Question: Let $v \in \mathbb{R}^3$ be a unit vector. If $\{v_1, v_2, v_3\}$ is an orthonormal basis of $\mathbb{R}^3$ and $v = a_1v_1 + a_2v_2 + a_3v_3$, can we have $|a_1| > 3$?
What I think:
Since the norm of $v$ is one, then
$\sqrt{v\cdot v} = 1$
$(a_1v_1)^2 + (a_2v_2)^2 + (a_3v_3)^2 = 1^2$
Since $3^2 = 9 $ and $v_1 \cdot v_1 = 1^2$ (since each vector's norm in an orthonormal basis is 1), then $(a_1v_1) \geq 9$. Since $(a_2v_2)^2 + (a_3v_3)^2 > 0,$ then $(a_1v_1)^2 + (a_2v_2)^2 + (a_3v_3)^2 > 1$. Hence, $|a_1| < 3.$
I'm pretty sure my assumption of $(a_1v_1) \geq 9$ is wrong but I don't know how else to solve this problem.
Best Answer
The norm squared $v\cdot v$ of $v = a_1v_1 + a_2v_2 + a_3v_3$ is $a_1^2+a_2^2+a_3^2$ because $[v_1,v_2,v_3]$ is orthonormal. Clearly $|a_1|>3$ would contradict that norm-squared being $1$ (it would have to be${}>9$ since squares are${}\geq0$).