Can someone help me with this proof:
Let $b_1, b_2$ be any orthonormal basis for $R^2$, and let
${e_1,e_2}$ be the standard basis.
I have vectors $$x = (x_1,x_2), y=(y_1,y_2)$$ in standard basis and
$$x^b = (x_1^b, x_2^b), y^b = (y_1^b, y_2^b)$$
in basis $\{b_1,b_2\}$.
I have to prove that if $x^b$ and $y^b$ are the same vectors as $x,y$ but in basis $\{b_1,b_2\}$ then this is true:
$$||x-y|| = ||x^b-y^b||_b$$
I tried this first:
We know that x and y can be expressed in basis $\{b_1, b_2 \}$ as:
$$
x_{[b]} = x_1b_1 + x_2b_2
$$
$$
y_{[b]} = y_1b_1 + y_2b_2
$$
then:
$$
||x-y||
= \sqrt{(x_1 – y_1)^2 + (x_2 – y_2)^2}
=\sqrt{(x_1b_1 – y_1b_1)^2 + (x_2b_2 – y_2b_2)^2}
$$
$$
= ||x^b – y^b||_b
$$
I also tried using the reverse triangle inequality, because I am allowed to assume that $||u|| = \sqrt{u \cdot u} = \sqrt{u^b \cdot u^b} = ||u^b||_b$.
Best Answer
Let me state it in another way. Let $A\colon \mathbb{R}^2 \longrightarrow \mathbb{R}^2$ be the base change transformation, then $(x_1^b, x_2^b) = A(x_1,x_2)$ and $(y_1^b, y_2^b) = A(y_1,y_2)$. The norm squared reads $$ \| (x_1^b, x_2^b) \|^2 = \left< (x_1^b, x_2^b) , (x_1^b, x_2^b)\right> = \left< A(x_1,x_2) , A(x_1,x_2)\right> = \left< (x_1,x_2) , A^TA(x_1,x_2)\right> $$ From $\{b_1,b_2\}$ and $\{e_1, e_2 \}$ being orthonormal bases, $A$ is a orthogonal transformation, $A^TA =Id$. Hence $$ \| (x_1^b, x_2^b) \|^2 = \left< (x_1,x_2) , A^TA(x_1,x_2)\right> = \left< (x_1,x_2) , (x_1,x_2)\right> = \| (x_1, x_2) \|^2. $$ Now the proof follows doing the same for $x-y$ instead of $x$.