1
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Also the eigenvectors should be complex. $\endgroup$ – Emilio Novati Jan 15 at 20:54
  • 1
    $\begingroup$ $D$ 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
2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.