I am currently doing a question to get all possible eigenvectors of a matrix, and while I believe I got the correct answer (and WolframAlpha also says that my answer is correct), I was marked wrong. I just want to know if I made some mistake, or perhaps the answer that the online system I had was wrong. The question is as follows:

Find all distinct (real or complex) eigenvalues of A. Then find the basic eigenvectors of A corresponding to each eigenvalue. For each eigenvalue, specify the number of basic eigenvectors corresponding to that eigenvalue, then enter the eigenvalue followed by the basic eigenvectors corresponding to that eigenvalue.

$$A = \begin{bmatrix} -6 & 5 \\ -10 & 8 \end{bmatrix}$$

I was able to determine the eigenvalues to be $1+i$ as well as $1-i$, which it did accept, except I got (as well as WolframAlpha) the Eigenvectors below:

$$1+i: \begin{bmatrix} \frac{7}{10}+\frac{i}{10} \\ 1 \end{bmatrix}$$ $$1-i: \begin{bmatrix} \frac{7}{10}-\frac{i}{10} \\ 1 \end{bmatrix}$$

While the correct answer it gave was as follows:

$$1+i: \begin{bmatrix} -2+i \\ -3+i \end{bmatrix}$$ $$1-i: \begin{bmatrix} -2-i\\ -3-i \end{bmatrix}$$

Would anyone be able to tell me if I am correct and there is some issue with my homework system, or I am wrong and I should be quadruple checking my work? This has been driving me crazy all night.

$$\left(\frac{7}{10}+\frac{i}{10}\right)(-3-i)=-2-i$$ do the same for the other eigenvector. $$\left(\frac{7}{10}-\frac{i}{10}\right)(-3+i)=-2+i$$
• Eigenvectors are not unique ! To any eigenvalue is connected an eigenspace, i.e. a linear space spanned by the eigenvectors, so if $v$ is an eigenvector also $\alpha v$ is an eigenvector of the same eigenspace, connected to the same eigenvalue. – Emilio Novati Dec 6 '15 at 17:37