Question Number 13942 by FilupS last updated on 25/May/17
![for v= [(y),(x) ], v∈R^2 v has basis vectors i^ and j^ Assume we apply a basis transform to obtain new basis vectors i^ ′ and j^ ′ What is the new v′?](Q13942.png)
$$\mathrm{for}\:\:\boldsymbol{{v}}=\begin{bmatrix}{{y}}\\{{x}}\end{bmatrix},\:\:\:\:\boldsymbol{{v}}\in\mathbb{R}^{\mathrm{2}} \\ $$$$\boldsymbol{{v}}\:\mathrm{has}\:\mathrm{basis}\:\mathrm{vectors}\:\hat {{i}}\:\mathrm{and}\:\hat {{j}} \\ $$$$\: \\ $$$$\mathrm{Assume}\:\mathrm{we}\:\mathrm{apply}\:\mathrm{a}\:\mathrm{basis}\:\mathrm{transform}\:\mathrm{to} \\ $$$$\mathrm{obtain}\:\mathrm{new}\:\mathrm{basis}\:\mathrm{vectors}\:\hat {{i}}'\:\mathrm{and}\:\hat {{j}}' \\ $$$$\: \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{new}\:\boldsymbol{{v}}'? \\ $$
Answered by ajfour last updated on 25/May/17
![v′= [((y′)),((x′)) ]= [((i^ .j^ ′),(j^ .j^ ′)),((i^ .i^ ′),(j^ .i^ ′)) ] [(y),(x) ] (may be..).](Q13957.png)
$$\boldsymbol{{v}}'=\begin{bmatrix}{{y}'}\\{{x}'}\end{bmatrix}=\begin{bmatrix}{\hat {{i}}.\hat {{j}}'}&{\hat {{j}}.\hat {{j}}'}\\{\hat {{i}}.\hat {{i}}'}&{\hat {{j}}.\hat {{i}}'}\end{bmatrix}\begin{bmatrix}{{y}}\\{{x}}\end{bmatrix} \\ $$$$\left({may}\:{be}..\right). \\ $$$$ \\ $$
Commented by FilupS last updated on 26/May/17
![[((i^ .j^ ′),(j^ .j^ ′)),((i^ .i^ ′),(j^ .i^ ′)) ] I have a feeling this is incorrect I don′t remember, but I think it should be something like [(i^ ,(i^ ′)),(j^ ,(j^ ′)) ]](Q13981.png)
$$\begin{bmatrix}{\hat {{i}}.\hat {{j}}'}&{\hat {{j}}.\hat {{j}}'}\\{\hat {{i}}.\hat {{i}}'}&{\hat {{j}}.\hat {{i}}'}\end{bmatrix}\:\:\mathrm{I}\:\mathrm{have}\:\mathrm{a}\:\mathrm{feeling}\:\mathrm{this}\:\mathrm{is}\:\mathrm{incorrect} \\ $$$$\: \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{remember},\:\mathrm{but}\:\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{should} \\ $$$$\mathrm{be}\:\mathrm{something}\:\mathrm{like}\:\:\begin{bmatrix}{\hat {{i}}}&{\hat {{i}}'}\\{\hat {{j}}}&{\hat {{j}}'}\end{bmatrix} \\ $$