Timeline for Matrix Inverse with Same Principal Minors
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 23, 2013 at 9:46 | answer | added | Dietrich Burde | timeline score: 2 | |
| May 22, 2013 at 15:22 | comment | added | Dietrich Burde | If $\det(A)=1$ then $a=d$, if $\det(A)=-1$, then $a=-d$ ? | |
| May 19, 2013 at 7:48 | comment | added | Sebastian Schlecht | @Dietrich. So $\textrm{det} A = \textrm{det} A^{-1} = (\textrm{det} A)^{-1} = \pm 1$. Hence, for $n = 2$ is $A^{-1} = \pm \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ Consequently: $a = d$. | |
| May 16, 2013 at 18:11 | comment | added | Dietrich Burde | What gives the case $n=2$ with $A=\begin{pmatrix} a & b \cr c & d \end{pmatrix}$ ? | |
| May 14, 2013 at 17:05 | comment | added | Sebastian Schlecht | ........... :-) | |
| May 14, 2013 at 16:29 | comment | added | Denis Serre | I eventually delete my answer. It seems that I described the set of involutory matrices! Fortunately, this was not a doctoral dissertation; see the MO question about urban legends... | |
| May 14, 2013 at 16:14 | comment | added | Sebastian Schlecht | Denis, I find your structural findings very interesting. Although my pure algebra knowledge might be too restricted to create an answer from this. I was coming more from a point of diagonally similar (with transpose) matrices, which do have the same corresponding principal minors under certain conditions. (Like it was investigated by D.J. Hartfiel and R. Leowy in On matrices having equal corresponding principal minors) So your answer went far off, what I expected and hopefully someone else can help out on this. | |
| May 14, 2013 at 15:41 | comment | added | Denis Serre | Sebastian, I changed deeply my answer, because there was a mistake in calculations. It is still nteresting, I hope, but in a different way. | |
| May 14, 2013 at 9:42 | history | asked | Sebastian Schlecht | CC BY-SA 3.0 |