为何Harris矩阵是半正定矩阵？技术推导疑问

Great question! Let's break this down step by step to clear up your confusion and prove the semi-positive definiteness of the Harris matrix.

First, let's recall what the Harris matrix (also called the structure tensor) actually is. For a local window in an image, it's defined as:
$$
M = \begin{bmatrix}
\sum_{(x,y) \in \text{window}} I_x(x,y)^2 & \sum_{(x,y) \in \text{window}} I_x(x,y)I_y(x,y) \
\sum_{(x,y) \in \text{window}} I_x(x,y)I_y(x,y) & \sum_{(x,y) \in \text{window}} I_y(x,y)^2
\end{bmatrix}
$$
Where $I_x$ and $I_y$ are the image's gradient values in the x and y directions at each pixel in the window.

Proving Positive Semi-Definiteness #

A matrix is positive semi-definite if for any non-zero vector $\mathbf{v} = [a, b]^T$, the quadratic form $\mathbf{v}^T M \mathbf{v}$ is non-negative (i.e., $\geq 0$). Let's compute this quadratic form directly:

$$
\begin{align*}
\mathbf{v}^T M \mathbf{v} &= a^2 \sum I_x^2 + 2ab \sum I_x I_y + b^2 \sum I_y^2 \
&= \sum_{(x,y) \in \text{window}} \left( a I_x(x,y) + b I_y(x,y) \right)^2
\end{align*}
$$

Notice that we've rewritten the sum as the sum of squared terms. Since squares are always non-negative ($z^2 \geq 0$ for any real $z$), the entire sum must also be non-negative. That's the core of the proof! No matter what values $a$ and $b$ you choose, this quadratic form can never be negative.

Correcting Your Feature Value Misconception #

You mentioned that the trace (sum of diagonal elements) is positive, leading you to think eigenvalues could be one positive and one negative—but that's impossible here! Because we just proved $M$ is positive semi-definite, all its eigenvalues must be non-negative.

The trace of $M$ is $\sum I_x^2 + \sum I_y^2$, which is positive unless every pixel in the window has zero gradient (a completely flat region, where $M$ is the zero matrix with both eigenvalues zero). So the two eigenvalues can only be:

• Both positive (corresponding to a corner, where gradients are strong in two perpendicular directions)
• One positive, one zero (corresponding to an edge, where gradients are strong in one direction only)
• Both zero (flat region, no gradients at all)

This aligns perfectly with how the Harris detector works: we look for windows where both eigenvalues are large (corners) by computing the response function $\text{det}(M) - k (\text{trace}(M))^2$.

内容的提问来源于stack exchange，提问作者justPassBy