In Harris corner detection, as I see, the goal is to find out those points $(x,y)$ which makes $S(x,y)$ a large value for any direction of shifting, $$ S(x,y) = [\Delta x , \Delta y]M\left[ \begin{array}{c} \Delta x \\ \Delta y \\ \end{array} \right]$$, while $$M = \sum_{(x,y)}\left[ \begin{array}{cc} I_x(x,y)^2,& I_x(x,y)I_y(x,y) \\ I_x(x,y)I_y(x,y),& I_y(x,y)^2 \\ \end{array} \right]$$. Here is my problem, as the wiki tells me that $M$ should have 2 large eigenvalues in order to make $S(x,y)$ large in any direction. WHY large eigenvalue? HOW?
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Basic linear algebra. M is a symmetric matrix, so there's a decomposition:
$M=Q^{\mathsf{T}} \Lambda Q$
where Q is a rotation and $\Lambda$ is a diagonal matrix.
We can rotate $[\text{$\Delta $x},\text{$\Delta $y}]$ by $Q^{\mathsf{T}}$
$Q^{\mathsf{T}} [\text{$\Delta $x},\text{$\Delta $y}]^{\mathsf{T}}=\left[\text{$\Delta $x}',\text{$\Delta $y}'\right]^{\mathsf{T}}$
giving
$S(x,y)=\left[\text{$\Delta $x}',\text{$\Delta $y}'\right]\Lambda \left[\text{$\Delta $x}',\text{$\Delta $y}'\right]^{\mathsf{T}}=\Lambda _{1,1} \left(\text{$\Delta $x}'\right)^2+\Lambda _{2,2} \left(\text{$\Delta $y}'\right)^2$
So $S(x,y)$ is large in both directions $\text{$\Delta $x}'$ and $\text{$\Delta $y}'$, if both eigenvalues are large.
