Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 46290

An orthogonal matrix is an invertible real matrix whose inverse is equal to its transpose.

2 votes
1 answer
187 views

Heuristics for counting degrees of freedom

I have recently learned about the representation theorem for isotropic, linear operators, which says the following: Defintion: Let $M_n$ be the vector space of $n \times n$ real matrices. We say a l …
Asaf Shachar's user avatar
  • 6,741
11 votes
1 answer
564 views

Is there a "formula" for the point in $\text{SO}(n)$ which is closest to a given matrix?

$\newcommand{\Sig}{\Sigma}$ $\newcommand{\dist}{\operatorname{dist}}$ $\newcommand{\distSO}[1]{\dist(#1,\SO)}$ $\newcommand{\distO}[1]{\text{dist}(#1,\On)}$ $\newcommand{\tildistSO}[1]{\operatorname{d …
Asaf Shachar's user avatar
  • 6,741
6 votes
1 answer
368 views

Can we choose smoothly the singular vectors of a matrix?

$\newcommand{\GLm}{\text{GL}_n^-}$Let $A$ be a real $n \times n$ matrix with non-positive determinant. Suppose that the smallest singular value of $A$ is strictly smaller than all the others (it has m …
Asaf Shachar's user avatar
  • 6,741
1 vote
Accepted

What is special in dimension $2$ (When characterizing isometries using the cofactor matrix)?

This is somewhat informal. My argument is in the spirit of dimensional analysis in physics. The essential point is that the cofactor matrix encapsulates $n-1$ dimensional volume, while the dimension …
Asaf Shachar's user avatar
  • 6,741
1 vote

Bounding the non-multiplicativity of isometric projection

This is a more detailed version of fedja's answer: We shall need the following preliminary results: Lemma 1: Let $\lambda \in \mathbb{S}^1$. Then, the distance of $\lambda$ from $1$ is not greater t …
2 votes
1 answer
164 views

The only rotation fields satisfying this PDE are constant

$\newcommand{\div}{\operatorname{div}}$$\newcommand{\SO}{\operatorname{SO(2)}}$$\newcommand{\R}{\operatorname{\mathbb{R}}}$$\newcommand{\bdx}{\partial_x}$$\newcommand{\bdy}{\partial_y}$$\newcommand{\t …
Asaf Shachar's user avatar
  • 6,741
5 votes
1 answer
213 views

When does isometric projection respect multiplication?

Every $A \in \text{GL}_n(\mathbb{R})$ has a unique orthogonal polar factor $O_A=A(\sqrt{A^TA})^{-1}$, ( $A=O_AP_A$, $O \in \operatorname{O}_n, P \in \operatorname{Psym}_n$see Polar decomposition). …
Asaf Shachar's user avatar
  • 6,741
6 votes
2 answers
234 views

Bounding the non-multiplicativity of isometric projection

Every $A \in \text{GL}_n(\mathbb{R})$ has a unique Polar decomposition: $A=O_AP_A$, $O \in \operatorname{O}_n, P \in \operatorname{Psym}_n$. In particular the orthogonal factor is given by $$O_A=A(\ …
Asaf Shachar's user avatar
  • 6,741
7 votes
3 answers
219 views

What is special in dimension $2$ (When characterizing isometries using the cofactor matrix)?

Let $A$ be a real $n \times n$ matrix. Denote by $\operatorname{cof} A$ The cofactor matrix of $A$. By definition, $A (\operatorname{cof} A)^T=\det A \cdot I$. Thus, it is immediate that $A \in \ope …
Asaf Shachar's user avatar
  • 6,741