0
$\begingroup$

Let $\{u_1,u_2,u_3,u_4\}$ be an orthonormal basis of $\mathbb C^4$. For a complex number $a$ we define the linear transformation $f_a: \mathbb C^4 \rightarrow \mathbb C^4$ as: $f_a(u_i)=au_{i+1}$ for $1\le i \le 3$ and $f_a(u_4)=au_1$. I need to find the $a \in \mathbb C$ so that the $f_a$ is unitary.

I believe that a unitary transformation will carry an orthonormal basis to an orthonormal basis, so actually $\{au_i\}$ will be orthonormal, which means that: $=\delta_{ij}$. But: $=a^2=a^2\delta_{ij}$. So: $a^2\delta_{ij}=\delta_{ij} \Rightarrow a^2=1 \Rightarrow a=\pm1$.

Is my solution correct?

  • 0
    I don't find anything wrong – 2017-02-08

1 Answers 1

0

In fact, we should have $$ \langle au_i,au_j \rangle = a\,\overline a \,\langle u_i,u_j\rangle = |a|^2 \langle u_i, u_j \rangle $$ So, it is only necessary that we have $|a| = 1$. For example, setting $a = i$ will still lead to a unitary transformation.

  • 0
    That is correct: what I was missing was this property: $=\overline{<\lambda y,x>}=\overline{\lambda}\overline{< y,x>}=\overline{\lambda}$. So, above $a=\pm i,\pm 1$. – 2017-02-08