Timeline for Should the formula for the inverse of a 2x2 matrix be obvious?
Current License: CC BY-SA 3.0
10 events
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| Oct 7, 2015 at 21:29 | comment | added | Allen Knutson | In particular, the adjugate has two weirdnesses: the size $n-1$ determinants, and the transpose. Those are coming from the $\Lambda^{n-1}$ and the ${}^*$, respectively. | |
| Mar 3, 2012 at 23:21 | comment | added | Allen Knutson | This is essentially how I teach Cramer's rule. | |
| Feb 23, 2012 at 13:40 | comment | added | Spiro Karigiannis | Very nice. I had been looking for an invariant construction of the adjugate. | |
| Feb 22, 2012 at 21:15 | comment | added | Martin Brandenburg | @Qiaochu: 1+. This seems to be a general definition for the adjugate of an endomorphism of a locally free object of rank $n$ in a symmetric monoidal cocomplete category. | |
| Feb 22, 2012 at 14:37 | comment | added | Vectornaut | This is fantastic! I always thought the adjugate was just another "playing with squares of numbers" trick... I'm pleasantly surprised to see that it has a "deeper meaning." | |
| Feb 21, 2012 at 18:26 | comment | added | Qiaochu Yuan | @Frank: I guess the geometric picture is something like this. Identifying $\Lambda^2(V)$ with $\mathbb{R}$ corresponds to choosing a volume form on $\mathbb{R}^2$, equivalently a symplectic form. So $\text{SL}_2(\mathbb{R})$ is isomorphic to the symplectic group and the inverse and symplectic adjoint coincide for matrices of determinant $1$. Now the symplectic adjoint satisfies $\langle Tv, w \rangle = \langle v, T^{\dagger} w \rangle$ where $\langle , \rangle$ denotes the symplectic form, and plugging $e_1, e_2$ into $v, w$ one can see what this condition means geometrically. | |
| Feb 21, 2012 at 15:11 | comment | added | Frank Thorne | I tried drawing parallelograms corresponding to various linear transformations. Although the intuition was clear for a subset of matrices generating $SL_2$ (such as those occurring in the Iwasawa decomposition, as mentioned by Daniel Litt), I was unable to "see" this for a general linear transformation. Do you have more to say about your last sentence? Thank you! | |
| Feb 21, 2012 at 5:22 | comment | added | William | Yeah, I think "$A^{-1} = \frac{1}{\det(A)}\adj (A)$" is the easiest way to remember, because for a 2x2 matrix computing the adjugate is trivial | |
| Feb 21, 2012 at 4:16 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
added 135 characters in body
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| Feb 21, 2012 at 4:02 | history | answered | Qiaochu Yuan | CC BY-SA 3.0 |