4

To understand homographies and how to find them, you will need a good dose of projective geometry. I will briefly describe some preliminary concepts that you need to know before trying to find the homography, but don't expect to understand all these concepts with one reading iteration and only by reading this answer, if you are not familiar with them, ...


3

In neural networks, the family of functions and the shapes that they can make for decision surfaces is determined by the activation function you use (in your case, tanh or hyperbolic tangent). Assuming at least one hidden layer, then the universal approximation theorem applies. How closely you can approximate any given function is limited by the number of ...


2

You might want to have a look at the wikipedia article of PCA, where it says: "The $k$th component can be found by subtracting the first $k − 1$ principal components from $\mathbf{X}$:" $$\hat{\mathbf{X}}_k = \mathbf{X} - \sum_{s=1}^{k-1}\mathbf{X}\mathbf{w}_s\mathbf{w}_s^T$$ Then you repeat the process to find the next component: $$\mathbf{w}_k = \...


1

You can sometimes exploit the structure of your matrix to perform faster matrix multiplication. For example, if your matrix is sparse (or dense), there are algorithms that exploit this fact. In your case, you can actually compute $A^n$ in less time than $\mathcal{O}(n^3$). For example, have a look at this question at CS SE and this one at Stack Overflow (...


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