# How to find eigenvector when the roots are complex

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 $$P\in \mathcal M_2(\mathbb{R})$$?

• Also the eigenvectors should be complex. – Emilio Novati Jan 15 at 20:54
• $D$ is not diagonal. What are you asking? – copper.hat Jan 15 at 20:57
• 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 $$\mathcal M(t)=e^{at}\begin{pmatrix}cos(2t)&sin(2t)\\sin(2t)&cos(2t)\end{pmatrix}$$ – C. Cristi Jan 15 at 20:57
• @copper.hat Sorry should've added "diagonal" – 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.

• Nice! And since $(2i,1)$ is the same eigenvector as $(2,-i)$, I believe we can also use $P=\begin{bmatrix}2&0\\0&-1\end{bmatrix}$. – 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.