0
$\begingroup$

I am trying to understand equation 26 given equation 25.

enter image description here

I know that generally, if we have an overdetermined system of linear equations of the form

$Ax = b$

the least squares solution is

$\hat{x} = (A^TA)^{-1}A^Tb$

Applying it to the above equation of

$D [R|t] = C$

where $[R|t]$ is the unknown matrix, we get

$\hat{[R|t]} = (D^TD)^{-1}D^TC$

but that's not exactly what's shown in equation 26. How did (26) come from (25) if they are "solving linearly"?

K is a 3x3 “camera matrix”, R is a 3x3 rotation matrix, and t is a 3x1 translation vector

Thanks!

$\endgroup$
5
  • $\begingroup$ Can you define what $K$, $R$, and $t$ are? $\endgroup$
    – Wolfy
    Commented Dec 20, 2018 at 6:34
  • $\begingroup$ Yes, sorry. K is a 3x3 “camera matrix”, R is a 3x3 rotation matrix, and t is a 3x1 translation vector $\endgroup$
    – Carpetfizz
    Commented Dec 20, 2018 at 6:35
  • $\begingroup$ Okay, please edit your question, so other people know as well :). $\endgroup$
    – Wolfy
    Commented Dec 20, 2018 at 6:36
  • $\begingroup$ These are two forms of the pseudo inverse matrix. We select one or the other depending on the dimensions of $D$, when $D$ is not square. When $D$ is square and full rank, no difference. An easy demonstration if $D$ is square: consider SVD ... $\endgroup$
    – Damien
    Commented Dec 20, 2018 at 9:12
  • $\begingroup$ A precision: if $D$ is not square, one of the two forms is invalid ... $\endgroup$
    – Damien
    Commented Dec 20, 2018 at 9:15

1 Answer 1

1
$\begingroup$

In fact, equation 25 is $ [R|t] D = C$ and not $D [R|t] = C$ , and indeed $ [R|t] DD^T = CD^T$ , yielding $\hat{[R|t]}= CD^T*(DD^T)^{-1}$

$\endgroup$
5
  • $\begingroup$ How did you get the second and third expressions? I know they match the text in the OP, but are you actually applying the least squares equation to get to it? $\endgroup$
    – Carpetfizz
    Commented Dec 20, 2018 at 8:31
  • $\begingroup$ I start from $[R|t]D=C$ and just right multiply by $D^T$ . The least squares is valid in exactly the same way, just consider $x^T A=b^T$ in your original definition. $\endgroup$
    – user619894
    Commented Dec 20, 2018 at 9:53
  • $\begingroup$ Thank you, do you mind pointing me to a reference that talks about how to solve $x^TA = b^T$? Does the proof work the same way as the proof for solving $Ax = b$ ? $\endgroup$
    – Carpetfizz
    Commented Dec 21, 2018 at 1:32
  • $\begingroup$ Also, did you mean "consider $x^TA^T = b^T$ ? $\endgroup$
    – Carpetfizz
    Commented Dec 21, 2018 at 2:11
  • 1
    $\begingroup$ The proof works the same way, since $x^T A-b^T = (A^T x-b)^T$ and the object to minimize, the norm squared, is the same. $\endgroup$
    – user619894
    Commented Dec 23, 2018 at 4:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .