How to perform Spatial derivative calculation?

Question

I'm trying to solve the following question about "Harris Corner Detection".

Consider the following image:

As the first step of the "Harris Corner Detection", we should compute the derivatives using the differentiation kernels shown above. No normalization (division by 2) is needed.

What I did: $$ I_{x}=\frac{\partial I(x,y)}{\partial x}=I(x+1,y)-I(x-1,y)=\begin{bmatrix}4 & 9 & 12\\ 8 & 11 & 14\\ 10 & 15 & 16 \end{bmatrix}-\begin{bmatrix}0 & 1 & 4\\ 0 & 5 & 7\\ 4 & 9 & 12 \end{bmatrix}=\begin{bmatrix}4 & 8 & 8\\ 8 & 6 & 7\\ 6 & 6 & 4 \end{bmatrix} $$ But the solution should be (source):

I'm a bit confused about spatial derivative calculation. I took the formula from here. What am I missing?

I quite don't get how to calculate: $$ M=\sum_{x,y}w(x,y)\begin{bmatrix}I_{x}^{2} & I_{x}I_{y}\\ I_{x}I_{y} & I_{y}^{2} \end{bmatrix} $$ I'm also a bit confused about terminology. Is $I_x$ actually that matrix or is it some value? function?

Is it possible to show how can I calculate $M$ using the following formula I provided above?

Answer 1

What you should do instead to calculate $I_x$:

$$ \begin{align} I_{x}(x,y)&=\frac{\partial I(x,y)}{\partial x}\\ &=I(x+1,y)-I(x-1,y)\\ &= \begin{bmatrix}5& 7 & 11\\ 9 & 12 & 16\\ 11 & 14 & 16 \end{bmatrix}-\begin{bmatrix}1 & 0 & 5\\ 1 & 4 & 9\\ 3 & 8 & 11 \end{bmatrix}\\ &=\begin{bmatrix} \bf\color{blue} 4 & \bf\color{blue}7 & \bf\color{blue}6\\ \bf\color{blue}8 & \bf\color{blue}8 & \bf\color{blue}7\\ \bf\color{blue}8 & \bf\color{blue}6 & \bf\color{blue}5 \end{bmatrix} \end{align} $$ which matches the expected output.

Checking on your source, it seems that $M$ is mis-transcribed:

Here, the $W$ is a window into the image, not a weighting as is written in the equation for $M$ in the original post. This just means that $x$ and $y$ are varied to capture the window and the sum of the four different items is computed.

To calculate this $H$, all you need is the $I_x$ above and the equivalent in the $y$ direction: $$ \begin{align} I_{y}(x,y)&=\frac{\partial I(x,y)}{\partial y}\\ &=I(x,y+1)-I(x,y+1)\\ \end{align} $$