why eigenvalues concerned in Harris corner detection? 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 $$M = \sum_{(x,y)}\left[ \begin{array}{cc}
I_x(x,y)^2,&amp; I_x(x,y)I_y(x,y) \\
I_x(x,y)I_y(x,y),&amp; I_y(x,y)^2 \\
\end{array} \right]$$.
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?
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.