---
title: "SU-CS205L JAN162025"
source: https://www.jemoka.com/posts/kbhsu_cs205l_jan162025/
date: 2025-01-16
---

Random properties of eigenthings&hellip;
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}
&ldquo;you are rotating the vector towards the large eigenvalues, and away from the vectors corresponding to the small eigenvalues&rdquo;
spacial deformation When one eigenvalue is very big, and the other is very small: you are squashing a unit sphere in the space into an &ldquo;ellipse&rdquo; which is dependent on one of the eigenvector&rsquo;s direction.
Hence, its possible for small perturbations in input to change the scales of outputs.
singular value decomposition see singular value decomposition
Factorization:
\begin{equation} A = U \Sigma V^{T} \end{equation}
for \(A: m \times n\), \(\Sigma: m \times n\), \(V: n \times n\)
Notice&mdash;we &ldquo;amplify errors&rdquo; given by \(\Sigma\), but \(U\) and \(V\) are both orthonormal so errors don&rsquo;t propagate there.
Columns of \(V\) are eigenvectors of \(A^{T}A\), columns of \(U\) are eigenvectors of \(A A^{T}\). Singular values are the non-negative square roots of the
You can approximate a matrix by throwing away rows/columns which has zero/small singular values. In this case, we can throw away stuff that&rsquo;s close enough for \(0\).
Notes on Rectangular Diagonal Matrix \begin{equation} \mqty(5 &amp; 0 \\ 0 &amp; 2 \\ 0 &amp; 0) \mqty(c_1 \\ c_2) = \mqty(10 \\ -1 \\ \alpha) \end{equation}
notice how this system has no solution at \(\alpha \neq 0\); when \(\alpha \approx 0\), we run into trouble numerically.
rank deficient If a matrix is rank-deficient, then there will be a \(0\) as a singular value.
notes tip an orthonormal matrix has determinant \(\pm 1\), if \(v\) had an reflection, the determinant would be \(-1\), so just flipping it until its \(+1\) we&rsquo;d be fine.
