Confused by a Least-Squares Linear Algebra Problem

Question

It's been a long time since I last took linear algebra and I'm confused by the formulation of a problem I ran into. The problem can be simplified into this:

$$\min_{\bar\alpha}\sum_i\lVert\bar\alpha\times\tilde a_i-\bar k_i\rVert^2$$ where I am given the sets of three-dimensional vectors $\{\tilde a_i\}$ and $\{\bar k_i\}$

I remember that via the properties of cross-multiplication I can change $\bar\alpha\times\tilde a_i = -\tilde a_i\times\bar\alpha$, and I can create a skew-symmetric matrix $A_i$ that satisfies $-A_i \bar\alpha=-\tilde a_i\times\bar a$. So now I can change my problem to look like:

$$\min_\bar\alpha \sum_i \lVert A_i \bar\alpha + \bar k_i \rVert^2$$

So looking at my old college notebooks this looks a bit like where I write that the least squares minimization of $\lVert A\bar c - \bar y\rVert^2 \; \forall \bar c$ is:

$$\bar c = (A^T A)^{-1}A^T \bar y$$

My problem is that in my notebooks there's no summation in this example, so I don't know how to apply it to my problem. How do I find $\bar \alpha$ that minimizes my expression?

Answer 1

The summation shouldn't be a problem, since $$\|a\|^2 + \|b\|^2 = \left\|\begin{bmatrix} a \\ b\end{bmatrix}\right\|^2,$$ i.e., you can simply stack them. Applied to your problem: $$ \sum_i \left\|{A}_i \bar{\alpha} + \bar{k}_i\right\|^2 = \left\|\tilde{A}\alpha + \tilde{k}\right\|^2,$$ where $\tilde{A} = \begin{bmatrix} A_1^T & A_2^T & \ldots & A_n^T\end{bmatrix}^T$ and $\tilde{k} = \begin{bmatrix} \bar{k}_1^T & \bar{k}_2^T & \ldots & \bar{k}_n^T\end{bmatrix}^T$. You can then apply Least Squares to the stacked problem as usual.

The answer should be $$\bar{\alpha} = - \tilde{A}^+ \tilde{k} = - (\tilde{A}^T\tilde{A})^{-1} \tilde{A}^T \tilde{k}.$$