Eigenvectors, eigenvalues and matrices

[EugP]

I have:

$$x' = \left(\begin{array}{cc}2&-5\\1&-2\end{array}\right) x$$

I found that the eigenvalues are $$r_1 = i$$ and $$r_2 = - i$$.
Also, I calculated the eigenvectors to be

$$\xi_1 = \left(\begin{array}{c}2 + i\\1\end{array}\right)$$

$$\xi_2 = \left(\begin{array}{c}2 - i\\1\end{array}\right)$$

The answer, however, is

$$\xi_1 = \left(\begin{array}{c}5\\2 - i\end{array}\right)$$

$$\xi_2 = \left(\begin{array}{c}5\\2 + i\end{array}\right)$$

I don't understand what I did wrong.
This is what I did for the eigenvector corresponding to $$r_1 = i$$:

$$\left(\begin{array}{cc}2 - i&-5\\1&-2 - i\end{array}\right)\left(\begin{array}{c}\xi_1\\\xi_2\end{array}\right) = 0$$

By multiplying row 2 and subtracting it from row 1, I got:

$$\left(\begin{array}{cc}0&0\\1&-2 - i\end{array}\right) \left(\begin{array}{c}\xi_1\\\xi_2\end{array}\right) = 0$$

So now I have:

$$\xi_1 + (-2 - i)\xi_2 = 0$$

$$\xi_1 = (2 + i)\xi_2$$

so:

$$\xi = \left(\begin{array}{c}\xi_1\\\xi_2\end{array}\right) = \left(\begin{array}{c}\(2 + i)\xi_2\\\xi_2\end{array}\right)$$

Now I let $$\xi_2 = 1$$:

$$\xi = \left(\begin{array}{c}2 + i\\1\end{array}\right)$$

I think I did everything right, but I get the wrong answer. Am I missing something?

 

[Timo]

If v is an eigenvector, then a*v (with scalar a) is an eigenvector too (by definition and matrices being linear maps). Multiply your solutions with 2-i and 2+i, respectively.

 

[EugP]

If v is an eigenvector, then a*v (with scalar a) is an eigenvector too (by definition and matrices being linear maps). Multiply your solutions with 2-i and 2+i, respectively.

Oh now I see what they did. But my solution is also correct, isn't it?

 

[HallsofIvy]

Yes. The set of all eigenvectors corresponding to a single eigenvalue forms a vector space. There are, necessarily, an infinite number of eigenvectors for any eigenvalue.

 

[EugP]

Yes. The set of all eigenvectors corresponding to a single eigenvalue forms a vector space. There are, necessarily, an infinite number of eigenvectors for any eigenvalue.

Thanks for clearing that up!