The context of this question is that I'm taking a convex optimization class and was studying about Lagrangians. I was looking at an example problem when I experienced some difficulty understanding something.

The optimization problem in question is a simple least-squares solution of linear equations:

$\text{minimize}\quad \ x^Tx$

$\text{subject to}\quad Ax = b$

The Lagrangian of this problem is

$$L(x, \nu) = x^Tx + \nu^T(Ax - b)$$

When we're finding the dual function $g(\nu)$, we use the gradient of the Lagrangian:

$$\nabla_xL(x, \nu) = 2x + A^T\nu=0$$

We can obtain the solution $x = -\frac{1}{2}A^Tv$ from this. Plugging this into the dual function I get:

$$ \begin{align} g(\nu)& = L(-\frac{1}{2}A^T\nu,\ \nu)\\ & = (-\frac{1}{2}A^T\nu)^T(-\frac{1}{2}A^T\nu) + \nu^T(A(-\frac{1}{2}A^T\nu) - b) \\ & = \frac{1}{4}\nu^TAA^T\nu - \frac{1}{2}\nu^TAA^T\nu - \nu^Tb \\ & = -\frac{1}{4}\nu^TAA^T\nu - \nu^Tb\end{align}$$

However, the dual function that's in the textbook is:

$$g(\nu) = -\frac{1}{4}\nu^TAA^T\nu - b^T\nu$$

I have two questions regarding this derivation:

  1. When we find the gradient of the Lagrangian, I initially thought the second term would be $\nu^TA$ because we find the derivative of $\nu^TAx$. This probably stems from my lack of understanding of vector differentiation but why is it $A\nu^T$?
  2. This is similar to question 1, but in the derivation of the dual function I got $\nu^Tb$ but apparently the correct term is $b^T\nu$. When we unfold the parentheses, do vector placements usually change?

The second is easily disposed of: $\nu^Tb$ is a scalar—the dot product of $\nu$ and $b$—therefore $$\nu^Tb=(\nu^Tb)^T=b^T\nu.$$ It doesn’t matter which one comes first.

As for $\nu^TAx$, its differential is indeed $\nu^TA$, but since you’re working with column vectors, $\nu^TA$ has the wrong shape: the gradient of a scalar-valued function is also a column vector. Expanding by coordinates, we have $\nu^TAx = \sum_{i,j} a_{ij} \nu_i x_j$ so that $\partial_j(\nu^TAx) = \sum_i a_{ij}\nu_i$, that is, the $j$th element of the column vector $\nabla(\nu^TAx)$ is the dot product of $\nu$ with the $j$th column of $A$, from which $\nabla(\nu^TAx) = A^T\nu$.

  • $\begingroup$ Hi, thanks for the answer and clearing things up. This may sound like a weird question, but is there a particular rule for this when performing multivariate calculus? Or do I just have to keep in mind the shapes of the vectors/matrices? $\endgroup$ – Seankala Jun 15 at 15:27
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    $\begingroup$ @Seankala Keep in mind that the gradient of a scalar-valued function is a vector of the same shape as the input to the function. When in doubt, expand by coordinates. It can be less confusing to compute the differential of the function instead, especially if it’s a composition of vector- and scalar-valued functions, then use the fact that the gradient is the transpose of the differential. $\endgroup$ – amd Jun 17 at 2:12
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    $\begingroup$ @Seankala I should note that the reason that one might choose to write the second term in $g(\nu)$ as $b^T\nu$ instead of $\nu^Tb$ is that a conventional way of writing the equation of a quadric in matrix form in which $\mathbf x$ is the variable is $\mathbf x^TQ\mathbf x+b^T\mathbf x+c=0$. Constant coefficients of a term are usually written before the variables. $\endgroup$ – amd Jun 17 at 2:16
  • $\begingroup$ Thanks for the explanation! $\endgroup$ – Seankala Jun 17 at 2:23

$\newcommand{\grad}{\nabla_x}$ 1) Remember, $\grad c^T x= c$, if $c$ is a constant (column) vector. An equivalent way to write this is that if $r$ is a row vector (and $x$ column vector), then $\grad rx = r\color{blue}{^T}$, i.e. we must remember to transpose the thing multiplying $x$ (put $r = c^T$ from above to see this, so $c = r^T$).

In your case, the row vector $r$ is $\nu^T A$, so $$\grad \left(\nu^T A x\right)=r^T = A^T \nu.$$

2) Recall that $\color{blue}{v^T w = w^T v}$ for all $v,w \in \Bbb{R}^n$ (since both equal $\sum\limits_{i=1}^{n} v_i w_i$). Hence $\nu^T b$ and $b^T \nu$ are both correct.


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