Timeline for Changing basis linear transformation
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 23, 2016 at 19:43 | comment | added | amd | Are you working with column vectors, in which case you left-multiply by the matrix, or with row vectors, in which case you right-multiply? It makes a difference, and unfortunately I can’t tell from $B$ since that’s symmetric. | |
| May 23, 2016 at 17:15 | vote | accept | Granger Obliviate | ||
| May 23, 2016 at 17:15 | comment | added | Tim Huijgens | The idea is that in an other basis we want a matrix $C$ which represents the same transformation as the matrix $A$ represents in the old basis. This can only be the case if $C$ maps all vectors in $\mathbb{R^3}$ to the same images as $A$. Therefore write down the image of $x$, this represents all images from $A$. Then write this down in the new basis-form, with new coördinates, calling it $y$. Finally we solve $Cx=y$ and we get $B$. If all works out correctly your answer should be $BAB^{-1}$. en.wikipedia.org/wiki/Change_of_basis | |
| May 23, 2016 at 17:11 | answer | added | windircurse | timeline score: 0 | |
| May 23, 2016 at 17:05 | comment | added | Granger Obliviate | How is that going to help me @Tim Huijgens? I'm sorry I'm really feeling blank with this I need a more clear explanation... I'm not getting there alone | |
| May 23, 2016 at 16:59 | comment | added | Tim Huijgens | Maybe you can try to write down the image of a random vector in $\mathbb{R^3}$, say $x=(x_1,x_2,x_3)$. Then ask yourself how you can write this vector in the new basis, call this one $y$. And solve the linear system of equations $Ax=y$. | |
| May 23, 2016 at 16:54 | history | asked | Granger Obliviate | CC BY-SA 3.0 |