# Derivative of squared Frobenius norm of a matrix

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?

• Don't be self-deprecating, your question is not dumb :) +1 – Andres Mejia Feb 4 '17 at 7:23

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 $\mathrm W$,

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

• I'm just saying what's the derivative of $\left\| X W - Y \right\|_{F}^{2}$ with respect to $X$. Just curious. – Royi Sep 8 '17 at 15:28
• I get it is $2 \left( X W - Y \right) {W}^{T}$. – Royi Sep 8 '17 at 15:30
• @Royi I just got the exact same. – Rodrigo de Azevedo Sep 8 '17 at 15:32
• @kong If you want a step-by-step derivation, use the directional derivative. – Rodrigo de Azevedo Dec 29 '17 at 12:48
• @kong You can always use the matrix cookbook. – Rodrigo de Azevedo Dec 29 '17 at 12:54

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 $$\frac{d\|XW-Y\|^2}{dw_{ij}}=\sum_{k}2x_{ki}(x_{ki}w_{ij}-y_{kj})=\left[2X^T(XW-Y)\right]_{i,j}.$$

• I found your answer very helpful! Thanks so much :). – Elchanan Solomon Sep 20 '18 at 10:04