Let $A=\begin{pmatrix}1&-4\\1&1\end{pmatrix}$ then I want to diagonalize this matrix. Doing it's characteristic polynomial I find out that $\lambda_{1,2}=1\pm2i$. Then it's diagonal matrix $$D=\begin{pmatrix}1&2\\-2&1\end{pmatrix}$$

Now how do I find $P$ in $A=PDP^{-1}$, basically how do I find the eigenvectors of this Matrix if the roots are complex?

I'm sorry... I see I didn't asked the right question.

Actually: I have an system of differential equations with that matrix and as you can see I can't diagonalize the matrix in that form, but however, I want to write it in the form of $A=PDP^{-1}$ with $D=\begin{pmatrix}a&b\\-b&a\end{pmatrix}$, how do I find $P\in \mathcal M_2(\mathbb{R})$?

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    $\begingroup$ Also the eigenvectors should be complex. $\endgroup$ – Emilio Novati Jan 15 at 20:54
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    $\begingroup$ $D$ is not diagonal. What are you asking? $\endgroup$ – copper.hat Jan 15 at 20:57
  • $\begingroup$ Definetely, but I'm trying to be real here, since I'm trying to solve a system of differential equation with this... I know that the fundamental matrix of solutions must be $$\mathcal M(t)=e^{at}\begin{pmatrix}cos(2t)&sin(2t)\\sin(2t)&cos(2t)\end{pmatrix}$$ $\endgroup$ – C. Cristi Jan 15 at 20:57
  • $\begingroup$ @copper.hat Sorry should've added "diagonal" $\endgroup$ – C. Cristi Jan 15 at 20:58

Suppose $A (u+iv) = (a+ib) (u+iv)$ where everything is real (except $i$, of course).

Then, if $u,v$ are linearly independent, choose the basis $u,v$ and note that $A$ has the form $\begin{bmatrix} a & b \\ -b & a\end{bmatrix}$ in this basis.

If $u,v,b$ are non zero then $u,v$ are linearly independent (because $u+iv, u-iv$ are linearly independent over $\mathbb{C}$).

With $P=\begin{bmatrix} 0 & 1 \\ {1 \over 2} & 0 \end{bmatrix}$ you will get the desired result.

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    $\begingroup$ Nice! And since $(2i,1)$ is the same eigenvector as $(2,-i)$, I believe we can also use $P=\begin{bmatrix}2&0\\0&-1\end{bmatrix}$. $\endgroup$ – Klaas van Aarsen Jan 15 at 21:43

You can use the complex diagonalization to find the complex solutions

$$ e^{(1+2i)t} \pmatrix{2i\cr 1\cr} \ \text{and its complex conjugate}\ e^{(1-2i)t} \pmatrix{-2i\cr 1\cr} $$ The real and imaginary parts of one of these complex solutions give you the real solutions you are looking for.


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