How to take the gradient of the quadratic form?

It's stated that the gradient of:

$$\frac{1}{2}x^TAx - b^Tx +c$$

is

$$\frac{1}{2}A^Tx + \frac{1}{2}Ax - b$$

How do you grind out this equation? Or specifically, how do you get from $x^TAx$ to $A^Tx + Ax$?

• did you understand the answer? Dec 4, 2017 at 1:54

The only thing you need to remember/know is that $$\dfrac{\partial (x^Ty)}{\partial x} = y$$ and the chain rule, which goes as $$\dfrac{d(f(x,y))}{d x} = \dfrac{\partial (f(x,y))}{\partial x} + \dfrac{d( y^T(x))}{d x} \dfrac{\partial (f(x,y))}{\partial y}$$ Hence, $$\dfrac{d(b^Tx)}{d x} = \dfrac{d (x^Tb)}{d x} = b$$

$$\dfrac{d (x^TAx)}{d x} = \dfrac{\partial (x^Ty)}{\partial x} + \dfrac{d (y(x)^T)}{d x} \dfrac{\partial (x^Ty)}{\partial y}$$ where $$y = Ax$$. And then, that is,

$$\dfrac{d (x^TAx)}{d x} = \dfrac{\partial (x^Ty)}{\partial x} + \dfrac{d( y(x)^T)}{d x} \dfrac{\partial (x^Ty)}{\partial y} = y + \dfrac{d (x^TA^T)}{d x} x = y + A^Tx = (A+A^T)x$$

• To help future generations: the full specification of the chain rule used here is $$\frac{df(g,h)}{dx} = \frac{d(g(x)^T)}{dx} \frac{\partial f(g,h)}{\partial g} + \frac{d(h(x)^T)}{dx} \frac{\partial f(g,h)}{\partial h}$$ The order of multiplication is very important since we're dealing with vectors! Sep 23, 2014 at 9:58
• the first statement seems wrong to me. Isn't the right statement $\nabla_x(x^Ty) = y$? @NeilTraft Oct 21, 2017 at 23:22
• Like for example how does the answerer know where $\dfrac{\partial y^T}{\partial x}$ goes on the left or on the right or if there is a transpose or not? Or maybe I just unfamiliar with the chain rule using gradients and I only know it using partial derivatives. Oct 21, 2017 at 23:37
• Where can someone learn about these differentiation rules? In my standard analysis and calculus courses, we didn't see the differentiation of matrices or vectors, only of multivariate functions Oct 15, 2018 at 9:44
• @Euler_Salter just write it as a multivariate function of the components of the vector $x$ and take the gradient. You will get the same result. Jan 20, 2019 at 21:49

Let $$f : \mathbb R^n \to \mathbb R$$ be defined by

$$f (\mathrm x) := \rm x^\top A \, x$$

Hence,

$$f (\mathrm x + h \mathrm v) = (\mathrm x + h \mathrm v)^\top \mathrm A \, (\mathrm x + h \mathrm v) = f (\mathrm x) + h \, \mathrm v^\top \mathrm A \,\mathrm x + h \, \mathrm x^\top \mathrm A \,\mathrm v + h^2 \, \mathrm v^\top \mathrm A \,\mathrm v$$

Thus, the directional derivative of $$f$$ in the direction of $$\rm v$$ at $$\rm x$$ is

$$\lim_{h \to 0} \frac{f (\mathrm x + h \mathrm v) - f (\mathrm x)}{h} = \mathrm v^\top \mathrm A \,\mathrm x + \mathrm x^\top \mathrm A \,\mathrm v = \langle \mathrm v , \mathrm A \,\mathrm x \rangle + \langle \mathrm A^\top \mathrm x , \mathrm v \rangle = \left\langle \mathrm v , \color{blue}{\left(\mathrm A + \mathrm A^\top\right) \,\mathrm x} \right\rangle$$

Lastly, the gradient of $$f$$ with respect to $$\rm x$$ is

$$\nabla_{\mathrm x} \, f (\mathrm x) = \color{blue}{\left(\mathrm A + \mathrm A^\top\right) \,\mathrm x}$$

• +1. What are that tags in your answer? Sep 30, 2020 at 14:31
• @C.F.G The tags are there to ensure I can find my answer even if the question is vandalized, tags are merged or renamed, etc. Insurance against others' misbehavior. Sep 30, 2020 at 14:34
• how were you able to swap the x^TA to A^Tx in the line with the inner products to arrive at <v, (A + A^T) x> Oct 17, 2020 at 20:53
• @Learningstatsbyexample Note that $\rm x^\top A v$ is a scalar, and a scalar is equal to its transpose. Hence, $$\rm x^\top A v = \left( \rm x^\top A v \right)^\top = \rm v^\top A^\top x$$ Does this clarify the matter? Oct 17, 2020 at 21:00
• < 7 minute response time; that is service :) Yes that helps, thanks! Oct 17, 2020 at 21:06

I am just writing this answer for future reference and for clarity because the accepted answer is not completely correct and may case confusion.

I will use a simple proof to better understand the multiplication rule in the calculation of $$\nabla\mathbf{x^TAx}$$.

For $$\mathbf{x}\in \mathbb{R}^n$$ and $$\mathbf{A}\in \mathbb{R}^{n\times n}$$ let: $$f(g(\mathbf{x}),h(\mathbf{x}))=\langle g(\mathbf{x}),h(\mathbf{x})\rangle=g^T(\mathbf{x})h(\mathbf{x})$$

Where: $$$$\begin{split} &g(\mathbf{x})=\mathbf{x}\\ &h(\mathbf{x})=\mathbf{Ax} \end{split}$$$$

From the definition of $$f$$, it is obvious that $$f(g(\mathbf{x}),h(\mathbf{x}))=\mathbf{x^TAx}$$.

In order to calculate the derivative, we will use the following fundamental properties, where $$\mathbf{I}$$ is the identity matrix:

$$$$\begin{split} &\dfrac{\partial \mathbf{A^Tx}}{\partial\mathbf{x}}=\dfrac{\partial \mathbf{x^TA}}{\partial\mathbf{x}}=\mathbf{A^T}\\ &\dfrac{\partial \mathbf{x}}{\partial\mathbf{x}}=\mathbf{I} \end{split}$$$$

Hence, from the multiplication rule (you can see the rule at wikipedia), we got: $$$$\begin{split} \dfrac{df(g(\mathbf{x}),h(\mathbf{x}))}{d\mathbf{x}}&=g^T(\mathbf{x})\dfrac{\partial h(\mathbf{x})}{\partial\mathbf{x}}+h^T(\mathbf{x})\dfrac{\partial g(\mathbf{x})}{\partial\mathbf{x}}=\\ &=\mathbf{x^T}\dfrac{\partial \mathbf{Ax}}{\partial\mathbf{x}}+(\mathbf{Ax})^T\dfrac{\partial \mathbf{x}}{\partial\mathbf{x}}=\\ &=\mathbf{x^TA}+\mathbf{x^TA^TI}=\\ &=\mathbf{x^TA}+\mathbf{x^TA^T}=\\ &=\mathbf{x^T}(\mathbf{A+A^T}) \end{split}$$$$

As a result, from the definition of gradient, we got: $$\nabla f=\Bigg(\dfrac{df}{d\mathbf{x}}\Bigg)^T=(\mathbf{x^T}(\mathbf{A+A^T}))^T=(\mathbf{A^T+A})\mathbf{x}$$

Note: The reason I did the proof this way is to be more generalizable. So you can can plug arbitrary functions $$g$$ and $$h$$ and use the above multiplication rule to derive the result.

• Thank you. Till date this is the simplest of all answers. However, reading this first might be helpful. Aug 3, 2020 at 7:56

There is another way to calculate the most complex one, $$\frac{\partial}{\partial \theta_k} \mathbf{x}^T A \mathbf{x}$$. It only requires nothing but partial derivative of a variable instead of a vector.

This answer is for those who are not very familiar with partial derivative and chain rule for vectors, for example, me. Therefore, although it seems long, it is actually because I write down all the details. :)

Firstly, expanding the quadratic form yields: \begin{align} f = \frac{\partial}{\partial \theta_k} \mathbf{x}^T A \mathbf{x} &= \frac{\partial}{\partial \theta_k} \sum_{i=1}^N \sum_{j=1}^N a_{ij}\frac{\partial}{\partial \theta_k}(\mathbf{x}_i \mathbf{x}_j) \end{align} Since $$\frac{\partial}{\partial \theta_k}(\mathbf{x}_i \mathbf{x}_j) = \begin{cases} \mathbf{x}_j, && \text{if } k = i \\ \mathbf{x}_i, && \text{if } k = j \\ 0, && \text{otherwise} \end{cases}$$ The equation is nothing but $$f = \sum_{j=1}^N a_{kj} \mathbf{x}_j + \sum_{i=1}^N a_{ik} \mathbf{x}_i$$ Almost done! Now we only need some simplification. Recall the very simple rule that $$\sum_{i=1}^N x_i y_i = \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix}^T \begin{bmatrix} y_1 \\ \vdots \\ y_n \end{bmatrix} = \mathbf{x}^T \mathbf{y}$$ Thus \begin{align} f &= \text{(k-th row of A) } \mathbf{x} + \text{(k-th column of A)}^T \mathbf{x} \end{align} Now it is time to compute the gradient from partial derivative! \begin{align} \nabla_\mathbf{x} \mathbf{x}^T A \mathbf{x} & = \begin{bmatrix} \frac{\partial \mathbf{x}^T A \mathbf{x}}{\partial x_1} \\ \vdots \\ \frac{\partial \mathbf{x}^T A \mathbf{x}}{\partial x_k} \\ \vdots \\ \frac{\partial \mathbf{x}^T A \mathbf{x}}{\partial x_N} \\ \end{bmatrix} = \begin{bmatrix} \vdots \\ \text{(k-th row of A) } \mathbf{x} + \text{(k-th column of A)}^T \mathbf{x} \\ \vdots \end{bmatrix} \\ &= \left( \begin{bmatrix} \vdots \\ \text{(k-th row of A) } \\ \vdots \end{bmatrix} + \begin{bmatrix} \vdots \\ \text{(k-th column of A) }^T \\ \vdots \end{bmatrix} \right) \mathbf{x} \\ &= (A + A^T)\mathbf{x} \end{align} So we are done!! The answer is: $$\nabla_\mathbf{x} \mathbf{x}^T A \mathbf{x} = (A + A^T)\mathbf{x}$$

Yet another approach.

We will utilize the following the identities

• Trace and Frobenius product relation $$\left\langle A, B \right\rangle={\rm tr}(A^TB) = A:B$$ or $$\left\langle A^T, B \right\rangle ={\rm tr}(AB) = A^T:B$$
• Cyclic property of Trace/Frobenius product \begin{align} \left\langle A, B C \right\rangle \equiv A:BC &= AC^T:B\\ &= B^TA:C\\ &= BC : A\\ &= {\text{etc.}} \cr \end{align}

Let $$f(x) := \left( \frac{1}{2} x^T A x - b^T x + c \right)$$.

We obtain the differential first, and then the gradient subsequently. \begin{align} d\,f(x) &= d\left( \frac{1}{2} x^T A x - b^T x + c \right) \\ &= d\left( \frac{1}{2} \left( x: A x \right) - \left( b : x \right) + c \right) \\ &= \frac{1}{2} \left[ \left( dx: A x \right) + \left( x: A dx \right) \right] - \left( b : dx \right) \\ &= \frac{1}{2} \left[ \left( A x : dx \right) + \left( A^Tx: dx \right) \right] - \left( b : dx \right) \\ &= \frac{1}{2} \left[ \left(A + A^T \right)x: dx \right] - \left( b : dx \right) \\ &= \left( \frac{1}{2} \left[ \left(A + A^T \right)x\right] - b \right): dx \\ \end{align}

Thus the gradient is \eqalign { \frac { \partial} {\partial x}f(x) &= \frac{1}{2} \left[ \left(A + A^T \right)x\right] - b \\ &= \frac{1}{2} A^T x + \frac{1}{2} A x - b . \cr }