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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-




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

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