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2 Answers
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$begingroup$
Given two symmetric $(A, B)$ consider these following the scalar functions and their gradients
$$eqalign{
alpha &= theta^TAtheta &implies frac{partialalpha}{partialtheta}=2Atheta cr
beta &= theta^TBtheta &implies frac{partialbeta}{partialtheta}=2Btheta cr
}$$
It's not terribly illuminating, but you can write the second gradient in terms of the first, i.e.
$$frac{partialbeta}{partialtheta} = BA^{-1}frac{partialalpha}{partialtheta}$$
For the purposes of your question, $A=I$ and $B=X^TX$.
answered Jan 13 at 21:30
greggreg
8,2751823
$endgroup$
add a comment |
$begingroup$
Given two symmetric $(A, B)$ consider these following the scalar functions and their gradients
$$eqalign{
alpha &= theta^TAtheta &implies frac{partialalpha}{partialtheta}=2Atheta cr
beta &= theta^TBtheta &implies frac{partialbeta}{partialtheta}=2Btheta cr
}$$
It's not terribly illuminating, but you can write the second gradient in terms of the first, i.e.
$$frac{partialbeta}{partialtheta} = BA^{-1}frac{partialalpha}{partialtheta}$$
For the purposes of your question, $A=I$ and $B=X^TX$.
answered Jan 13 at 21:30
greggreg
8,2751823
$endgroup$
add a comment |
$begingroup$
Given two symmetric $(A, B)$ consider these following the scalar functions and their gradients
$$eqalign{
alpha &= theta^TAtheta &implies frac{partialalpha}{partialtheta}=2Atheta cr
beta &= theta^TBtheta &implies frac{partialbeta}{partialtheta}=2Btheta cr
}$$
It's not terribly illuminating, but you can write the second gradient in terms of the first, i.e.
$$frac{partialbeta}{partialtheta} = BA^{-1}frac{partialalpha}{partialtheta}$$
For the purposes of your question, $A=I$ and $B=X^TX$.
answered Jan 13 at 21:30
greggreg
8,2751823
$endgroup$
Given two symmetric $(A, B)$ consider these following the scalar functions and their gradients
$$eqalign{
alpha &= theta^TAtheta &implies frac{partialalpha}{partialtheta}=2Atheta cr
beta &= theta^TBtheta &implies frac{partialbeta}{partialtheta}=2Btheta cr
}$$
It's not terribly illuminating, but you can write the second gradient in terms of the first, i.e.
$$frac{partialbeta}{partialtheta} = BA^{-1}frac{partialalpha}{partialtheta}$$
For the purposes of your question, $A=I$ and $B=X^TX$.
answered Jan 13 at 21:30
greggreg
8,2751823
answered Jan 13 at 21:30
greggreg
8,2751823
answered Jan 13 at 21:30
greggreg
8,2751823
answered Jan 13 at 21:30
greggreg
8,2751823
8,2751823
add a comment |
add a comment |
$begingroup$
Although that equality is true, it does not give insight into why it is true.
There are many ways to compute that gradient, but here is a direct approach that simply computes all the partial derivatives individually.
Let $A$ be a symmetric matrix. (In your context, $A= X^top X$.)
The partial derivative of $theta^top A theta = sum_i sum_j A_{ij} theta_i theta_j$ with respect to $theta_k$ is
$$frac{partial}{partial theta_k} theta^top A theta = sum_i sum_j A_{ij} frac{partial}{partial theta_k}(theta_i theta_j) = A_{kk} cdot 2 theta_k + sum_{i ne k} A_{ik} cdot theta_ i + sum_{j ne k} A_{kj} theta_j = 2sum_i A_{ki} theta_i = 2 (A theta)_k$$
Stacking the partial derivatives into a vector gives you the gradient, so
$$nabla_theta theta^top A theta = 2 A theta.$$
answered Jan 13 at 19:29
angryavianangryavian
41k23380
$endgroup$
add a comment |
$begingroup$
Although that equality is true, it does not give insight into why it is true.
There are many ways to compute that gradient, but here is a direct approach that simply computes all the partial derivatives individually.
Let $A$ be a symmetric matrix. (In your context, $A= X^top X$.)
The partial derivative of $theta^top A theta = sum_i sum_j A_{ij} theta_i theta_j$ with respect to $theta_k$ is
$$frac{partial}{partial theta_k} theta^top A theta = sum_i sum_j A_{ij} frac{partial}{partial theta_k}(theta_i theta_j) = A_{kk} cdot 2 theta_k + sum_{i ne k} A_{ik} cdot theta_ i + sum_{j ne k} A_{kj} theta_j = 2sum_i A_{ki} theta_i = 2 (A theta)_k$$
Stacking the partial derivatives into a vector gives you the gradient, so
$$nabla_theta theta^top A theta = 2 A theta.$$
answered Jan 13 at 19:29
angryavianangryavian
41k23380
$endgroup$
add a comment |
$begingroup$
Although that equality is true, it does not give insight into why it is true.
There are many ways to compute that gradient, but here is a direct approach that simply computes all the partial derivatives individually.
Let $A$ be a symmetric matrix. (In your context, $A= X^top X$.)
The partial derivative of $theta^top A theta = sum_i sum_j A_{ij} theta_i theta_j$ with respect to $theta_k$ is
$$frac{partial}{partial theta_k} theta^top A theta = sum_i sum_j A_{ij} frac{partial}{partial theta_k}(theta_i theta_j) = A_{kk} cdot 2 theta_k + sum_{i ne k} A_{ik} cdot theta_ i + sum_{j ne k} A_{kj} theta_j = 2sum_i A_{ki} theta_i = 2 (A theta)_k$$
Stacking the partial derivatives into a vector gives you the gradient, so
$$nabla_theta theta^top A theta = 2 A theta.$$
answered Jan 13 at 19:29
angryavianangryavian
41k23380
$endgroup$
Although that equality is true, it does not give insight into why it is true.
There are many ways to compute that gradient, but here is a direct approach that simply computes all the partial derivatives individually.
Let $A$ be a symmetric matrix. (In your context, $A= X^top X$.)
The partial derivative of $theta^top A theta = sum_i sum_j A_{ij} theta_i theta_j$ with respect to $theta_k$ is
$$frac{partial}{partial theta_k} theta^top A theta = sum_i sum_j A_{ij} frac{partial}{partial theta_k}(theta_i theta_j) = A_{kk} cdot 2 theta_k + sum_{i ne k} A_{ik} cdot theta_ i + sum_{j ne k} A_{kj} theta_j = 2sum_i A_{ki} theta_i = 2 (A theta)_k$$
Stacking the partial derivatives into a vector gives you the gradient, so
$$nabla_theta theta^top A theta = 2 A theta.$$
answered Jan 13 at 19:29
angryavianangryavian
41k23380
answered Jan 13 at 19:29
angryavianangryavian
41k23380
answered Jan 13 at 19:29
angryavianangryavian
41k23380
answered Jan 13 at 19:29
angryavianangryavian
41k23380
41k23380
add a comment |
add a comment |
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