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In linear regression, the loss function is expressed as

$$\frac1N \left\|XW-Y\right\|_{\text{F}}^2$$

where $X, W, Y$ are matrices. Taking derivative w.r.t $W$ yields

$$\frac 2N \, X^T(XW-Y)$$

Why is this so?

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Let

$$\begin{array}{rl} f (\mathrm W) &:= \| \mathrm X \mathrm W - \mathrm Y \|_{\text{F}}^2 = \mbox{tr} \left( (\mathrm X \mathrm W - \mathrm Y)^{\top} (\mathrm X \mathrm W - \mathrm Y) \right)\\ &\,= \mbox{tr} \left( \mathrm W^{\top} \mathrm X^{\top} \mathrm X \mathrm W - \mathrm Y^{\top} \mathrm X \mathrm W - \mathrm W^{\top} \mathrm X^{\top} \mathrm Y + \mathrm Y^{\top} \mathrm Y \right)\end{array}$$

Differentiating with respect to $\mathrm W$,

$$\nabla_{\mathrm W} f (\mathrm W) = 2 \, \mathrm X^{\top} \mathrm X \mathrm W - 2 \, \mathrm X^{\top} \mathrm Y = \color{blue}{2 \, \mathrm X^{\top} \left( \mathrm X \mathrm W - \mathrm Y \right)}$$


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    $\begingroup$ I'm just saying what's the derivative of $ \left\| X W - Y \right\|_{F}^{2} $ with respect to $ X $. Just curious. $\endgroup$ – Royi Sep 8 '17 at 15:28
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    $\begingroup$ @kong You can always use the matrix cookbook. $\endgroup$ – Rodrigo de Azevedo Dec 29 '17 at 12:54
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    $\begingroup$ @kong The derivatives of the linear terms are easy. Just use the properties of the trace and the definition of the Frobenius inner product. The derivative of the quadratic term is not so easy, but one can use the definition of the directional derivative. $\endgroup$ – Rodrigo de Azevedo Dec 29 '17 at 13:03
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    $\begingroup$ @kong Section 2.5.2, equation 108. $\endgroup$ – Rodrigo de Azevedo Dec 29 '17 at 13:40
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    $\begingroup$ @kong No, because $(X^T X)^T = X^T X$. When you transpose a product of matrices, the order is reversed. $\endgroup$ – Rodrigo de Azevedo Dec 29 '17 at 14:17
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Let $X=(x_{ij})_{ij}$ and similarly for the other matrices. We are trying to differentiate $$ \|XW-Y\|^2=\sum_{i,j}(x_{ik}w_{kj}-y_{ij})^2\qquad (\star) $$ with respect to $W$. The result will be a matrix whose $(i,j)$ entry is the derivative of $(\star)$ with respect to the variable $w_{ij}$.

So think of $(i,j)$ as being fixed now. Only some of the terms in $(\star)$ depend on $w_{ij}$. Taking their derivative gives $$ \frac{d\|XW-Y\|^2}{dw_{ij}}=\sum_{k}2x_{ki}(x_{ki}w_{ij}-y_{kj})=\left[2X^T(XW-Y)\right]_{i,j}. $$

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  • $\begingroup$ I found your answer very helpful! Thanks so much :). $\endgroup$ – Elchanan Solomon Sep 20 '18 at 10:04
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Just want to have more details on the process. The process should be Denote $X = [x_{ij}], W = [w_{ij}], Y = [y_{ij}]$, then we have $$ \left \| XW - Y \right \|^{2} = \sum_{k, j} (\sum_{i} x_{ki} w_{ij} - y_{kj})^{2}, $$ This is a scalar and by taking the derivative w.r.t. the matrix $W$ we get a matrix. By taking $i, j$ as the known number, we get $$ \frac{d \left \| XW - Y \right \|^{2}}{d w_{ij}} = \sum_{k} 2x_{ki} (\sum_{i} x_{ki} w_{ij} - y_{kj})\\ = \sum_{k} 2x_{ki} (XW - Y)_{kj} \\ = [2 X^{T} (XW - Y)]_{ij} $$ Thus we have $$ \frac{d \left \| XW - Y \right \|^{2}}{d W} = 2 X^{T} (XW - Y) $$ First time answering a question, hope it is right, thanks!

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