I am finding a vector $\mathbf{c}$ that minimizes $J(\mathbf{c})$, that is $J(\mathbf{c}) = \text{E}\left\{\left\vert\mathbf{c}^H\mathbf{r} - a_{t}\right\vert^2\right\}$.

$\mathbf{c}$ and $\mathbf{r}$ are both vector and $a_t$ is some scalar value.

My professor taught us:

We need to find $\mathbf{c}$ such that $$\frac{\partial J(\mathbf{c}) }{ \partial \mathbf{c}^H} = \text{E}\left\{ \mathbf{r}(\mathbf{r}^H\mathbf{c} - a_t^*) \right\} = 0$$

I agree with the direction of solving the above; since $f(t)=\vert t \vert^2$ is convex, it is enough to find the point such that $df(t)/dt=0$.

However, I do not understand how $$ \frac{\partial \left\vert\mathbf{c}^H\mathbf{r} - a_{t}\right\vert^2 }{ \partial \mathbf{c}^H} = \frac{\partial}{\partial \mathbf{c}^H}\left[ (\mathbf{c}^H\mathbf{r}-a_t)(\mathbf{r}^H\mathbf{c}-a_t^*)\right] $$ becomes $$ \mathbf{r}(\mathbf{r}^H\mathbf{c} - a_t^*) $$


First, see here to see why $ \frac{\partial \bar{z}}{\partial z} = 0 $, which we will use shortly.

Define: $$ \mathbf{v} = \frac{\partial}{\partial \mathbf{c}^H} \left\vert \mathbf{c}^H \mathbf{r} - a_t\right\vert^2 = \frac{\partial}{\partial \mathbf{c}^H} \left[ (\mathbf{c}^H\mathbf{r} - a_t)(\mathbf{r}^H\mathbf{c} -a_t^*) \right] $$ where $\mathbf{c},\mathbf{r}$ are vectors and $a_t$ is a scalar.

For simplicity, just consider the $i$th component: \begin{align} v_i &= \frac{\partial}{\partial c^*_i} \left[ (\mathbf{c}^H\mathbf{r} - a_t)(\mathbf{r}^H\mathbf{c} -a_t^*) \right] \\ &= (\mathbf{c}^H\mathbf{r} - a_t) r_i^* \underbrace{\frac{\partial c_i}{\partial c_i^*}}_{0} + (\mathbf{r}^H\mathbf{c} - a_t^*)r_i \\ &= (\mathbf{r}^H\mathbf{c} -a_t^*)r_i \end{align}

Since each component looks like that, the full vector can be written: $$ \mathbf{v} = (\mathbf{r}^H\mathbf{c} -a_t^*)\mathbf{r} $$ Let me know if that makes sense.

  • $\begingroup$ Really nice impressed answer!!! Thank you so much. component-wise differentiation is very understandable. However, I need to memorize $\frac{\partial x}{\partial x^*}=0$ for now. Imaginary calculus is so difficult for me, not a mathematician but an engineer. $\endgroup$ – Danny_Kim May 1 '17 at 3:47
  • $\begingroup$ @Danny_Kim Glad to help :) $\endgroup$ – user3658307 May 1 '17 at 3:50

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