Let $p=(x-(a+bi))(x-(a-bi))$ be the characteristic polynomial of linear operator $T$ ($T$ in $\Bbb C$) and its basis of eigenvector is $A=\{u+iv,u-iv\}$. Find a basis $B$ in $\Bbb R^2$ such that $[T]_B$ is as follows:
$$ \left(\begin{array}{ccc} a & -b \\ b & a \\ \end{array}\right)$$
So, I started writing $B=\{x+iy,w+iz\}$ and then I wrote $x+iy$ and $w+iz$ as linear combination of $A$:
$$(x,y)=(x/2u + y/2v)(u,v) + (x/2u - y/2v)(u,-v)$$
and similarly:
$$(w,z)=(w/2u + z/2v)(u,v) + (w/2u - z/2v)(u,-v)$$
I know that $T(u,v)=(a,b).(u,v)$ and $T(u,-v)=(a,-b)(u,-v)$ because are eigenvectors. Now I want:
$$T(x,y)=a(x,y)+b(w,z) \quad \mbox{and} \quad T(w,z)=-b(x,y)+a(w,z)$$
gathering the facts, I need:
$$T(x,y)=a(x,y)+b(w,z) = (x/2u + y/2v)T(u,v) + (x/2u - y/2v)T(u,-v) = (x/2u + y/2v)(a,b).(u,v) + (x/2u - y/2v)(a,-b)(u,-v)$$
at this point I want find $x,y,z,w$ right? But I have two equations, then this seems impossible...
Am I on the right way?
We have that $T(u+iv)=(a+bi)(u+iv)$, so $T(u)+iT(v)=(au-bv)+i(bu+av)$
and $\;\;\;\;\;\;\;\;\;\;\;T(u-iv)=(a-bi)(u-iv)$, so $T(u)-iT(v)=(au-bv)-i(bu+av).$
Adding these two equations and dividing by 2 gives $T(u)=au-bv$,
and subtracting them and then dividing by $2i$ gives $T(v)=bu+av$.
Now try to find a basis $\{w_1,w_2\}$ for $\mathbb{R^2}$ such that $T(w_1)=aw_1+bw_2$ and $T(w_2)=-bw_1+aw_2$.