29 - Orthonormal Basis for Unitary Maps

Orthonormal Basis for Unitary Maps

Assume $V$ is finite-dimensional and $T:V\to V$ is unitary then

$$\text{ then there exists an orthonormal basis of eigenvectors}$$

Proof

As $\mathbb{K}=\mathbb{C}$ is algebraically closed then
There exists a $\lambda$ and $v\ne 0\in V$ such that

$$Tv=\lambda v$$

Then $U=⟨v⟩$ is $T-$invariant so the complement $U^{\perp }$ is also $T$-invariant
Therefore restriction $T{\mid }_{U^{\perp }}$ is a map of $U^{\perp }$ to itself which satisfies the hypothesis

By induction on dimension $n:=\dim (V)$ and noting that $\dim U^{\perp }=n-1$
Then there exist orthonormal basis $\{e_2,\cdots ,e_n\}$ of $U^{\perp }$
Putting

$$e_1=\frac{v}{||v||}$$

then

$$\{e_1,e_2,\cdots ,e_n\}\text{ is an orthonormal basis of eigenvectors for }V$$