Proving Unitary Matrix in M2ℝ: Orthonormal Basis

[SMA_01]

Let u be a unitary matrix in M₂(ℝ).
Prove that if {b₁, b₂} is an orthonormal basis of ℝ², then u(b₂) is determined up to a negative sign by u(b₁).

Can anyone provide some intuition that will help me understand the question (don't really understand it)? Any tips/hints appreciated.


Thanks.

 

[ShayanJ]

u is unitary so $u u^{\dagger}=I$
$\{b_1,b_2\}$ is an orthonormal basis of $\mathbb{R}^2$ so $b_i \cdot b_j =\delta_{ij}$
The point is:
$(ub_1)\cdot(ub_2)=(ub_1)^{\dagger}(ub_2)=b_1^{\dagger}u^{\dagger}ub_2=b_1^{\dagger}b_2=b_1\cdot b_2=\delta_{ij}$
So because $\{b_1,b_2\}$ is an orthonormal basis,so is $\{ub_1,ub_2\}$ which means $ub_2$ is orthogonal to $ub_1$ which means once $ub_1$ is determined,$ub_2$ is determined up to a negative sign!

 

[SMA_01]

Thank you! I'm confused by the meaning of "up to a negative sign"...what does that meant exactly?

 

[HallsofIvy]

That knowing what u(b1) is will tell you what the absolute value of u(b2) is but not whether it is positive or negative.