0
$\begingroup$

Suppose we have a function $f: \mathbb{R}^{d-1} \to \mathbb{R}$ given by

$$ f(x_1, \ldots, x_{d-1}) = \operatorname{trace} \left( A^T \mathbf{1}_{n\times 1} [ x_1, x_2, \ldots, x_{d-1}] B\right)$$

where $A,B$ are $n\times d$ and $(d-1)\times d$ matrices respectively. I'm trying to compute the gradient of this function. Currently I am just trying to expand out the entire expression inside the trace and compute the gradient that way, but it's quite messy and I'm not able to push it through. Can anyone point out a better approach? Thank you.

$\endgroup$
  • $\begingroup$ Is $\mathbf{1}_{n \times 1}$ the column vector with all entries equal to $1$? $\endgroup$ – Nick May 30 '19 at 18:11
  • $\begingroup$ @Nick Yes. Sorry, I should have mentioned that above. $\endgroup$ – Katie Dobbs May 30 '19 at 18:13
1
$\begingroup$

If you know about the Frechet derivative and its relation to the gradient of a function, this question becomes almost a triviality. For this question, the key result is the following:

Theorem:

If $V$ and $W$ are finite-dimensional real vector spaces, and $f: V \to W$ is a linear transformation, then for every $x \in V$, $f$ is differentiable at $x$ and its derivative at $x$ (which is a linear map from $V$ into $W$) is given by \begin{equation} df_x(\cdot) = f(\cdot) \end{equation}

What this theorem says is that a linear transformation is its own best linear approximation (i.e it is its own derivative). Now, note that the function $f: \mathbb{R}^{d-1} \to \mathbb{R}$ you have defined is linear (because trace is linear, and matrix multiplication is also linear). So, for every $x = (x_1, \dots, x_{d-1}) \in \mathbb{R}^{d-1}$, by the theorem above, we have \begin{equation} df_x(\cdot) = f(\cdot) \end{equation} In general, the gradient of $f$ at $x$ is the matrix of $df_x$ relative to the standard basis. So, \begin{align} \nabla f(x) &= \text{matrix of $df_x$ wrt standard basis} \\ &= \text{matrix of $f$ wrt standard basis} \\ &= \begin{bmatrix} f(e_1), & \dots &, f(e_{d-1}) \end{bmatrix} \end{align}

In other words, $\nabla f(x)$ is the $1 \times (d-1)$ matrix whose $i^{th}$ entry is $f(e_i)$; I'll leave it to you to compute what $f(e_i)$ is.


As you mentioned, expanding everything out in terms of components and taking partial derivatives is a nightmare in this case. This is why, if you're not already familiar with the Frechet derivative, I highly recommend you learn more about it... it simplifies computations like this immensely. As a reference, I would highly recommend Loomis and Sternberg's book Advanced Calculus (section 3.6 in particular) to learn about this.

$\endgroup$
0
$\begingroup$

I may have hit submit on this question too early!

We could use the cyclic property of traces to make the function appear in the form $$ \operatorname{trace} \left([x_1, x_2, \ldots, x_{d-1}] C\right) $$

where $C$ is a $(d-1) \times 1$ matrix (i.e. a column vector with $d-1$ entries). In terms of the variables above, $C = BA^T \mathbf{1}_{n\times 1}.$ The product inside the trace is now just a $1 \times 1$ matrix, i.e. we have

$$f(x_1, x_2, \ldots, x_{d-1}) = [x_1,x_2,\ldots, x_{d-1}] C$$

so $\nabla f(x) = C.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.