find the coordinates of the vectors $(u,v)$ in the ordered basis $B=\{a,b\}$?

Question

let $a= (x_1,x_2)$ and $b=(y_1,y_2)$ be vectors in $\mathbb{R}^2$ such that $$x_1y_1 +x_2y_2=0$$ , $$x_1^2+ x_2^2 =y_1^2 +y_2^2 =1$$

It is given that $B=\{a, b\}$ is basis for $\mathbb{R}^2 $

find the coordinates of the vectors $(u,v)$ in the ordered basis $B=\{a,b\}$

My attempt : I take $(u,v)= c_1 a + c_2b= (c_1x_1 +c_2y_1 , c_1x_2 +c_2y_2)$

so the coordinates are $c_1x_1 +c_2y_1=u$ and $v= c_1x_2 +c_2y_2$

Answer 1

That basis is an orthnormal basis, and therefore the coordinates of $(u,v)$ with respect to that basis are $\bigl\langle(u,v),a\bigr\rangle(=ux_1+vx_2)$ and $\bigl\langle(u,v),b\bigr\rangle(=uy_1+vy_2)$. That is$$(u,v)=(ux_1+vx_2)a+(uy_1+vy_2)b.$$

Answer 2

Lets $z^T$ be the transpose of the column vector $z$ and set $(u,v)=z$. Since the basis is orthonormal the projection matrix onto each vector is $aa^T$ and $bb^T$ and so we have $z = (aa^T)z + (bb^T)z = a(a^Tz) + b(b^Tz)$ and since $a^Tz$ is the dot product $a \cdot z$, it is a scalar and commutes with $a$ and $b$ so we can write this as $(a \cdot z)a + (a \cdot z)b = (ux_1 + vx_2)a + (uy_1 + vy_2)b$ and we are done.

Answer 3

Your first equality $x_1y_1 + x_2y_2$ is exactly the definition of $(x_1, x_2)$ and $(y_1, y_2)$ are orthogonal.

Your second equality tells you that both vectors have a norm equal to 1.

Hence, they form an orthonormal basis, and the formula giving the coordinates of any vector in this basis is very simple: you simply have to compute the scalar product of your vector and of the vectors of the basis. No "correction" is needed since your basis is orthogonal, and no "renormalization" is needed since your basis vectors have already normalized.

Finally;

$(u, v) = \left<(u, v), a\right> a + \left<(u, v), b\right> b = (ux_1 + vx_2)a + (uy_1 + vy_2)b$.

As a result, the coordinates of $(u, v)$ in the new basis are $(ux_1 + vx_2, uy_1 + vy_2)$