How to normalize an eigenvector when it has $\sqrt{i}$ as an entry?

Question

If $V=\mathbb{C}^2$ and $T(a,b)=(2a+ib,a+2b)$ I found $$[T]_\beta=\begin{bmatrix}2 & i \\ 1 & 2\end{bmatrix}$$ hence the eigenvalues are $2+\sqrt{i},2-\sqrt{i}$.

So using this, eigenvectors are $(\sqrt{i},1);(-\sqrt{i},1)$. Now to find orthonormal basis we need to use Gram-Schmidt and then normalize the vector. But my doubt is how to take norm of $(\sqrt{i},1)$? What will be conjugate of $\sqrt{i}$? And is my method correct?

This question is from book Linear algebra by Steven Friedberg

Answer 1

I'm not sure, but I tried doing it like $z \bar z = |z|^2$, so the norm comes out to be $|\sqrt{\Bbb i}| ^2 =1$.

Hence, $\langle (\sqrt{\Bbb i}, 1), (\sqrt{\Bbb i}, 1)\rangle = 2$.

Answer 2

(1)For any complex number $x$ we have $x=r(\cos A+i\sin A)$ with $r\geq 0$ and real number $A$.And $|x|=r.$ For $x\neq 0$ the complex square roots of $ x$ are $r^{1/2}(\cos A/2+i\sin A/2)$ and $r^{1/2}(\cos (A/2+\pi)+i\sin (A/2+\pi))$. Observe that if $y^2=x$ then $|y|=|x|^{1/2}.$ (2) The conjugate of $r(\cos A+i\sin A)$ for real $r,A$ is $r(\cos A-i\sin A)$.