SU-CS205L JAN092025
Last edited: August 8, 2025Key Sequence
Review
Notation
New Concepts
Important Results / Claims
Questions
Interesting Factoids
- “optimization is a wrong way to train a neural network” (because you can’t hope to sample a trajectory in \(\mathbb{R}^{n}\) at large \(n\))
SU-CS205L JAN142025
Last edited: August 8, 2025Key Sequence
Notation
New Concepts
- using domain knowledge: skinning
- Linear Systems
- Elimination Matricies
- sparsity
Important Results / Claims
Questions
Interesting Factoids
SU-CS205L JAN162025
Last edited: August 8, 2025Random properties of eigenthings…
Hermitian Matrix
A matrix such that: \(A^{*^{T}} = A\)
\begin{equation} v^{*^{T}} A = \lambda^{*} v^{*^{T}} \end{equation}
Meaning; \(Av = \lambda v \implies \qty(v^{*^{T}} A v = v^{*^{T}} \lambda v) \implies \lambda^{*} = \lambda\)
Symmetric matrices have this property.
vector deformation
Suppose \(v\) are the eigenvectors of \(A\), some vector \(c = \sum_{k}^{} a_{k} v_{k}\) (as long as they are from different eigenvalues), and so, applying \(A\):
\begin{equation} Ac = \sum_{k}^{} a_{k} A v_{k} = \sum_{k}^{} \qty(a_{k} \lambda_{k}) v_{k} \end{equation}
SU-CS205L JAN212025
Last edited: August 8, 2025Insights to SVD: “ever matrix is a diagonal matrix, when viewed in the right space”
We can solve a linear system by moving it around:
\begin{align} Ac = b \\ \Rightarrow\ & U \Sigma V^{T} c = b \\ \Rightarrow\ & U \qty(\Sigma V^{T} c) = b \\ \Rightarrow\ & \Sigma V^{T} c = U^{T} b \end{align}
(since \(U\) is orthonormal, we can just flip it to invert it)
Call \(U^{T} b = \hat{b}\), call \(V^{T} c = \hat{c}\). We now have:
SU-CS205L JAN232025
Last edited: August 8, 2025Issues with Direct Methods
- for instance, direct solvers have problems at numerical stability issues (for instance numerically stable quadratic formula); for cubics, there maybe unacceptable errors since there’s no such fix
Continuous Collision Detection
Implementing collision detection: three points, generally, are three \(v_1, v_2, v_3\) in \(\mathbb{R}^{3}\); yet, if they become linearly dependent, we know collision happened. In particular if the rank of \(\mqty(v_1, v_2, v_3)\) < 3, we have collided.
Problem! Solving this (taking the determinant of our matrix to figure out when collisions happened) will result in a cubic polynomial! This is numerically quite unstable.
