# $P_{1c} = AP$ , $P_{2c} = BP$. How to find $P$? (being that $A$ and $B$ are $3\times 4$ matrices and $P$ is a $4\times 1$ vector)

This problem arose in my stereo vision project. $$P_{1c} = A*P$$

$$P_{2c} = B*P$$ where: $P_{1c}$ and $P_{2c}$ are $3\times1$ vectors, $A$ and $B$ are $3 \times 4$ matrices and $P$ is a $4\times1$ vector.

Or, more verbosely:

$$\left( \begin{array}{ccc} a_1 \\ b_1 \\ w_1 \end{array} \right) = \left( \begin{array}{ccc} A_{11} & A_{12} & A_{13} & A_{14} \\ A_{21} & A_{22} & A_{23} & A_{24} \\ A_{31} & A_{32} & A_{33} & A_{34} \\ \end{array} \right) * \left( \begin{array}{ccc} X \\ Y \\ Z \\ 1 \\ \end{array} \right)$$

$$\left( \begin{array}{ccc} a_2 \\ b_2 \\ w_2 \end{array} \right) = \left( \begin{array}{ccc} B_{11} & B_{12} & B_{13} & B_{14} \\ B_{21} & B_{22} & B_{23} & B_{24} \\ B_{31} & B_{32} & B_{33} & B_{34} \\ \end{array} \right) * \left( \begin{array}{ccc} X \\ Y \\ Z \\ 1 \\ \end{array} \right)$$ $P_{1c}, P_{2c}, A$ and $B$ are known, find $P$.

An explicit solution is a nice start, and in fact I already have it:

(Explicit Solution - page 6)

but from the nature of the problem, I can tell that it will not always be possible to find $P$ using the explicit solution (since I only have estimates for $P_{1c},P_{2c},A$ and $B$), so I was trying to tackle the problem like this:

$f(P) = (A*P-P_{1c})^2+(B*P-P_{2c})^2$

Minimize $f(P)$ with respect to $P$.

I believe this notation is not as rigorous as one would expect, since $P$ is a vector. So, to put it in another way:

Knowing $P_{1c},P_{2c},A$ and $B$, find the vector $P$ wich gives the "best-fit" to $$P_{1c} = A*P$$

$$P_{2c} = B*P$$ (in a least squares sense).

Any help would be great, and please feel free to approach this problem with a different strategy than the one I'm trying, but I would appreciate if someone could answer this without "opening up" the matrices (instead of using the $A_{ij}$'s and working out with the explicit equations, only use the matrices, and matrix operations like $A^T, A^{-1}$), if that's even possible.

You have one matrix equation:

$$\pmatrix{P_{1c} \\ P_{2c}} = \pmatrix{A \\ B}\cdot P \tag{1}$$

It is an overdetermined system (unless there are too many dependencies in the data).

Here is what you want for the least squares solution to it:

$$P=\left[\pmatrix{A^\top & B^\top}\pmatrix{A \\ B}\right]^{-1}\pmatrix{A^\top & B^\top}\pmatrix{P_{1c} \\ P_{2c}}$$

There are methods that are more efficient than this explicit formula here, that avoids the extra calculation of squaring then inverting. You may notice that I got this using what is called a left inverse of $\pmatrix{A \\ B}$ on the equation (1). There are possibly more than one left inverse in general, but this one is the specific left inverse that gives the least squares solution.

Let $M = \pmatrix{A \\ B}$. Using user7530's helpful comment, it may be factored as $M=QR$, which is a standard form available in any computer package, simply called the QR factorization. I will show how to use that form later, but first the derivation of the formula I gave you.

To solve for $P$ in the equation $$X= MP \tag{2}$$ where $M$ has more rows than columns, consider a left inverse $\hat{M}$ such that $\hat{M}M=I$. At first glance, assuming the existence of $\hat{M}$, this seems to be the solution since it gives $$\hat{M}X = \hat{M}MP = IP = P$$ and $$P=\hat{M}X$$ gives the solution. But the left inverse may not be unique. So the question is which left inverse do we want. Consider equation (2). We are looking for some $P$ to solve it, but it may not be possible. So instead we look at the closest possible solution. Any $P$ when right multiplied with $M$ will necessarily give a result that is in the column span of $M$, since a right multiply can only mix the columns of $M$. This along with one-sided inverses turns out to be all the considerations necessary to find the unique solution.

Since the result of $\hat{M}X$ should have the same column-span as $M$, this gives a unique $\hat{M}$ and it is

$$\hat{M} = \left[M^\top M\right]^{-1}M^\top$$

To use the QR factorization, we have $M=QR$. Then \begin{align} P&=\hat{M} X \\ &= \left[M^\top M\right]^{-1}M^\top X\\ & = \left[R^\top Q^\top Q R\right]^{-1} M^\top X\\ & = \left[R^\top R\right]^{-1} M^\top X\\ \end{align}

Where the last line there is easier to compute since $R$ is triangular with zero rows.

• +1 One good "more efficient" approach is to compute the QR decomposition of $\left(\begin{array}{c}A\\B\end{array}\right)$. Jan 23, 2013 at 17:03
• Could you post a derivation for your answer? Or maybe cite a reference that would guide me to a better understanding of how you got there? Thanks! Jan 23, 2013 at 17:04
• I am more of a self study type of person and am not too good at references, and I personally find the usual Wikipedia links to be difficult; the articles never explain terribly well in my opinion. But I would be happy to expand on my answer, and will add to it now. Jan 23, 2013 at 17:16
• Just to be clear, I think I understood how you got to P=[(A⊤B⊤)(AB)]−1(A⊤B⊤)(P1cP2c) (multiply both sides by the transpose and then by the inverse of the product between the transpose and AB). But why is this the least squares answer? Jan 23, 2013 at 17:22
• It has to do with the row-span: the least squares solution lies in the row-span of $\pmatrix{A \\ B}$ Jan 23, 2013 at 17:38