0
$\begingroup$

As far as I understand, the structure tensor is:

$$ M = \sum_{(x,y) \in W} \begin{bmatrix} I_x^2 & I_xI_y \\ I_xI_y & I_y^2 \end{bmatrix} = \begin{bmatrix} \sum_{(x,y) \in W} I_x^2 & \sum_{(x,y) \in W} I_xI_y \\ \sum_{(x,y) \in W} I_xI_y & \sum_{(x,y) \in W} I_y^2 \end{bmatrix} $$

How do I compute $I_x^2, I_y^2, I_xI_y$. I used Hadamard product for compute $I_x^2, I_y^2, I_xI_y$. But if I use Hadamard product, I will get $\det(M) = 0$. I'm stuck in here for too long. Can anyone help?

Improve this question
$\endgroup$
16
  • $\begingroup$ Do you know what $I_x,I_y$ are? If so, I don't see any difficulty, beyond programming, which is off-topic here. $\endgroup$
    – Yuval Filmus
    May 20, 2020 at 14:25
  • 1
    $\begingroup$ Where does Hadamard product come in? It's not there in the formulas. $\endgroup$
    – Yuval Filmus
    May 20, 2020 at 14:26
  • 1
    $\begingroup$ Sorry,my misunderstanding. $I_x, I_y$ are just scalars. But if $I_x,I_y$ are scalars. I'll get $det(M)=0$, $tr(M) = \sum_{(x,y)\in W } (I_x^2 + I_y^2)$, and $W$ contains more than single point. Why when $W$ contains more than contains more than single point $det(M) \nequiv 0$? $\endgroup$
    – Blind
    May 20, 2020 at 18:38
  • 1
    $\begingroup$ I agree with your trace formula, but not with your formula for the determinant. Suppose for example that the first point has $(I_x,I_y) = (1,0)$, and the second one has $(I_x,I_y) = (0,1)$ (and there are no other points). Then $M$ is the identity matrix, which has nonzero determinant. $\endgroup$
    – Yuval Filmus
    May 20, 2020 at 18:40
  • 2
    $\begingroup$ The formula to the determinant is just not what you wrote. You're putting the sum in the wrong place. It's not true in general that $\det(A+B) = \det A + \det B$. $\endgroup$
    – Yuval Filmus
    May 20, 2020 at 19:52

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.